Dynamics HW

Dynamics Assignment #1

Dr. Peyman Honarmandi: The City College of New York

Chapter 11, Problem 1

The motion of a particle is defined by the relation 4 21.5 30 5 10,x t t t where x and t are expressed inmeters and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle when

4 s.t


The motion of a particle is defined by the relation 4 3 26 2 12 3 3,x t t t t where x and t are expressed inmeters and seconds, respectively. Determine the time, the position, and the velocity when 0.a


The acceleration of a particle is directly proportional to the square of the time t. When 0,t the particle isat 24 m.x Knowing that at 6 s, 96 mt x and 18 m/s,v express x and v in terms of t.


The acceleration of a particle is defined by the relation / .a k x It has been experimentally determined that15 ft/sv when 0.6 ftx and that 9 ft/sv when 1.2 ft.x Determine (a) the velocity of the particle

when 1.5 ft,x (b) the position of the particle at which its velocity is zero.


Based on experimental observations, the acceleration of a particle is defined by the relation (0.1a sin x/b),where a and x are expressed in m/s2 and meters, respectively. Knowing that 0.8 mb and that 1 m/sv when 0,x determine (a) the velocity of the particle when 1 m,x (b) the position where the velocity ismaximum, (c) the maximum velocity.

Solution of Dynamics Assignment #1


Chapter 11, Solution 1

Given: 4 2

3

2

1.5 30 5 10

6 60 5

18 60

x t t t

dxv t t

dtdv

a tdt

Evaluate expressions at 4 s.t

4 21.5(4) 30(4) 5(4) 10 66 mx 66.0 mx

36(4) 60(4) 5 149 m/sv 149.0 m/sv

2 218(4) 60 228 m/sa 2228.0 m/sa


We have 4 3 26 2 12 3 3x t t t t

Then 3 224 6 24 3dx

v t t tdt

and 272 12 24dv

a t tdt

When 0:a 2 272 12 24 12(6 2) 0t t t t

or (3 2)(2 1) 0t t

or2 1

s and s (Reject)3 2

t t 0.667 st

At2

s:3

t 4 3 2

2/32 2 2 2

6 2 12 3 33 3 3 3

x

2/3or 0.259 mx

3 2

2/32 2 2

24 6 24 33 3 3

v

2/3or 8.56 m/sv


We have 2 constanta kt k

Now 2dva kt

dt

At 6 s, 18 m/s:t v 2

18 6

v tdv kt dt

or 3118 ( 216)

3v k t

or 3118 ( 216)(m/s)

3v k t

Also 3118 ( 216)

3

dxv k t

dt

At 0, 24 m:t x 3

24 0

118 ( 216)

3

x tdx k t dt

or 41 124 18 216

3 4x t k t t

Now

At 6 s, 96 m:t x 41 196 24 18(6) (6) 216(6)

3 4k

or 41 m/s

9k

Then 41 1 124 18 216

3 9 4x t t t

or 41( ) 10 24

108x t t t

and 31 118 ( 216)

3 9v t

or 31( ) 10

27v t t


vdv ka

dx x

Separate and integrate using 0.6 ft, 15 ft/s.x v

15 0.6

2

15 0.6

2 2

1ln

2

1 1(15) ln

2 2 0.6

v x

xv

dxvdv k

x

v k x

xv k

(1)

When 9 ft/s, 1.2 ftv x

2 21 1 1.2(9) (15) ln

2 2 0.6k

Solve for k.2 2103.874 ft /sk

(a) Substitute 2 2103.874 ft /s and 1.5 ftk x into (1).

2 21 1 1.5(15) 103.874 ln

2 2 0.6v

5.89 ft/sv (b) For 0,v

210 (15) 103.874 ln

2 0.6

ln 1.0830.6

x

x

1.772 ftx


We have 0.1 sin0.8

dv xv a

dx

When 0, 1 m/s:x v 1 0

0.1 sin0.8

v x xvdv dx

or 2

0

1( 1) 0.1 0.8 cos

2 0.8

xx

v x

or 210.1 0.8 cos 0.3

2 0.8

xv x

(a) When 1 m:x 21 10.1( 1) 0.8 cos 0.3

2 0.8v

or 0.323 m/sv

(b) When max, 0:v v a 0.1 sin 00.8

x

or 0.080134 mx 0.0801 mx

(c) When 0.080134 m:x

2max

2 2

1 0.0801340.1( 0.080134) 0.8 cos 0.3

2 0.8

0.504 m /s

v

or max 1.004 m/sv

Dynamics Assignment #2



A truck travels 220 m in 10 s while being decelerated at aconstant rate of 0.6 m/s2. Determine (a) its initial velocity,(b) its final velocity, (c) the distance traveled during thefirst 1.5 s.


A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runswith constant velocity. If the sprinter’s time for the first 35 m in 5.4 s, determine(a) his acceleration, (b) his final velocity, (c) his time for the race.


As relay runner A enters the 20-m-long exchange zone with aspeed of 12.9 m/s, he begins to slow down. He hands the baton torunner B 1.82 s later as they leave the exchange zone with thesame velocity. Determine (a) the uniform acceleration of each ofthe runners, (b) when runner B should begin to run.


Automobiles A and B are traveling in adjacent highway lanesand at 0t have the positions and speeds shown. Knowingthat automobile A has a constant acceleration of 21.8 ft/s andthat B has a constant deceleration of 21.2 ft/s , determine(a) when and where A will overtake B, (b) the speed of eachautomobile at that time.


Boxes are placed on a chute at uniform intervals of time Rt andslide down the chute with uniform acceleration. Knowing that asany box B is released, the preceding box A has already slid 6 mand that 1 s later they are 10 m apart, determine (a) the valueof ,Rt (b) the acceleration of the boxes.


Slider block A moves to the left with a constant velocity of 6 m/s.Determine (a) the velocity of block B, (b) the velocity of portion D ofthe cable, (c) the relative velocity of portion C of the cable with respectto portion D.

Solution of Dynamics Assignment #2



(a) Initial velocity. 20 0

1

2x x v t at

00

1

2220 1

( 0.6)(10)10 2

x xv at

t

0 25.9 m/sv

(b) Final velocity. 0v v at

25.0 ( 0.6)(10)v 19.00 m/sfv

(c) Distance traveled during first 1.5 s.

20 0

2

1

21

0 (25.0)(1.5) ( 0.6)(1.5)2

x x v t at

36.8 mx


Given: 0 35 m,# # constanta

35 m 100 m, constantv , #

At 0, 0 when 35 m, 5.4 st v x t

Find:

(a) a

(b) v when 100 mx

(c) t when 100 mx

(a) We have 210 0

2x t at for 0 35 mx# #

At 5.4 s:t 2135 m (5.4 s)

2a

or 22.4005 m/sa

22.40 m/sa

(b) First note that maxv v for 35 m 100 m.x# #

Now 2 0 2 ( 0)v a x for 0 35 mx# #

When 35 m:x 2 2max 2(2.4005 m/s )(35 m)v

or max 12.9628 m/sv max 12.96 m/sv

(c) We have 1 0 1( )x x v t t for 35 m 100 mx, #

When 100 m:x 2100 m 35 m (12.9628 m/s)( 5.4) st

or 2 10.41 st


(a) For runner A: 20

10 ( )

2A A Ax v t a t

At 1.82 s:t 2120 m (12.9 m/s)(1.82 s) (1.82 s)

2 Aa

or 22.10 m/sAa

Also 0( )A A Av v a t

At 1.82 s:t 21.82( ) (12.9 m/s) ( 2.10 m/s )(1.82 s)

9.078 m/sAv

For runner B: 2 0 2 0B B Bv a x

When 220 m, : (9.078 m/s) 2 (20 m)B B A Bx v v a

or 22.0603 m/sBa

22.06 m/sBa

(b) For runner B: 0 ( )B B Bv a t t

where Bt is the time at which he begins to run.

At 1.82 s:t 29.078 m/s (2.0603 m/s )(1.82 )sBt

or 2.59 sBt

Runner B should start to run 2.59 s before A reaches the exchange zone.


2 21.8 ft/s 1.2 ft/sA Ba a

0

0

| | 24 mi/h 35.2 ft/s

| | 36 mi/h 52.8 ft/sA

B

v

v

Motion of auto A:

0( ) 35.2 1.8A A Av v a t t (1)

2 20 0

1 1( ) ( ) 0 35.2 (1.8)

2 2A A A Ax x v t a t t t (2)

Motion of auto B:

0( ) 52.8 1.2B B Bv v a t t (3)

2 20 0

1 1( ) ( ) 75 52.8 ( 1.2)

2 2B B B Bx x v t a t t t (4)

(a) 1 overtakes at .A B t t

2 21 1 1: 35.2 0.9 75 52.8 0.6A Bx x t t t t

21 11.5 17.6 75 0t t

1 13.22 s and 15.0546t t 1 15.05 st

Eq. (2): 235.2(15.05) 0.9(15.05)Ax .734 ftAx

(b) Velocities when 1 15.05 st

Eq. (1): 35.2 1.8(15.05)Av

62.29 ft/sAv 42.5 mi/hAv

Eq. (3): 52.8 1.2(15.05)Bv

34.74 ft/sBv 23.7 mi/hBv

A overtakes B


Let 1 s be the time when the boxes are 10 m apart. St =

Let 0 0 0 0; ( ) ( ) 0; ( ) ( ) 0A B A B A Ba a a x x v v .= = = = = =

(a) For 210,2At x a> = t

For 2 21 1, ( ) ( )2 2R B R St t x a t t a t a> = − = =

12

At 21, 6 62R At t x m at= = → = R (1)

At , 10 mR S A Bt t t x x= + − =

2 2

2 2

1 110 ( ) ( )2 21 1102 2

R S S

R R S S

a t t a t

at at t at

= + −

= + + 212 Sat−

.(1) 10 6EqR Sat t⎯⎯⎯→ = +

10 6 4 m/s 4 m/s1R

Sat at

t R−

= = → = (2)

Dividing Equation (1) by Eq. (2),

21

2 1 62 4

RR

R

att

at= = 3.00 sRt =

(b) Solving Eq. (2) for a,

2 24 1.333 m/s 4 ft/s3

a = = = 21.33 m/sa =


From the diagram, we have

3 constantA Bx y

Then 3 0A Bv v (1)

and 3 0A Ba a (2)

(a) Substituting into Eq. (1) 6 m/s + 3 0Bv

or 2 m/sB v

(b) From the diagram constantB Dy y

Then 0B Dv v

2 m/sD v

(c) From the diagram constantA Cx y

Then 0A Cv v 6 m/sCv

Now / ( 6 m/s) (2 m/s) = 8 m/sC D C Dv v v

/ 8 m/sC D v

Dynamics Assignment #3 ME 247



Slider block B moves to the right with a constantvelocity of 300 mm/s. Determine (a) the velocityof slider block A, (b) the velocity of portion C ofthe cable, (c) the velocity of portion D of thecable, (d) the relative velocity of portion C of thecable with respect to slider block A.


At the instant shown, slider block B is movingwith a constant acceleration, and its speed is150 mm/s. Knowing that after slider block Ahas moved 240 mm to the right its velocity is60 mm/s, determine (a) the accelerations of Aand B, (b) the acceleration of portion D of thecable, (c) the velocity and change in position ofslider block B after 4 s.


An airplane used to drop water on brushfires is flyinghorizontally in a straight line at 315 km/h at an altitude of80 m. Determine the distance d at which the pilot shouldrelease the water so that it will hit the fire at B.


The nozzle at A discharges cooling water with an initialvelocity v0 at an angle of 6° with the horizontal onto a grindingwheel 350 mm in diameter. Determine the range of values ofthe initial velocity for which the water will land on thegrinding wheel between Points B and C.


As slider block A moves downward at a speed of 0.5 m/s, the velocitywith respect to A of the portion of belt B between idler pulleys C and Dis / 2 m/sCD A v . Determine the velocity of portion CD of the beltwhen (a) 45 , (b) 60 .


The velocities of skiers A and B areas shown. Determine the velocity ofA with respect to B.

Solution of Dynamics Assignment #3 ME 247



From the diagram ( ) 2 constantB B A Ax x x x

Then 2 3 0B Av v (1)

and 2 3 0B Aa a (2)

Also, we have constantB Ax x

Then 0D Av v (3)

(a) Substituting into Eq. (1) 2(300 mm/s) 3 0Av

or 200 mm/sA v

(b) From the diagram ( ) constantB B Cx x x

Then 2 0B Cv v

Substituting 2(300 mm/s) 0Cv

or 600 mm/sC v

(c) From the diagram ( ) ( ) constantC A B Ax x x x

Then 2 0C A Dv v v

Substituting 600 mm/s 2(200 mm/s) 0Dv

or 200 mm/sD v

(d) We have /

600 mm/s 200 mm/sC A C Av v v

or / 400 mm/sC A v


From the diagram ( ) 2 constantB B A Ax x x x

Then 2 3 0B Av v (1)

and 2 3 0B Aa a (2)

(a) First observe that if block A moves to the right, A v and Eq. (1) .B v Then, using

Eq. (1) at 0t

02(150 mm/s) 3( ) 0Av

or 0( ) 100 mm/sAv

Also, Eq. (2) and constant constantB Aa a

Then 2 20 0( ) 2 [ ( ) ]A A A A Av v a x x

When 0( ) 240 mm:A Ax x

2 2(60 mm/s) (100 mm/s) 2 (240 mm)Aa

or 240 mm/s

3Aa

or 213.33 mm/sA a

PROBLEM 11.54 (Continued)

Then, substituting into Eq. (2)

2402 3 mm/s 0

3Ba

or 220 mm/sBa 220.0 mm/sB a

(b) From the solution to Problem 11.53

0D Av v

Then 0D Aa a

Substituting 240 mm/s 0

3Da

or 213.33 mm/sD a

(c) We have 0( )B B Bv v a t

At 4 s:t 2150 mm/s ( 20.0 mm/s )(4 s)Bv

or 70.0 mm/sB v

Also 20 0

1( ) ( )

2B B B By y v t a t

At 4 s:t 0

2 2

( ) (150 mm/s)(4 s)

1( 20.0 mm/s )(4 s)

2

B By y

or 0( ) 440 mmB B y y


First note 0 315 km/h

87.5 m/s

v

Vertical motion. (Uniformly accelerated motion)

210 (0)

2y t gt

At B: 2 2180 m (9.81 m/s )

2t

or 4.03855 sBt

Horizontal motion. (Uniform)

00 ( )xx v t

At B: (87.5 m/s)(4.03855 s)d

or 353 md


First note

0 0

0 0

( ) cos 6

( ) sin 6x

y

v v

v v


0 0( )xx x v t


2 20 0

1( ) ( 9.81 m/s )

2yy y v t gt g

At Point B: (0.175 m) sin 10

(0.175 m) cos 10

x

y

0: 0.175 sin 10 0.020 ( cos 6 )x v t

or0

0.050388

cos 6Bt v

20

1: 0.175 cos 10 0.205 ( sin 6 )

2B By v t gt

Substituting for Bt

2

00 0

0.050388 1 0.0503880.032659 ( sin 6 ) (9.81)

cos 6 2 cos 6v

v v

or21

2 20 2

(9.81)(0.050388)

cos 6 (0.032659 0.050388 tan 6 )v

or 0( ) 0.678 m/sBv

At Point C: (0.175 m) cos 30

(0.175 m) sin 30

x

y

0: 0.175 cos 30 0.020 ( cos 6 )x v t

or0

0.171554

cos 6Ct v

20

1: 0.175 sin 30 0.205 ( sin 6 )

2C Cy v t gt

Substituting for Ct

2

00 0

0.171554 1 0.1715540.117500 ( sin 6 ) (9.81)

cos 6 2 cos 6v

v v


or21

2 20 2

(9.81)(0.171554)

cos 6 (0.117500 0.171554 tan 6 )v

or 0( ) 1.211 m/sCv

00.678 m/s 1.211 m/sv# #


We have /CD A CD A v v v

where (0.5 m/s)( cos 65 sin 65 )

(0.21131 m/s) (0.45315 m/s)A

v i ji j

and / (2 m/s)(cos sin )CD A v i j

Then [( 0.21131 2 cos ) m/s] [( 0.45315 2 sin ) m/s]CD v i j

(a) We have ( 0.21131 2 cos 45 ) ( 0.45315 2 sin 45 )

(1.20290 m/s) (0.96106 m/s)CD

v i ji j

or 1.540 m/sCD v 38.6°

(b) We have ( 0.21131 2 cos 60 ) ( 0.45315 2 sin 60 )

(0.78869 m/s) (1.27890 m/s)CD

v i ji j

or 1.503 m/sCD v 38.3°


We have /A B A B v v v

The graphical representation of this equation is then as shown.

Then 2 2 2/ 10 14 2(10)(14) cos 15A B v

or / 5.05379 m/sA Bv

and10 5.05379

sin sin 15

or 30.8°

/ 5.05 m/sA B v 55.8°

Alternative solution.

/

10 cos 10 10 sin 10 (14 cos 25 14 sin 25 )

2.84 4.14

A B A B

v v vi j i j

i j

5.05 m/s 55.8°




Three children are throwingsnowballs at each other. Child Athrows a snowball with ahorizontal velocity 0.v If thesnowball just passes over thehead of child B and hits child C,determine (a) the value of v0,(b) the distance d.


A model rocket is launched from Point A with an initialvelocity v0 of 250 ft/s. If the rocket’s descent parachutedoes not deploy and the rocket lands 400 ft from A,determine (a) the angle that v0 forms with the vertical,(b) the maximum height above Point A reached by therocket, and (c) the duration of the flight.


Knowing that the velocity of block B with respect to block Ais / 5.6 m/sB A v 70°, determine the velocities of A and B.


Conveyor belt A, which forms a 20°angle with the horizontal, moves ata constant speed of 4 ft/s and isused to load an airplane. Knowingthat a worker tosses duffel bag Bwith an initial velocity of 2.5 ft/s atan angle of 30° with the horizontal,determine the velocity of the bagrelative to the belt as it lands on thebelt.


Determine the required velocity of the belt B if the relativevelocity with which the sand hits belt B is to be (a) vertical,(b) as small as possible.


Determine the peripheral speed of the centrifuge test cab A forwhich the normal component of the acceleration is 10g.


Determine the maximum speed that the cars of the roller-coastercan reach along the circular portion AB of the track if the normalcomponent of their acceleration cannot exceed 3 g.




(a) Vertical motion. (Uniformly accelerated motion)

210 (0)

2y t gt

At B: 2 211 m (9.81 m/s ) or 0.451524 s

2 Bt t

Horizontal motion (Uniform)

00 ( )xx v t

At B: 07 m (0.451524 s)v

or 0 15.5031 m/sv 0 15.50 m/sv

(b) Vertical motion: At C: 2 213 m (9.81 m/s )

2t

or 0.782062 sCt

Horizontal motion.

At C: (7 )m (15.5031 m/s)(0.782062 s)d

or 5.12 md


Set the origin at Point A. 0 00, 0x y

Horizontal motion: 00

sin sinx

x v tv t

(1)

Vertical motion: 20

1cos

2y v t gt

2

0

1 1cos

2y gt

v t

(2)

22 2 2 2

20

1 1sin cos 1

2( )x y gt

v t

2 2 2 2 4 2 20

1

4x y gyt g t v t

2 4 2 2 2 20

1( ) 0

4g t v gy t x y (3)

At Point B, 2 2 400 ft, 400 cos 30 ft

400 sin 30 200 ft

x y x

y

2 4 2 2 2

4 2

1(32.2) [250 (32.2)( 200)] 400 0

4

259.21 68940 160000 0

t t

t t

2 2 2263.62 s and 2.34147 s

16.2364 s and 1.53019 s

t

t

Restrictions on : 0 120 # #

2 212

400 cos 30tan 0.085654

200 (16.1)(16.2364)

4.8957

x

y gt

and2

400 cos 302.13435

200 (16.1)(1.53019)

115.105


Use 4.8957 corresponding to the steeper possible trajectory.

(a) Angle . 4.90

(b) Maximum height. max0 atyv y y

0

0

2 20

max 0

2 2

cos 0

cos

cos1cos

2 2

(250) cos 4.8957

(2)(32.2)

yv v gt

vt

g

vy v t gt

g

max 963 fty

(c) Duration of the flight. (time to reach B)

16.24 st


From the diagram

2 3 constantA Bx x

Then 2 3 0A Bv v

or2

| |3B Av v

Now /B A B A v v v

and noting that Av and Bv must be parallel to surfaces A and B, respectively, the graphical representation ofthis equation is then as shown. Note: Assuming that Av is directed up the incline leads to a velocity diagramthat does not “close.”

First note 180 (40 30 )

110B

B

Then23 5.6

sin (110 ) sin 40 sin (30 )AA

B B

vv

or2

sin 40 sin (110 )3A A Bv v

or sin (110 ) 0.96418B

or 35.3817B

and 4.6183B

For 35.3817 :B 2 5.6 sin 40

3 sin (30 35.3817 )B Av v

or 5.94 m/s 3.96 m/sA Bv v

5.94 m/sA v 30°

3.96 m/sB v 35.4°

For 4.6183 :B 2 5.6 sin 40

3 sin (30 4.6183 )B Av v

or 9.50 m/sAv

6.34 m/sBv

9.50 m/sA v 30°

6.34 m/sB v 4.62°


First determine the velocity of the bag as it lands on the belt. Now

0 0

0 0

[( ) ] ( ) cos 30

(2.5 ft/s)cos 30

[( ) ] ( ) sin 30

(2.5 ft/s)sin 30

B x B

B y B

v v

v v


00 [( ) ]B xx v t 0( ) [( ) ]B x B xv v

(2.5 cos 30 ) t 2.5 cos 30


20 0

1[( ) ]

2B yy y v t gt 0( ) [( ) ]B y B yv v gt

211.5 (2.5 sin 30 )

2t gt 2.5 sin 30 gt

The equation of the line collinear with the top surface of the belt is

tan 20y x

Thus, when the bag reaches the belt

211.5 (2.5 sin 30 ) [(2.5 cos 30 ) ]tan 20

2t gt t

or 21(32.2) 2.5(cos 30 tan 20 sin 30 ) 1.5 0

2t t

or 216.1 0.46198 1.5 0t t

Solving 0.31992 s and 0.29122 st t (Reject)

The velocity Bv of the bag as it lands on the belt is then

(2.5 cos 30 ) [2.5 sin 30 32.2(0.319 92)]

(2.1651 ft/s) (9.0514 ft/s)B

v i j

i j

Finally /B A B A v v v

or / (2.1651 9.0514 ) 4(cos 20 sin 20 )

(1.59367 ft/s) (10.4195 ft/s)B A

v i j i j

i j

or / 10.54 ft/sB A v 81.3°


A grain of sand will undergo projectile motion.

0constant 5 ft/s

x xs sv v

y-direction. 22 (2)(32.2 ft/s )(3 ft) 13.90 ft/sysv gh

Relative velocity. /S B S B v v v (1)

(a) If /S Bv is vertical,

/ 5 13.9 ( cos 15 sin 15 )

5 13.9 cos 15 sin 15S B B B

B B

v v v

v v

j i j i ji j i j

Equate components.5

: 0 5 cos 15 5.176 ft/scos 15B Bv v

i

5.18 ft/sB v 15°

(b) /S Cv is as small as possible, so make / toS B Bv v into (1).

/ /sin 15 cos 15 5 13.9 cos 15 sin 15S B S B B Bv v v v i j i j i j

Equate components and transpose terms.

/

/

(sin 15 ) (cos 15 ) 5

(cos 15 ) (sin 15 ) 13.90S B B

S B B

v v

v v

Solving, / 14.72 ft/sS Bv 1.232 ft/sBv

1.232 ft/sB v 15°


2 2

22 2 2 2

10 10(9.81 m/s ) 98.1 m/s

(8 m)(98.1 m/s ) 784.8 m /s

n

n n

a g

va v a

28.0 m/sv


We have2

nv

a

Then 2 2max( ) (3 32.2 ft/s )(80 ft)ABv

or max( ) 87.909 ft/sABv

or max( ) 59.9 mi/hABv




At a given instant in an airplane race, airplane A is flyinghorizontally in a straight line, and its speed is being increasedat the rate of 28 m/s . Airplane B is flying at the same altitudeas airplane A and, as it rounds a pylon, is following a circularpath of 300-m radius. Knowing that at the given instant thespeed of B is being decreased at the rate of 23 m/s , determine,for the positions shown, (a) the velocity of B relative to A,(b) the acceleration of B relative to A.


From a photograph of a homeowner using a snowblower, itis determined that the radius of curvature of the trajectoryof the snow was 8.5 m as the snow left the discharge chuteat A. Determine (a) the discharge velocity Av of the snow,(b) the radius of curvature of the trajectory at its maximumheight.


A projectile is fired from Point A with an initialvelocity 0v which forms an angle with thehorizontal. Express the radius of curvature ofthe trajectory of the projectile at Point C interms of 0, , ,x v and g.


The rotation of rod OA about O is defined by the relation 2(4 8 ),t t where and t are expressed in radians and seconds, respectively. Collar B slides along therod so that its distance from O is 10 6sin ,r t where r and t are expressed ininches and seconds, respectively. When 1 s,t determine (a) the velocity of thecollar, (b) the total acceleration of the collar, (c) the acceleration of the collarrelative to the rod.


The motion of a particle on the surface of a right circularcylinder is defined by the relations , 2 ,R A t andz 2/4,At where A is a constant. Determine the magnitudesof the velocity and acceleration of the particle at any time t.




First note 450 km/h 540 km/h 150 m/sA Bv v

(a) We have /B A B A v v v

The graphical representation of this equation is then as shown.

We have 2 2 2/ 450 540 2(450)(540) cos 60°B Av

or / 501.10 km/hB Av

and540 501.10

sin sin 60

or 68.9

/ 501 km/hB A v 68.9

(b) First note 2 28 m/s ( ) 3 m/sA B t a a 60

Now2 2(150 m/s)

( )300 m

BB n

B

va

or 2( ) 75 m/sB n a 30

Then

2 2

( ) + ( )

3( cos60 sin 60 ) + 75( cos 30 sin 30 )

= (66.452 m/s ) (34.902 m/s )

B B t B n

a a ai j i j

i j

Finally /B A B A a a a

or /

2 2

( 66.452 34.902 ) (8 )

(74.452 m/s ) (34.902 m/s )

B A

a i j i

i j

or 2/ 82.2 m/sB A a 25.1


(a) We have2

( ) AA n

A

va

or 2

2 2

(9.81 cos 40°)(8.5 m)

63.8766 m /s

Av

or 7.99 m/sA v 40

(b) We have2

( ) BB n

B

va

where Point B is the highest point of the trajectory, so that

( ) cos 40B A x Av v v

Then2 2 2

2

(63.8766 m /s )cos 40

9.81 m/sB

or 3.82 mB


We have2

( ) CC n

C

va

or2

cosC

C

v

g

Noting that the horizontal motion is uniform, we have

0 0( ) ( ) 0 + ( ) ( cos )A x C x xv v x v t v t

where 0( ) cos ( ) cosA x C x Cv v v v

Then 0 0cos cos and ( ) cosC C xv v v v (1)

or 0cos cosC

v

v

so that3

0 cosC

C

v

g v

For the uniformly accelerated vertical motion have

0 0( ) ( ) sinC y yv v gt v gt

From above 00

( cos ) orcos

xx v t t

v

Then 00

( ) sincosC yx

v v gv

(2)

Now 2 2 2( ) ( )C C x C yv v v

Substituting for ( )C xv [Eq. (1)] and ( )C yv [Eq. (2)]

2

2 20 0

0

2 220 2 4 2

0 0

( cos ) sincos

2 tan1

cos

Cx

v v v gv

gx g xv

v v

or

3/ 22 23 3

0 2 4 20 0

2 tan1

cosC

gx g xv v

v v

Finally, substituting into the expression for ,C obtain

3/ 22 2 20

2 4 20 0

2 tan1

cos cos

v gx g x

g v v


We have 210 6sin (4 8 )r t t t

Then 6 cos ( 1)r t t

and 26 sinr t

At 1 s:t 10 in.r 4 rad

6 in./sr 0

0r 28 rad/s

(a) We have B rr r v e e

so that (6 in./s)B r v e

(b) We have 2( ) ( 2 )

(10)(8 )B rr r r r

a e ee

or 2(80 in./s )B a e

(c) We have /B OA ra

so that / 0B OA a


We have 212

4R A t z At

Then1

0 22

R z At

and1

0 02

R z A

Now 2 2 2 2

02 2 2

22

2 2 2

( ) ( ) ( )

10 ( 2 )

2

14

4

R zv v v v

R R z

A At

A t

or 2 2116

2v A t

and 2 2 2 2

0 0 02 2 2 2( ) ( 2 ) ( )

R za a a a

R R R R z

22 2

2 4

1[ (2 ) ] 0

2

116

4

A A

A

or 4164 1

2a A

Dynamics Assignment #6ME 247



A hockey player hits a puck so that it comes to rest in 9 s after sliding 30 m on the ice. Determine (a) theinitial velocity of the puck, (b) the coefficient of friction between the puck and the ice.


A 20-kg package is at rest on an incline when a force P is applied to it.Determine the magnitude of P if 10 s is required for the package totravel 5 m up the incline. The static and kinetic coefficients of frictionbetween the package and the incline are both equal to 0.3.


The two blocks shown are originally at rest. Neglecting the masses ofthe pulleys and the effect of friction in the pulleys and between block Aand the horizontal surface, determine (a) the acceleration of each block,(b) the tension in the cable.


The coefficients of friction between the load and the flat-bedtrailer shown are 0.40s and 0.30.k Knowing that thespeed of the rig is 45 mi/h, determine the shortest distance inwhich the rig can be brought to a stop if the load is not to shift.


Each of the systems shown is initially at rest.Neglecting axle friction and the masses of thepulleys, determine for each system (a) theacceleration of block A, (b) the velocity of blockA after it has moved through 10 ft, (c) the timerequired for block A to reach a velocity of 20 ft/s.


A spring AB of constant k is attached to a support at A and to acollar of mass m. The unstretched length of the spring is .Knowing that the collar is released from rest at 0x x andneglecting friction between the collar and the horizontal rod,determine the magnitude of the velocity of the collar as itpasses through Point C.


The coefficients of friction between blocks A and C and thehorizontal surfaces are 0.24s and 0.20.k Knowingthat 5 kg,Am 10 kg,Bm and 10 kg,Cm determine(a) the tension in the cord, (b) the acceleration of eachblock.

Solution of Dynamics Assignment #6ME 247



(a) Assume uniformly decelerated motion.

Then 0v v at

At 9 s:t 00 (9)v a

or 0

9

va

Also 2 20 2 ( 0)v v a x

At 9 s:t 200 2 (30)v a

Substituting for a 2 000 2 (30) 0

9

vv

or 0 6.6667 m/sv 0or 6.67 m/sv

and 26.66670.74074 m/s

9a

(b) We have

0: 0yF N W or N W mg

Sliding: k kF N mg

:xF ma F ma or k mg ma

or

2

2

0.74074 m/s

9.81 m/sk

a

g

or 0.0755k


Kinematics: Uniformly accelerated motion. 0 0( 0, 0)x v

20 0

1,

2x x v t at

or 22 2

2 (2)(5)0.100 m/s

(10)

xa

t

0: sin 50 cos 20 0

sin 50 cos 20

yF N P mg

N P mg

: cos 50 sin 20xF ma P mg N ma

or cos 50 sin 20 ( sin 50 cos 20 )P mg P mg ma

(sin 20 cos 20 )

cos 50 sin 50

ma mgP

For motion impending, set 0 and 0.4sa

(20)(0) (20)(9.81)(sin 20 0.4 cos 20 )

cos 50 0.4 sin 50

419 N

P

For motion with 20.100 m/s , use 0.3.ka

(20)(0.100) (20)(9.81)(sin 20 0.3 cos 20 )

cos 50 0.3 sin 50P

301 NP


A:

B:

From the diagram

3 constantA Bx y

Then 3 0A Bv v

and 3 0A Ba a

or 3A Ba a (1)

(a) A: : x A AF m a A BT m a

Using Eq. (1) 3 A BT m a

B: : 3y B B B B BF m a W T m a

Substituting for T

3(3 )B A B B Bm g m a m a

or2

29.81 m/s0.83136 m/s

30 kg1 91 9

25 kg

BA

B

ga

m

m

Then 22.49 m/sA a

and 20.831 m/sB a

(b) We have 23 30 kg 0.83136 m/sT

or 74.8 NT


Load: We assume that sliding of load relative to trailer is impending:

m

s

F F

N

Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration max .a

0: 0yF N W N W

0.40m sF N W

max:x mF ma F ma

max0.40W

W ag

2max 12.88 ft/sa

2max 12.88 ft/sa

Uniformly accelerated motion.

2 20 2 with 0v v ax v 0 45 mi/h 66ft/sv

2max

2

12.88 ft/s

0 (66) 2( 12.88)

a a

x

169.1 ftx


Let y be positive downward for both blocks.

Constraint of cable: constantA By y

0A Ba a or B Aa a

For blocks A and B, :F ma

Block A: AA A

WW T a

g or A

A AW

T W ag

Block B: B BB B A

W WP W T a a

g g

A BB A A A

W WP W W a a

g g

Solving for aA, A BA

A B

W W Pa g

W W

(1)

2 20 0 0( ) 2 [ ( ) ] with ( ) 0A A A A A Av v a y y v

02 [ ( ) ]A A A Av a y y (2)

0 0( ) with ( ) 0A A A Av v a t v

A

A

vt

a (3)

(a) Acceleration of block A.

System (1): 200 lb, 100 lb, 0A BW W P

By formula (1), 1200 100

( ) (32.2)200 100Aa

2

1( ) 10.73 ft/sA a

System (2): 200 lb, 0, 50 lbA BW W P

By formula (1), 2200 100

( ) (32.2)200Aa

22( ) 16.10 ft/sA a

System (3): 2200 lb, 2100 lb, 0A BW W P

By formula (1), 32200 2100

( ) (32.2)2200 2100Aa

2

3( ) 0.749 ft/sA a


(b) 0 at ( ) 10 ft. Use formula (2).A A Av y y

System (1): 1( ) (2)(10.73)(10)Av 1( ) 14.65 ft/sAv

System (2): 2( ) (2)(16.10)(10)Av 2 17.94 ft/sAv

System (3): 3( ) (2)(0.749)(10)Av 3( ) 3.87 ft/sAv

(c) Time at 20 ft/s. Use formula (3).Av

System (1): 120

10.73t 1 1.864 st

System (2): 220

16.10t 2 1.242 st

System (3): 320

0.749t 3 26.7 st


Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider Band 0x is its initial value. Let L be the stretched length of the spring. Then, from the right triangle

2 2L x

The elongation of the spring is ,e L and the magnitude of the force exerted by the spring is

2 2( )sF ke k x

By geometry,2 2

cosx

x

: cosx x sF ma F ma

2 2

2 2( )

xk x ma

x

2 2

0

0 0

v

x

k xa x

m x

v dv a dx

02 2 2 2

2 20

2 2 2 2 20 0

2 2 2 2 20 0

2 2 2 2 20 0

0

00

1 1

2 2

1 10

2 2

2 2

2

v

xx

k x kv x dx x x

m mx

kv x x

m

kv x x

mk

x xm

2 20

answer:k

v xm

Chapter 12, Solution 28We first check that static equilibrium is not maintained:

( ) ( ) ( )

0.24(5 10)

3.6

A m C m s A CF F m m g

g

g

Since 10g 3.6g,B BW m g . equilibrium is not maintained.

Block A: :y A AF N m g

0.2A k A AF N m g

: 0.2A A A A AF m a T m g m a (1)

Block C: :y C CF N m g

0.2C k C CF N m g

: 0.2x C C C C CF m a T m g m a (2)

Block B: y B BF m a

2B B Bm g T m a (3)

From kinematics:1

( )2B A Ca a a (4)

(a) Tension in cord. Given data: 5 kg

10 kgA

B C

m

m m

Eq. (1): 0.2(5) 5 AT g a 0.2 0.2Aa T g (5)

Eq. (2): 0.2(10) 10 CT g a 0.1 0.2Ca T g (6)

Eq. (3): 10 2 10 Bg T a 0.2Ba g T (7)

Substitute into (4):

2

10.2 (0.2 0.2 0.1 0.2 )

224 24

1.2 0.35 (9.81 m/s )7 7

g T T g T g

g T T g

33.6 NT

(b) Substitute for T into (5), (7), and (6):

2240.2 0.2 0.4857(9.81 m/s )

7Aa g g

24.76 m/sA a

2240.2 0.3143(9.81 m/s )

7Ba g g

23.08 m/sB a

2240.1 0.2 0.14286(9.81 m/s )

7Ca g g

21.401 m/sC a




Block B of mass 10 kg rests as shown on the upper surface of a 22-kgwedge A. Knowing that the system is released from rest and neglectingfriction, determine (a) the acceleration of B, (b) the velocity of Brelative to A at 0.5 s.t


A 450-g tetherball A is moving along a horizontal circular pathat a constant speed of 4 m/s. Determine (a) the angle that thecord forms with pole BC, (b) the tension in the cord.


During a hammer thrower’s practice swings, the 7.1-kg head A ofthe hammer revolves at a constant speed v in a horizontal circle asshown. If 0.93 m and 60 , determine (a) the tension inwire BC, (b) the speed of the hammer’s head.


A single wire ACB of length 80 in. passes through a ring at C that isattached to a sphere which revolves at a constant speed v in thehorizontal circle shown. Knowing that 1 60 and 2 30 and thatthe tension is the same in both portions of the wire, determine the speed v.


A 54-kg pilot files a jet trainer in a half vertical loop of 1200-mradius so that the speed of the trainer decreases at a constant rate.Knowing that the pilot’s apparent weights at Points A and C are1680 N and 350 N, respectively, determine the force exerted onher by the seat of the trainer when the trainer is at Point B.


Rod OA rotates about O in a horizontal plane. The motion of the300 g collar B is defined by the relations r 300 100 cos (0.5t)and (t2 3t), where r is expressed in millimeters, t in seconds,and in radians. Determine the radial and transverse components ofthe force exerted on the collar when ( ) 0, ( ) 0.5 s.a t b t




(a) : sin 30 cos 40x A A A AB A AF m a W N m a

or

122

2cos 40

A

AB

a gN

Now we note: / , B A B Aa a a where /B Aa is directed along the top surface of A.

: cos 20 sin 50 y B y AB B B AF m a N W m a

or 10 ( cos 20 sin 50 )AB AN g a

Equating the two expressions for ABN

122

210( cos 20 sin 50 )

cos 40

A

A

a gg a

or2(9.81)(1.1 cos 20 cos 40 )

6.4061 m/s2.2 cos 40 sin 50Aa

/: sin 20 cos50 x B x B B B A B AF m a W m a m a

or /

2

2

sin 20 cos50

(9.81sin 20 6.4061cos50 ) m/s

7.4730 m/s

B A Aa g a


Finally /B A B A a a a

We have 2 2 26.4061 7.4730 2(6.4061 7.4730)cos50Ba

or 25.9447 m/sBa

and7.4730 5.9447

sin sin 50

A:

B:

or 74.4

25.94 m/sB a 75.6

(b) Note: We have uniformly accelerated motion, so that

0v at

Now / /B A B A B A B At t t v v v a a a

At 0.5 s:t 2/ 7.4730 m/s 0.5 sB Av

or / 3.74 m/sB A v 20


First we note2

A

A nv

a a

where sinABl

(a) 0: cos 0y AB AF T W

or cosA

ABm g

T

2

: sin Ax A A AB A

vF m a T m

Substituting for ABT and

22 2

22

2

sin sin 1 coscos sin

(4 m/s)1 cos cos

1.8 m 9.81 m/s

A AA

AB

m g vm

l

or2cos 0.906105cos 1 0

Solving cos 0.64479

or 49.9

(b) From above20.450 kg 9.81 m/s

cos 0.64479A

ABm g

T

or 6.85 NABT


First we note2A

A nv

a a

(a) 0: sin 60 0y BC AF T W

or27.1 kg 9.81 m/s

sin 6080.426 N

BCT

80.4 NBCT

(b)2

: cos60 Ax A A BC A

vF m a T m

or 2 (80.426 N)cos60 0.93 m

7.1 kgAv

or 2.30 m/sAv


First we note2C

C N

va a

where sin 30 sin 60AC BCL L

Now AC BC ABCL L L

or1 1

80 in.sin 30 sin 60

or 25.359 in.

0: cos30 cos60 0y CA CB CF T T W

or 0.73205cos30 cos60

CC

m gT m g

2

: sin 30 sin 60 Cx C C CA CB C

vF m a T T m

or2

0.73205 (sin 30 sin 60 ) CC C

vm g m

or2 2 25.359

0.73205 (32.2 ft/s ) ft (sin 30 sin 60 )12Cv

or 8.25 ft/sCv


First we note that the pilot’s apparent weight is equal to the vertical force that she exerts on theseat of the jet trainer.

At A:2

: An n A

vF ma N W m

or 2 2

2 2

1680 N(1200 m) 9.81 m/s

54 kg

25,561.3 m /s

Av

At C:2

: Cn n C

vF ma N W m

or 2 2

2 2

350 N(1200 m) 9.81 m/s

54 kg

19,549.8 m /s

Cv

Since constant,ta we have from A to C

2 2 2C A t ACv v a s

or 2 2 2 219,549.8 m /s 25,561.3 m /s 2 ( 1200 m) ta

or 20.79730 m/sta

Then from A to B

2 2

2 2 2

2 2

2

25,561.3 m /s 2( 0.79730 m/s ) 1200 m2

22,555 m /s

B A t ABv v a s

At B:2

: Bn n B

vF ma N m

or2 222,555 m /s

54 kg1200 mBN

or 1014.98 NB N

: | |t t B tF ma W P m a

or 2(54 kg)(0.79730 9.81) m/sBP

or 486.69 NB P

Finally, 2 2 2 2pilot( ) (1014.98) (486.69)

1126 N

B B BF N P

or pilot( ) 1126 NB F 25.6°


Polar coordinates and their derivatives.

0.3 0.1 cos (0.5 ) mr t 2( 3 ) radt t

0.05 sin (0.5 ) m/sr t (2 3) rad/st

2 2.025 cos (0.5 ) m/sr t 22 rad/s

(a) For t 0. 0.4 mr 0

0r 3 rad/s

2 20.025 m/sr 22 rad/s

Components of acceleration.

2

2 2

2

2

0.025 0.4( 3 )

35.777 m/s

2

(0.4)(2 ) 2( 3 )(0)

2.513 m/s

ra r r

a r r

Components of force.

(0.3)( 35.777)r rF ma 10.73 NrF

(0.3)(2.513)F ma 0.754 NF

(b) For t 0.5 s. 0.37071 mr 3.927 rad

0.11107 m/sr 2 rad/s

20.17447 m/sr 22 rad/s

Components of acceleration.

2

2

2

14.809 m/s

2

3.725 m/s

ra r r

a r r

Components of force.

(0.3)( 14.809)r rF ma 4.44 NrF

(0.3)(3.725)F ma 1.118 NF




A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h.(See sample Problem 12.6 for the definition of rated speed). Knowing that aracing car starts skidding on the curve when traveling at a speed of 180 mi/h,determine (a) the banking angle , (b) the coefficient of static friction betweenthe tires and the track under the prevailing conditions, (c) the minimum speed atwhich the same car could negotiate that curve.

Page 1 of 4


A small, 300-g collar D can slide on portion AB of a rod which is bent as shown.Knowing that 40 and that the rod rotates about the vertical AC at a constant rate of5 rad/s, determine the value of r for which the collar will not slide on the rod if the effectof friction between the rod and the collar is neglected.

Page 2 of 4

Problem 3

The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface. The coefficient of kinetic friction between the surfaces is 0.1 and g=32.2 ft/s2. Determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge.

Page 3 of 4

Problem 4

A 54-kg pilot files a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer increases at a constant rate. Knowing that the pilot’s apparent weights at Points A and C are 1682 N and 962 N, respectively, determine the force exerted on her by the seat of the trainer when the trainer is at Point B.

Page 4 of 4




Weight W mg

Acceleration2v

a

: sin cosx xF ma F W ma

2

cos sinmv

F mg

(1)

: cos siny yF ma N W ma

2

sin cosmv

N mg

(2)

(a) Banking angle. Rated speed 120 mi/h 176 ft/s.v 0F at rated speed.

2

2 2

0 cos sin

(176)tan 0.96199

(1000)(32.2)

43.89

mvmg

v

g

43.9

(b) Slipping outward. 180 mi/h 264 ft/sv 2

2

2

2

cos sin

sin cos

(264) cos 43.89 (1000)(32.2)sin 43.89

(264) sin 43.89 (1000)(32.2)cos 43.89

0.39009

F v gF N

N v g

0.390

(c) Minimum speed.2

2

2

cos sin

sin cos

(sin cos )

cos sin

F N

v g

v g

gv

2 2

(1000)(32.2) (sin 43.89 0.39009 cos 43.89 )

cos 43.89 0.39009 sin 51.875

13.369 ft /s

115.62 ft/sv

78.8 mi/hv


First note D ABCv r

0: sin 40 0yF N W

orsin 40

mgN

2

: cos 40 Dn n

vF ma N m

r

or2( )

cos 40sin 40

ABCrmgm

r

or2

2

2

1

tan 40

9.81 m/s 1

tan 40(5 rad/s)

0.468 m

ABC

gr

or 468 mmr

ds

4- The 12-lb block B starts from rest and slides on the 30-lb wedge,4, which is supported by ahorizontal surface. The coefficient of kinetic friction between the surfaces is 0.f . Determine(a) the acceleration of the wedge, and ft) the acceleration of the block relative to the wedge.(10 Points)

F B.D,*+y

tLh

'r

=rYt QNAt

-'-=7

ZeFo,

wu= )Zlb = ^n3 n-B=0.frz

wh =3otb = %tr -- frA--o.B2

Fo" Rbch B i

?'f F,-= -Ba^ ----P

-----r> rnB Cer3o

= aQn -= N,gti^sa ' fn - 4tGEo

N6 slnio -,/ N,o -,/Nn b3o = 14 Qo (r)

ryh- h='30 - Nucn3o : o

2&+7tMr*30 + No C*'3o e)

,oht-so - /B : ,B , a B/n - a^ Gso )

aa + -o&Ej3P -.qNR -- -R a B/ft (zl

n?

(fu=l Nn

I\k=,tro

tru =t o.-> Nn

Nrt=

"\ Zry *oot -p NB - -olA-'so = - -B aoEi3o

=2 mo3;3s \^'*""-o&Gso -

MB

A

14t

G,l ..

No,.t, //"rL a'rcooroo%e %e'r)

NB (Si^io -/C'io) *l2h -/

NR (si^to4bro7at r.;to) :

0.321f, /\/R : -e an +,/%

? L5) ;h 6t J rolvz 4n

lmns)n3o ) qfr --

qn = 0.35

,(o^ oI. 6)-2t

NR= t0 '34 lb

rol,,u- f". A R /ft i

--(o 3?Z) q uA-o,t (1o.34)

4 eXui,^, "'-') 4 un/t'o''-t ('s

tA^/ cm L. oLfo"'*of '

suL. ? Lzl & Cr) o,^J tt/'o i/"' NB I

srL NR & qn ;6 ? t9 *'1"

79 0.3?L C-,so lo'35) + tz 5"^ 3o

louh : t3.65 ryr, \

woc,so - ffio^%" ,'.

" ,o6LSo -l t't,, cn3o

*a4n n,/w

8)

//" a

.// wa

o3216

-m 0rtn

Problem 4

First we note that the pilot's apparent weight is equal to the vertical force that she exerts on the

seat ofthe jet trainer.

X,a: +luFn=ms,' Nn-W =-&p

rt=e2oo"[##-r.rr nvs'z)

=25605.78 m2 /s2

AtC: +llFn =*on, N, +W =*t

$=ftr"tg =1rzoo"'r[ffi.e.8r mis'?) I

=33149.78 m2 ls2

Since a, = cotstant, we have ftom A to C

or

or

Then froml to B

fc =fi +2a, Lsn,

33149.78 m2 ls2 =25605.78 m2 /s2 + 2a,(n x1200 m)

AtB: I-EF,

at =1 ttils2

G=ln+2a, Nnu

=25605.78 m2/s2 +2(t rIVs1)( +x t2oo ml' '\.2 )=29377.78 mzlsz

.,2

= lilan\ N, = m'np

29377.78 m2 h2Nn =54 *g

r2oo,n

N, = 1l!! |ri .-

or

or

$r, = *o,t Pu -w - ffi ot

or

or

Finally,

or

Pu =(54 kg)(l+9.81) m/s2

Pr = 583'74 Nt

({ir",)a =JN|;E =

= 1445 N

1n

Q32Dz +683.74)2

($no,)r =1445 N \23'8" <




A 4-kg stone is dropped from a height h and strikes the ground with a velocity of 25 m/s. (a) Find the kineticenergy of the stone as it strikes the ground and the height h from which it was dropped. (b) Solve Part a,assuming that the same stone is dropped on the moon. (Acceleration of gravity on the moon 1.62 m/s2.)


In an ore-mixing operation, a bucket full of ore issuspended from a traveling crane which moves along astationary bridge. The bucket is to swing no more than 4 mhorizontally when the crane is brought to a sudden stop.Determine the maximum allowable speed v of the crane.


A package is projected 10 m up a 15º incline so that it just reachesthe top of the incline with zero velocity. Knowing that thecoefficient of kinetic friction between the package and the inclineis 0.12, determine (a) the initial velocity of the package at A,(b) the velocity of the package as it returns to its original position.


The subway train shown is traveling at a speed of 30 mi/hwhen the brakes are fully applied on the wheels of cars Band C, causing them to slide on the track, but are notapplied on the wheels of car A. Knowing that thecoefficient of kinetic friction is 0.35 between the wheelsand the track, determine (a) the distance required to bringthe train to a stop, (b) the force in each coupling.


Two identical blocks are released from rest. Neglecting the mass of thepulleys and the effect of friction, determine (a) the velocity of Block Bafter it has moved 2 m, (b) the tension in the cable.


A 3-kg block rests on top of a 2-kg block supported by, but not attached to a springof constant 40 N/m. The upper block is suddenly removed. Determine (a) themaximum speed reached by the 2-kg block, (b) the maximum height reached by the2-kg block.




(a) On the earth.

2 21 1(4 kg)(25 m/s) 1250 N m

2 2T mv 1250 JT

2

1 1 2 2 1 1 2 2

2

(4 kg)(9.81 m/s ) 39.240 N

0 39.240 N

(1250 N m)31.855 m

(39.240 N)

W mg

T U T T U Wh T

Th

W

31.9 mh

(b) On the moon.

Mass is unchanged, 4 kg.m

Thus T is unchanged. 1250 JT

Weight on the moon is 2(4 kg)(1.62 m/s )

6.48 Nm m

m

W mg

W

(1250 N m)192.9 m

(6.48 N)mm

Th

W

192.9 mmh


1

2

21

2

0

1

20

v v

v

T mv

T

1 2

2 2 2 2 2 2

2

1 2

1 1 2 2

4 m

(10 m) (4 m)

100 16 84 84

10 10 84 0.8349 m

(9.81)(0.8349) 0.8190 m

U mgh d

AB d y y

y y

h y

U m

T U T

2

2

10.8190 m 0

2

(2)(0.8190) 16.38

mv

v

4.05 m/sv


(a) Up the plane, from A to C, 0.Cv

21, 0

2( sin15 )(10 m)

A A C

A C

T mv T

U W F

0: cos1 5 0F N W

cos1 5N W

2

2

2

0.12 cos1 5

(sin15 0.12cos15 )(10 m)

1(sin15 0.12cos15 )(10 m)

2

(2)(9.81)(sin15 0.12cos15 )(10 m)

73.5

k

A C

A A C C A

A

A

F N W

U W

WT U T v W

g

v

v

8.57 m/sA v 15

(b) Down the plane from C to A.

2

2

2

2

10 ( sin15 )10

2( reverses direction.)

10 (sin15 0.12cos15 )(10 m)

2

(2)(9.81)(sin15 0.12cos15 )(10 m)

28.039

C A A C A

C C A A A

A

A

T T mv U W F

F

T U T W mv

v

v

5.30 m/sA v 15°


0.35 (0.35)(100 kips) 35 kips

(0.35)(80 kips) 28 kipsk B

C

F

F

1 30 mi/h 44 ft/sv 2 20 0v T

(a) Entire train: 1 1 2 2T U T

22

1 (80 kips 100 kips 80 kips)(44 ft/s) (28 kips 35 kips) 0

2 32.2 ft/sx

124.07 ftx 124.1 ftx

(b) Force in each coupling: Recall that 124.07 ft.x

Car A: Assume ABF to be in tension.

1 1 2 2

21 80 kips(44) (124.07 ft) 0

2 32.219.38 kips

AB

AB

T V T

F

F

19.38 kips (tension)ABF

Car C: 1 1 2 2T U T

21 80 kips(44) ( 28 kips)(124.07 ft) 0

2 32.228 kips 19.38 kips

BC

BC

F

F

8.62 kipsBCF 8.62 kips (tension)BCF


(a) Kinematics: 2

2B A

B A

x x

v v

A and B. Assume B moves down.

1

1

2 22

22

22

1 2

1 2

1 2

0

0

1 1

2 2

1(2 kg)

2 4

5

4(cos30 )( ) (cos30 )

2 m

1 m

3(2)(9.81) [ 1 2]

2

16.99 J

A A B B

BB

B

A A B B

B

A

v

T

T m v m v

vv

T v

U m g x m g x

x

x

U

U

Since work is positive, block B does move down.

1 1 2 2

2

2

50 16.99

4

13.59

B

B

T U T

v

v

3.69 m/sB v 60°

(b) B alone.

1

1

2

0

0

3.69 m/s, (from ( ))

v

T

v a

2 22 2

1 2

21 2

1 2

1 1(2)(3.69) 13.59 J

2 2( )(cos30 )( ) ( )( )

3(2 kg)(9.81 m/s ) ( ) (2 m)

2

33.98 2

B

B B B

T m v

U m g x T x

U T

U T


1 1 2 2 0 33.98 2 13.59

2 33.98 13.59 20.39

T U T T

T

10.19 NT


(a) At the initial Position , the force in the spring equals the weight of both blocks, i.e., 5g N, thusat a distance x, the force in the spring is

5

5 40s

s

F g kx

F g x

Maximum velocity of the 2 kg block occurs while the spring is still in contact with the block.

2 2 21 2

21 2

0

1 10 (2 kg)( )

2 2

(5 40 ) 2 3 20x

T T mv v v

U g x dx gx gx x

2 2

1 1 2 2 0 3 20T U T gk x v (1)

Maximum v when 0

3 40

3 (max ) m

40

dv

dxg x

gx v

Substituting in (1) 2 (max ) 0.7358 mx v

2 2max (3)(9.81)(0.7358) (20)(0.7358)

21.65 10.83

10.82

v

max 3.29 m/sv


(b) 0

0

1 3

initial compression of the spring

(2 3 )m

40 85 40

0 0s

x

g g gx

F g x

T T

/8

1 30

2 2

1 3

2

1 1 3 3

(5 40 ) 2

5 202

8 64

200 2 0

6410 (10)(9.81)

64 64

gU g x dx gh

g gU gh

gT U T gh

gh

1.533 mh




A 10-lb collar B can slide without friction along a horizontal rodand is in equilibrium at A when it is pushed 5 in. to the right andreleased. The underformed length of each spring is 12 in. and theconstant of each spring is k 1.6 1b/in. Determine (a) themaximum speed of the collar, (b) the maximum acceleration ofthe collar.


A 3-kg collar can slide without friction on a vertical rod and is resting in equilibriumon a spring. It is pushed down, compressing the spring 150 mm, and released.Knowing that the spring constant is 2.6k kN/m, determine (a) the maximumheight h reached by the collar above its equilibrium position, (b) the maximum speedof the collar.


A spring is used to stop a 50-kg package, which is moving down a20º incline. The spring has a constant 30 kN/mk and is held bycables so that it is initially compressed 50 mm. Knowing that thevelocity of the package is 2 m/s when it is 8 m from the spring andneglecting friction, determine the maximum additional deformationof the spring in bringing the package to rest.


A 1-lb collar is attached to a spring and slides without friction alonga circular rod in a vertical plane. The spring has an undeformedlength of 5 in. and a constant 10 lb/ft.k Knowing that the collar isreleased from being held at A, determine the speed of the collar andthe normal force between the collar and the rod as the collar passesthrough B.


A 1200-kg automobile is moving at a speed of 90 km/h when the brakes are fully applied, causing all fourwheels to skid. Determine the time required to stop the automobile (a) on dry pavement (k 0.75), (b) on anicy road (k 0.10).


A light train made of two cars travels at 45 mi/h. Car A weighs18 tons, and car B weighs 13 tons. When the brakes are applied,a constant braking force of 4300 lb is applied to each car.Determine (a) the time required for the train to stop after thebrakes are applied, (b) the force in the coupling between thecars while the train is slowing down.


The system shown is released from rest. Determine the time it takes forthe velocity of A to reach 1 m/s. Neglect friction and the mass of thepulleys.


At an intersection, car B was traveling south and car A wastraveling 30° north of east when they slammed into eachother. Upon investigation, it was found that after the crash,the two cars got stuck and skidded off at an angle of 10°north of east. Each driver claimed that he was going at thespeed limit of 50 km/h and that he tried to slow down butcouldn’t avoid the crash because the other driver was goinga lot faster. Knowing that the masses of cars A and B were1500 kg and 1200 kg, respectively, determine (a) which carwas going faster, (b) the speed of the faster of the two cars ifthe slower car was traveling at the speed limit.




(a) Maximum velocity occurs at A where the collar is passing through its equilibrium position.

Position .

1 0T 2 2

(1.6 lb/in.) (12 in./ft) 19.2 lb/ft

5 12 13 in.

113 in. 12 in. 1 in. ft.

125

5 in. ft12

OC

OC

AC

k

L

L

L

2 21

2 2

1

1 1( ) ( )

2 2

19.2 1 5lb/ft ft ft

2 12 12

1.733 lb ft

OC ACV k L k L

V

Position .

2 2 22 2 max max

1 1 10 5= =

2 2 g gT mv v v

2 0 (Both springs are unstretched.)V

21 1 2 2 max

50 1.733 0T V T V v

g

22 2 2max

(1.733 lb ft) (32.2 ft/s )11.16 ft /s

(5 lb)v

max 3.34 ft/sv

(b) Maximum acceleration occurs at C where the horizontal force on the collar is at maximum.

1 2 max

max

max

cos

cos

1 5 5 10 lb(19.2 lb/ft) ft ft

12 13 12

108.615

OC AC

F ma F F ma

k L k L ma

ag

ag

2

max(8.615 lb) (32.2 ft/s )

(10 lb)a 2

max 27.7 ft/sa


(a) Maximum height is reached when

2 0v

Thus, 1 2 0T T

g eV V V

Position . 1( ) 0gV

Total spring deflection from undeflected spring position 1x

1

2

1 3

1

2 3 21 1

/ 0.150

(3 kg)(9.81 m/s )/ 0.150 0.150 m

(2.6 10 N/m)

0.01132 0.150 0.1613 m

1 1( ) (2.6 10 N/m)(0.1613 m) 33.83 J

2 2e

x mg k

x mg k

x

V k x

Position .1

2

0 33.83 33.83 J

( ) (0.150 ) 3 (0.150 )g

V

V mg h g h

2( ) 0eV (spring is not attached to the collar)

1 1 2 2 1 1 2 2

2

0 ( ) ( ) 0 ( ) ( )

0 0 33.83 0 3 (0.150 ) 0

33.83 J(0.150 m)

(3 kg)(9.81 m/s )

0.9995 m

g e g eT V T V V V V V

g h

h

1000 mmh


(b) Maximum velocity occurs when the acceleration 0, i.e., at equilibrium.

At Position

2 2 23 3 max max

23 3 3 1

2 3 23

3

21 1 3 3 max

2max

1 1(3) 1.5

2 21

( ) ( ) (0.150) ( 0.150)2

1(3 kg)(9.81 m/s )(0.150 m) (2.6 10 N/m)(0.1613 0.150 m)

24.415 J 0.1660 J 4.581 J

0 33.83 1.5 4.581

(29.

g e

T mv v v

V V V mg k x

V

V

T V T V v

v

2 2249)/1.5 19.50 m /s max 4.42 m/sv


Let Position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by thecable. Let Position 2 be the position of maximum compression. Let x be the additional compression of thespring. Use the principle of conservation of energy. 1 1 2 2 .T V T V

Position 1: 2 21 1

1 1

2 3 21 1

1 1(50)(2) 100 J

2 2(50)(9.81) (8 sin 20 ) 1342.09 J

1 1(30 10 )(0.050) 37.5 J

2 2

g

e

T mv

V mgh

V ke

Position 2: 22 2 2

2 2

2 3 2 22 2

10 since 0.

2(50)(9.81)( sin 20 ) 167.76

1 1(30 10 )(0.05 ) 37.5 1500 15,000

2 2

g

e

T mv v

V mgh x x

V ke x x x

Principle of conservation of energy.

2

2

100 1342.09 37.5 167.61 37.5 1500 15,000

15,000 1332.24 1442.09 0

x x x

x x

Solving for x,

0.26882 and 0.35764x 0.269 mx


For the collar, 210.031056 lb s /ft

32.2

Wm

g

For the spring, 10 lb/ft 5 in.Ok

At A: 7 5 5 17 in.

12 in. 1 ftA

O

At B: 2 2(7 2) 5 13 in.

21.8 in. ft

3

B

B O

Velocity of the collar at B.

Use the principle of conservation of energy.

A A B BT V T V

where 210

2A AT mv

2

2

2 2 2

2

2

2

2 2 2

1( ) (0)

21

(10)(1) 0 5 ft lb21 1

(0.031056) 0.0155282 21

( )2

1 2 5(10) (1)

2 3 12

1.80556 ft lb

0 5 0.015528 1.80556

205.72 ft /s

A A O

B B B B

B B O

B

B

V k W

T mv v v

V k Wh

v

v

14.34 ft/sB v


Forces at B.

2

2( ) (10) 6.6667 lb.

3

5sin

135

5 in. ft12

(0.031056)(205.72)

511215.3332 lb

s B O

Bn

F k

mvma

: siny y s nF ma F W N ma

sin

515.3332 1 (6.6667)

13

13.769 lb

n sN ma W F

N

13.77 lbN


1

1 1 1

0k

k k k

mv Wt

mv mv vt

W mg g

(a) For 0.75k

2

25 m/s

(0.75)(9.81 m/s )t 3.40 st

(b) For 0.10k

2

25 m/s

(0.10)(9.81 m/s )t 25.5 st


(a) Entire train: 1 45 mi/h 66 ft/sv

18 13 31 tons 62,000 lbA BW W

1 262,000

0 (4300 4300) (66)tg

1 2 2

(62,000 lb)(66 ft/s)

(32.2 ft/s )(8600 lb)t 1 2 14.78 st

(b) Car A: 1 218 tons 36,000 lb, 14.78 sAW t

2

(36,000 lb)0 [(4300 lb) ][14.78 s] (66 ft/s)

(32.2 ft/s )CF

692.5 lbCF 693 lb (tension)CF


Kinematics: Dependent motion

Cable length: constantA A BL x x x

2 0A BdL

v vdt

Here velocities are defined as positive if downward.

2B Av v (1)

Let T be the tension in the cable.

Use the principle of impulse and momentum.

Collar A, components : 2 A A ATt m gt m v (2)

Block B, components : B B BTt m gt m v (3)

Subtract twice Eq. (3) from Eq. (2) to eliminate Tt.

(2 ) 2

2

(2 )

B A B B A A

B B A A

B A

m m gt m v m v

m v m vt

m m g

Data: 20 kg 30 kg 2 10 kgA B B Am m m m

1.0 m/s, i.e., 1.0 m/sAv

From Eq. (1) 2.0 m/sBv

[(2)(15)(2.0) (20)( 1.0)]

(10)(9.81)t

0.815 st


(a) Total momentum of the two cars is conserved.

, :mv x cos 30 ( ) cos 10A A A Bm v m m v (1)

, :mv y sin 30 ( ) sin 10A A B B A Bm v m v m m v (2)

Dividing (1) into (2),

sin 30 sin 10

cos 30 cos 30 cos 10B B

A A

m v

m v

(tan 30 tan 10 )( cos 30 )

(1500)(0.4010) cos 30

(1200)

0.434 2.30

B A

A B

B

A

BA B

A

v m

v m

v

v

vv v

v

Thus, A was going faster.

(b) Since Bv was the slower car,

50 km/hBv

(2.30)(50)Av 115.2 km/hAv




A 6-lb block is attached to a cable and to a spring as shown. The constant of the spring isk 8 lb/in. and the tension in the cable is 3 lb. If the cable is cut, determine (a) themaximum displacement of the block, (b) the maximum speed of the block.


A 1.2-lb collar can slide without friction along the semicircularrod BCD. The spring is of constant 1.8 lb/in. and its undeformedlength is 8 in. Knowing that the collar is released from rest at B,determine (a) the speed of the collar as it passes through C, (b) theforce exerted by the rod on the collar at C.


The initial velocity of the block in position A is 30 ft/s. Knowingthat the coefficient of kinetic friction between the block and theplane is 0.30,k determine the time it takes for the block toreach B with zero velocity, if (a) 0, (b) 20 .


Solve Problem 13.129, assuming that a constant brakingforce of 4300 lb is applied to car B, but the brakes on car Aare not applied.

PROBLEM 13.129 A light train made of two cars travelsat 45 mi/h. Car A weighs 18 tons, and car B weighs 13 tons.When the brakes are applied, a constant braking force of4300 lb is applied to each car. Determine (a) the timerequired for the train to stop after the brakes are applied,(b) the force in the coupling between the cars while thetrain is slowing down.


A 125-g ball moving at a speed of 3 m/s strikes a 250-g plate supported bysprings. Assuming that no energy is lost in the impact, determine (a) the velocityof the ball immediately after impact, (b) the impulse of the force exerted by theplate on the ball.




8 lb/in. 96 lb/ftk

1

2 21 1 2 2 2

0: ( ) 6 3 3 lb

1 6 lb0, 0: 0.09317

2 32.2

y sF F C

v T T v v

For weight: 1 2 (6 lb) 6U x x

For spring: 21 2

0(3 96 ) 3 48

xU x dx x x

2 21 1 2 2 2

2 22

: 0 6 3 48 0.09317

3 48 0.09317

T U T x x x v

x x v

(1)

(a) For 2, 0:mx v 23 48 0x x

3 10, ft

48 16mx x 0.75 in.m x

(b) For ,mv we seek maximum of 21 2 3 48v x x

1 2 3 13 96 0 ft ft

96 32

dUx x

dx

Eq. (1):2

21 13 ft 48 ft 0.09317

32 32 mv

2 0.5031 0.7093 ft/sm mv v 8.51 in./sm v

Note: 1 2U for the spring may be computed using sF x curve

1 2

2

area

13 96

2

U

x x


2 2 2

2 2

12 6 3 13.748 in.

12 3 12.369 in.

1.8 lb/in. 21.6 lb/ft

AB

AC

L

L

k

(a) Speed at C.

At B: 0 0B Bv T

2

2

( ) ( )

13.748 in. 8 in.

5.748 in. 0.479 ft

1( ) ( )

21

( ) (21.6 lb/ft)(0.479 ft)2

( ) 2.478 lb ft

B B e B g

AB

AB

B e AB

B e

B e

V V V

L

L

V k L

V

V

6( ) (1.2 lb) ft 0.600 lb ft

12( ) ( ) ( ) 2.478 0.600 3.078 lb ft

B g

B B e B g

V Wr

V V V

At C: 2 22

2

2

1 1 1.2 lb

2 2 32.2 ft/s

0.018634

1( ) ( )

212.369 in. 8 in.

4.369 in. 0.364 ft

C C C

C C

C e AC

AC

T mv v

T v

V k L

L

2

2

2 2 2

1( ) = (21.6 lb/ft)(0.364) 1.4316 lb ft

2( ) 0

( ) ( ) ( ) 1.4316 0 1.4316 lb ft

0 3.078 0.018634 1.4316

(3.078 1.4316)88.36 ft /s

(0.018634)

C e

C g

C C e C g

B B C C

C

C

V

V

V V V

T V T V

v

v

9.40 ft/sCv


(b) Force of rod on collar at C.

1

0 (no friction)

3tan 14.04°

12( )(cos sin )

(21.6)(0.364)(cos 14.04 sin 14.04 )

7.628 1.9069 (lb)

z

x y

e AC

e

e

F

F F

k L

F i j

F i k

F i k

F i k

2

2 2

2

( 7.628) ( 1.2) 1.906

(1.2 lb)(88.36 ft /s )7.628 lb 0 1.2 lb

(32.2 ft/s )(0.5 ft)

7.628 lb

7.78 lb

x y

x y

x

y

F F

mvma

r

F F

F

F

F i j k

j k

(7.63 lb) (7.78 lb) F i j


(a) 0

2

30 ft/s0

(32.2 ft/s )(0.30)A

A kk

vW v Wt tg g

3.11st

(b) 20

Impulse-momentum in x direction

cos 20 sin 20 0A kW

v Wt Wtg

2

( cos 20 sin 20 )

30 ft/s

(32.2 ft/s )(0.30cos 20 sin 20 )

A

k

vt

g

1.493 st


(a) Entire train: 1 45 mi/h 66 ft/sv

18 13 31 tons 62,000 lbA BW W

1 2 2

(62,000 lb)0 (4300 lb) (66 ft/s)

(32.2 ft/s )t

1 2 29.55 st 1 2 29.6 st

(b) Car A:

1 2 1 1 20 ( ) 29.55 sC AF t m v t

2

(36,000 lb)(66 ft/s)2497 lb

(32.2 ft/s )(29.55 s)CF 2500 lb (tension)CF


(a) For the system which is the ball and the plate, momentum is conserved.

All forces are non-impulsive, except the equal and opposite forces between the plate and the ball.

( ) ( ) ( )B B pmv mv mv

(0.125 kg)(3 m/s) ( 0.125 kg)( ) (0.250 kg)

0.5 1.5

B p

p B

v v

v v

(1)

Since there is no energy lost, the kinetic energy of the system is conserved.

Before impact, 2

2

1

21

(0.125 kg)(3 m/s)20.563 J

B BT m v

After impact, 2 21 1

2 2B B P PT m v m v

Substituting for pv from (1)

2 2

2

2

1 1(0.125 kg)( ) (0.250 kg)[0.5 1.5]

2 2

0.09375( ) 0.1875 0.2813

0.563 0.09375( ) 0.1875 0.2813

B B

B B

B B

T v v

T v v

T T v v

2 2 3 0

2 4 121 2 3 , 1

2

B B

B

v v

v

( 3 m/s before impact)Bv 1.000 m/sB v

(b) Ball alone.

(0.125 kg)( 3 m/s) (0.125 kg)(1 m/s)F t 0.500 N st F




Two steel blocks are sliding on a frictionless horizontalsurface with the velocities shown. Knowing that afterimpact the velocity of B is observed to be 10.5 ft/s to theright, determine the coefficient of restitution between thetwo blocks.


A 600-g ball A is moving with a velocity of magnitude 6 m/s when it ishit as shown by a 1-kg ball B, which has a velocity of magnitude 4 m/s.Knowing that the coefficient of restitution is 0.8 and assuming nofriction, determine the velocity of each ball after impact.


A girl throws a ball at an inclined wall from a heightof 1.2 m, hitting the wall at A with a horizontalvelocity 0v of magnitude 15 m/s. Knowing that thecoefficient of restitution between the ball and thewall is 0.9 and neglecting friction, determine thedistance d from the foot of the wall to the Point Bwhere the ball will hit the ground after bouncing offthe wall.


A 1-kg block B is moving with a velocity v0 of magnitude 0 2 m/sv as it hits the 0.5-kg sphere A, which is at rest and hanging from a cordattached at O. Knowing that 0.6k between the block and thehorizontal surface and 0.8e between the block and the sphere,determine after impact (a) the maximum height h reached by thesphere, (b) the distance x traveled by the block.


A 1.3-lb sphere A is dropped from a height of 1.8 ft onto a 2.6-lbplate B, which is supported by a nested set of springs and is initiallyat rest. Knowing that the coefficient of restitution between the sphereand the plate is 0.8,e determine (a) the height h reached by thesphere after rebound, (b) the constant k of the single springequivalent to the given set if the maximum deflection of the plate isobserved to be equal to 3h.


A 700-g sphere A moving with a velocity v0 parallel to theground strikes the inclined face of a 2.1-kg wedge B which canroll freely on the ground and is initially at rest. After impact, thesphere is observed from the ground to be moving straight up.Knowing that the coefficient of restitution between the sphereand the wedge is 0.6,e determine (a) the angle that theinclined face of the wedge makes with the horizontal, (b) theenergy lost due to the impact.




The total momentum is conserved.

A A B B A A B Bm v m v m v m v

(1.5 lb) (0.9 lb) (1.5 lb) (0.9 lb)(10 ft/s) (6 ft/s) ( ) (10.5 ft/s)

15 5.4 9.45

1.57.30 ft/s

A

A

vg g g g

v

Coefficient of restitution.

10.5 7.30

10 60.800

B A

A B

v ve

v v

0.800e


Before After

6 m/s

( ) (6)(cos 40 ) 4.596 m/s

( ) 6(sin 40°) 3.857 m/s

( ) 4 m/s

( ) 0

A

A n

A t

B B n

B t

v

v

v

v v

v

t-direction.

Total momentum conserved.

( ) ( ) ( ) ( )

(0.6 kg)( 3.857 m/s) 0 (0.6 kg)( ) (1 kg)( )

2.314 m/s 0.6 ( ) ( )

A A t B B t A B t B B t

A t B

A t B t

m v m v m v m v

v v t

v v

(1)

Ball A alone.

Momentum conserved.

3.857 ( )

( ) 3.857 m/s

A A A A A tt t

A t

m v m v v

v

(2)

Replacing ( )A tv in (2) in Eq. (1)

2.314 (0.6)( 3.857) ( )

2.314 2.314 ( )

( ) 0

B t

B t

B t

v

v

v

n-direction

Relative velocities.

[( ) ( ) ] ( ) ( )

[(4.596) ( 4)](0.8) ( ) ( )

6.877 ( ) ( )

A n B n B n A n

B n A n

B n A n

v v e v v

v v

v v

(3)


Total momentum conserved.

( ) ( ) ( ) ( )

(0.6 kg)(4.596 m/s) (1 kg)( 4 m/s) (1 kg)( ) (0.6 kg)( )

1.2424 ( ) 0.6 ( )

A A n B B n A A n B B n

B n A n

B n A n

m v m v m v m v

v v

v v

(4)

Solving Eq. (4) and (3) simultaneously,

( ) 5.075 m/s

( ) 1.802 m/sA n

B n

v

v

Velocity of A.

2 2

|( ) |tan

|( ) |

3.857

5.07537.2 40 77.2

(3.857) (5.075)

6.37 m/s

A t

A n

A

v

v

v

6.37 m/sA v 77.2

Velocity of B. 1.802 m/sB v 40


Momentum in t direction is conserved.

sin 30

(15)(sin 30 )

7.5 m/s

t

t

mv mv

v t

v

Coefficient of restitution in n-direction.

( cos30 )

(15)(cos30 )(0.9)

11.69 m/s

n

n

n

v e v

v

v

Writing v in terms of x and y components

0

0

0

0

( ) cos30 sin 30

( ) (11.69)(cos30 ) (7.5)(sin 30 ) 6.374 m/s

( ) sin 30 cos30

( ) (11.69)(sin 30 ) (7.5)(cos30 ) 12.340 m/s

x n t

x

y n t

y

v v v

v

v v v

v

Motion of a projectile. (origin at 0)2

0 0

22

( )( )

2

1.2 (12.340 m/s) (9.81 m/s )2

ygt

y y v t

ty t


Time to reach Point B ( 0)By

29.810 1.2 12.340

2B Bt t

2.610 sBt

0 0( )

0 6.374

(6.374)( )

(6.374 m/s)(2.610 s)

16.63 m

x

B B

B

x x v t

x t

x t

x

1.2 cot 60°

15.94 mBd x

15.94 md


Velocities just after impact

Total momentum in the horizontal direction is conserved.

0 (1 kg) (2 m/s) (0.5 kg) ( ) (1 kg) ( )A A B B A A B B

A B

m v m v m v m v

v v

4 2A Bv v (1)

Relative velocities.

( ) ( )

(0 2) (0.8)

1.6

A B B A

B A

B A

v v e v v

v v

v v

(2)

Solving Eqs. (1) and (2) simultaneously,

0.8 m/s

2.4 m/sB

A

v

v

(a) Conservation of energy.

21 1 1

21

10

21

(2.4 m/s) 2.882

A

A A

T m v V

T m m

2

2

0

A

T

V m gh

1 1 2 2 2.88T V T V m 0 0A m (9.81)A

0.294 mh


(b) Work and energy.

2 21 1 2

1 2

1 2

1 1 2 2

1 1(0.8 m/s) 0.32 0

2 2(0.6)( )(9.81)

5.886

0.32 5.886 0

B B B

f k x B B

B

B B

T m v m m T

U F x Nx m gx m x

U m x

T U T m m x

0.0544 mx 54.4 mmx


Velocity of A and B after impact

2

2

1.30.04037 lb s /ft

32.2

2.60.08075 lb sec /ft

32.2

AA

BB

Wm

g

Wm

g

Sphere A falls. Use conservation of energy to find ,Av the speed just before impact. Use the plate surface asthe datum.

21 1 0 2 2

21 1 2 2 0

10, , , 0

21

0 02

A A A

A A A

T V m gh T m v V

T V T V m gh m v

With 0

0

1.8 ft,

2 (2)(32.2)(1.8)A

h

v gh

10.767 ft/sA v

Analysis of the impact. Conservation of momentum.

withA A B B A A B Bm m m m v v v v 0B v

Dividing by Am and using y-components with ( / 2)B Am m

10.767 0 ( ) 2( )A y B yv v (1)

Coefficient of restitution. ( ) ( ) [( ) ( ) ]B y A y A y B yv v e v v

( ) ( ) ( ) 10.767B y A y A yv v e v e (2)

Solving Eqs. (1) and (2) simultaneously with 0.8e gives

( ) 2.153 ft/s

( ) 6.460 ft/s

2.153 ft/s ,

6.460 ft/s

A y

B y

A

B

v

v

v

v

(a) Sphere A rises. Use conservation of energy to find h.

21 1 2 2

21 1 2 2

1( ) , 0, 0,

21

: ( ) 0 02

A A A

A A A

T m v V T V m gh

T V T V m v m gh

2 2( ) (2.153)

2 (2)(32.2)Av

hg

0.0720 fth


(b) Plate B falls and compresses the spring. Use conservation of energy.

Let 0 be the initial compression of the spring and be the additional compression of the spring afterimpact. In the initial equilibrium state,

0 00: 0 ory B BF k W k W (3)

Just after impact: 2 21 1 0

1 1( ) ,

2 2B BT m v V k

At maximum deflection of the plate, 2 0T

22 2 2 0

1( ) ( ) ( )

2g e BV V V W k

Conservation of energy: 1 1 2 2T V T V

2 2 2 20 0 0

1 1 1 1( ) 0

2 2 2 2B B Bm v k W k k k

Invoking the result of Eq. (3) gives

2 21 1( )

2 2B Bm v k (4)

Data: 20.08075 lb s /ft, 6.460 ft/sB Bm v

3 (3)(0.072) 0.216 fth 2 2

2 2

( ) (0.08075)(6.460)

(0.216)B Bm v

k

72.2 lb/ftk


(a) Momentum of sphere A alone is conserved in the t-direction.

0

0

cos sin

tanA A A

A

m v m v

v v

(1)

Total momentum is conserved in the x-direction.

0 ( )B B A B B A xm v m v m v v

0, ( ) 0B A xv v

0

0

0 0.700 2.1 0

/B

B

v v

v v B

(2)

Relative velocities in the n-direction.

0

0

( sin 0) sin cos

( )(0.6) cotB A

B A

v e v v

v v v

(3)

Substituting Bv from Eq. (2) into Eq. (3)

0 0

0

0.6 0.333 cot

0.267 cotA

A

v v v

v v

(4)

Dividing (4) into (1)

21 tantan

0.267 cot

tan 1.935 62.7

(b) From (1) 0 tan (1.935)A Av v v

0 00.5168 , /A Bv v v v B (2)

2 2 2lost

1 1( )

2 2A A A A B BT m v m v m v

2 2 2lost 0 0 0

2lost 0

2lost 0

1 1(0.7)( ) [(0.7)(0.5168 ) (2.1)( /3) ]

2 21

[0.7 0.1870 0.2333]2

0.1400 J

T v v v

T v

T v

2lost 00.1400T v




An airline employee tosses two suitcases, of mass 15 kg and20 kg, respectively, onto a 25-kg baggage carrier in rapidsuccession. Knowing that the carrier is initially at rest and thatthe employee imparts a 3-m/s horizontal velocity to the 15-kgsuitcase and a 2-m/s horizontal velocity to the 20-kg suitcase,determine the final velocity of the baggage carrier if the firstsuitcase tossed onto the carrier is (a) the 15-kg suitcase, (b)the 20-kg suitcase.


A 180-lb man and a 120-lb woman stand side by side at thesame end of a 300-lb boat, ready to dive, each with a 16-ft/svelocity relative to the boat. Determine the velocity of theboat after they have both dived, if (a) the woman dives first,(b) the man dives first.


For the system of particles of Problem 14.13, the velocity of particle B in y direction is known as vy = 2 ft/s. Determine (a) the components vx and vz of the velocity of particle B for which the angular momentum HO of the system about O is parallel to the z axis, (b) the value of HO.

PROBLEM 14.13 A system consists of three particles A, B, and C. We know that 5lb,AW = 4 lb,BW = and 3 lbCW = and that the velocities of the particles expressed in ft/s are, respectively, 2 3A = + −v i j 2k,

,B x y zv v v= + +v i j k and 3 2 .C = − − +v i j k Determine (a) the components vx and vz of the velocity of particle B for which the angular momentum HO of the system about O is parallel to the x axis, (b) the value of HO.


A small airplane of mass 1500 kg and a helicopter of mass3000 kg flying at an altitude of 1200 m are observed tocollide directly above a tower located at O in a woodedarea. Four minutes earlier, the helicopter had been sighted8.4 km due west of the tower and the airplane 16 km westand 12 km north of the tower. As a result of the collision,the helicopter was split into two pieces, H1 and H2, ofmass m1 = 1000 kg and m2 = 2000 kg, respectively; theairplane remained in one piece as it fell to the ground.Knowing that the two fragments of the helicopter werelocated at Points H1 (500 m, –100 m) and H2 (600 m,– 500 m), respectively, and assuming that all pieces hit theground at the same time, determine the coordinates of thePoint A where the wreckage of the airplane will be found.


In a game of pool, ball A is moving with a velocity v0 when it strikesballs B and C, which are at rest and aligned as shown. Knowing thatafter the collision the three balls move in the directions indicated, andthat 0 12 ft/sv and 6.29 ft/s,Cv determine the magnitude of thevelocity of (a) ball A, (b) ball B.


In Problem 14.3, determine the work done by thewoman and by the man as each dives from theboat, assuming that the woman dives first.


In a game of pool, ball A is moving with the velocity 0 0vv iwhen it strikes balls B and C, which are at rest side by side.Assuming frictionless surfaces and perfectly elastic impact (i.e.,conservation of energy), determine the final velocity of eachball, assuming that the path of A is (a) perfectly centered andthat A strikes B and C simultaneously, (b) not perfectly centeredand that A strikes B slightly before it strikes C.


Two small spheres A and B, weighing 5 lb and 2 lb, respectively, areconnected by a rigid rod of negligible weight. The two spheres areresting on a horizontal, frictionless surface when A is suddenly giventhe velocity 0 (10.5 ft/s) .v i Determine (a) the linear momentum of thesystem and its angular momentum about its mass center G, (b) thevelocities of A and B after the rod AB has rotated through 180 .


Two small disks A and B, of mass 2 kg and 1 kg,respectively, may slide on a horizontal and frictionlesssurface. They are connected by a cord of negligible massand spin about their mass center G. At 0,t G is movingwith the velocity 0v and its coordinates are 0x 0,

0 1.89 m.y Shortly thereafter, the cord breaks and disk Ais observed to move with a velocity (5 m/s)A v j in astraight line and at a distance 2.56 ma from the y axis,while B moves with a velocity (7.2 m/s) (4.6 m/s)B v i jalong a path intersecting the x axis at a distance 7.48 mb from the origin O. Determine (a) the initial velocity 0v ofthe mass center G of the two disks, (b) the length of thecord initially connecting the two disks, (c) the rate in rad/sat which the disks were spinning about G.




There are no horizontal external forces acting during the impacts. Thebaggage carrier is to coast between impacts.

(a) 15-kg suitcase tossed on carrier first:

Let 1v be the common velocity of suitcase A and the carrier afterthe first impact and 2v be the common velocity of the twosuitcases and the carrier of the second impact.

Conservation of momentum:

1(15)(3) (40)v 1 1.125 m/sv

20-kg suitcase tossed next:

2(20)(2) (40)(1.125) 60v 2 1.417 m/sv

(b) 20-kg suitcase tossed on carrier first:

Let 1v be the common velocity of suitcase B and the carrier afterthe first impact and 2v be the common velocity of all after thesecond impact.


1(20)(2) (45)v 1 0.8889 m/s v

15-kg suitcase tossed next:

2(15)(3) (45)(0.8889) 60v

2 1.417 m/s v


(a) Woman dives first.


1 1120 300 180

(16 ) 0v vg g

1(120)(16)

3.20 ft/s600

v

Man dives next. Conservation of momentum:

1 2 2300 180 300 180

(16 )v v vg g g

12

480 (180)(16)9.20 ft/s

480

vv

2 9.20 ft/sv

(b) Man dives first.


1 1180 300 120

(16 ) 0v vg g

1(180)(16)

4.80 ft/s600

v

Woman dives next. Conservation of momentum:

1 2 2300 120 300 120

(16 )v v vg g g

12

420 (120)(16)9.37 ft/s

420

vv

2 9.37 ft/s v


( ) ( ) ( )O i i i i i i

i x i y i z

m m z y z

v v v

i j k

H r v

5 4 30 5 4 4 4 3 8 6 0

2 3 2 2 3 2 1x z

g g gv v

i j k i j k i j k

1[5( 10 12) 4(4 6) 3(6 0)]zv

g i

1[5(8 0) 4(3 4 ) 3(0 8)]x zv v

g j

1[5(0 10) 4(8 4 ) 3( 16 18)]xv

g k

1[(16 116) (12 16 16) ( 16 12) ]O z z z xv v v v

g H i j k (1)

(a) For to be parallel to theO zH axis, we must have 0:z yH H

0: 16 116 0z zH v 7.25 ft/szv

0: 12 16(7.25) 16 0y xH v 8.33 ft/sxv

(b) Substituting into Eq. (1):

1[ 16(8.33) 12]

32.2O H k (4.51 ft lb s)O H k


Motion of mass center G.

At collision:(8400 m)

(35.00 m/s)4(60 s)H v i i

(16,000 m) (12,000 m)

4(60 s)

(66.67 m/s) (50 m/s)

A

i jv

i jVelocity of mass center:

( )H A H H A Am m m m v v v

4500 3000(35.00 ) 1500(66.67 50 ) v i i j

(45.556 m/s) (16.667 m/s) v i j

Vertical motion of G:

22

1 2 2(1200 m)15.641s

2 9.81 m/s

hh gt t

g

Position of G at time of ground impact:(45.556 16.667 )(15.641)t r v i j

(712.55 m) (260.69 m) r i j (1)

From Eq. (14.12):

1 1 2 2( )H A H H H H A Am m m m m r r r r (2)

4500(712.55 260.69 ) 1000(500 100 ) 2000(600 500 ) 1500 A i j i j i j r

1.5 (4.5 712.55 500 2 600) ( 4.5 260.69 100 2 500)A r i j

(1004 m) (48.7 m)A r i j


Conservation of linear momentum. In x direction:

(12 ft/s)cos 45 sin 4.3 sin 37.4

(6.29)cos 30

0.07498 0.60738 3.0380

A B

A B

m mv mv

m

v v

(1)

In y direction:

(12 ft/s)sin 45 cos 4.3 cos 37.4

(6.29)sin 30

0.99719 0.79441 5.3403

A B

A B

m mv mv

m

v v

(2)

(a) Multiply (1) by 0.79441, (2) by 0.60738, and add:

0.66524 5.6570Av 8.50 ft/sAv

(b) Multiply (1) by 0.99719, (2) by –0.07498, and add:

0.66524 2.6290Bv 3.95 ft/sBv


Woman dives first.


1 1

1

120 300 180(16 ) 0

(120)(16)3.20 ft/s

600

v vg g

v

116 12.80 ft/sv

Kinetic energy before dive: 0 0T

Kinetic energy after dive: 2 21

1 300 180 1 120(3.20) (12.80)

2 32.2 2 32.2381.61 ft lb

T

Work of woman: 1 0 381.61 ft lbT T 1 0 382 ft lbT T

Man dives next. Conservation of momentum:

1 2 2

12

300 180 300 180(16 )

480 (180)(16)9.20 ft/s

480

v v vg g g

vv

16 9.20 6.80 ft/s

Kinetic energy before dive: 21

1 300 180(3.20)

2 32.276.323 ft lb

T

Kinetic energy after dive: 2 22

1 300 1 180(9.20) (6.80)

2 32.2 2 32.2523.53 ft lb

T

Work of man: 2 1 447.2 ft lbT T 2 1 447 ft lbT T


(a) A strikes B and C simultaneously.

During the impact, the contact impulses make 30 angles with the velocity v0.

Thus, (cos30 sin 30 )

(cos30 sin 30 )B B

C C

v

v

v i j

v i j

By symmetry, A Avv i

Conservation of momentum: 0 A B Cm m m m v v v v

y component: 0 0 sin 30 sin 30B C C Bmv mv v v

x component: 0 cos30 cos30A B Cmv mv mv mv

00

0

2( )

cos30 3

3

AB C A

AB C

v vv v v v

v vv v

Conservation of energy: 2 2 2 20

1 1 1 1

2 2 2 2A B Cmv mv mv mv

2 2 20 0

2 2 20 0 0 0

2( )

32

( )( ) ( )3

A A

A A A A

v v v v

v v v v v v v v

0 0 0 0

0 0

2 1 5 1( )

3 3 3 5

6 2 3

55 3

A A A A

B C

v v v v v v v v

v v v v

00.200A vv

00.693B vv 30

00.693C vv 30

(b) A strikes B before it strikes C.

First impact: A strikes B.

During the impact, the contact impulse makes a 30 angle with the velocity 0.v


Thus, (cos30 sin 30 )B Bv v i j

Conservation of momentum: 0 A Bm m m v v v

y component: 0 ( ) sin 30 ( ) sin 30A y B y BAm v mv v v

x component: 0 0( ) cos30 ( ) cos30A x B A x Bv m v mv v v v

Conservation of energy:

2 2 2 20

2 2 20

2 2 2 2 2 20 0

1 1 1 1( ) ( )

2 2 2 21 1 1

( cos30 ) ( sin 30 )2 2 21

2 cos 30 sin 302

A x A y B

B B B

B B B B

mv m v m v mv

m v v v v

m v v v v v v

20 0 0 0

0 0

3 1cos30 , ( ) sin 30 ,

2 4

3( ) cos30 sin 30

4

B A x

A y

v v v v v v

v v v

Second impact: A strikes C.During the impact, the contact impulse makes a 30 angle with the velocity 0.v

Thus, (cos30 sin 30 )C Cv v i j

Conservation of momentum: A A Cm m m v v v

x component:

0

( ) ( ) cos30 ,

1( ) ( ) cos30 cos30

4

A x A x C

A x A x C C

m v m v mv

v v v v v

y component:

0

( ) ( ) sin 30

3( ) ( ) sin 30 sin 30

4

A y A y C

A y A y C C

m v m v mv

v v v v v



2 2 2 2 2

222 2 20 0 0 0

2 2 20 0

2 2 20 0

1 1 1 1 1( ) ( ) ( ) ( )

2 2 2 2 2

1 1 3 1 1 3cos30 sin 30

2 16 16 2 4 4

1 1 1cos30 cos 30

2 16 2

3 3sin 30 sin 30

16 2

A x A y A x A y C

C C C

C C

C C

m v m v m v m v mv

m v v m v v v v v

m v v v v

v v v v v

2C

20

0 0

0 0 0

1 30 cos30 sin 30 2

2 2

1 3 3cos30 sin 30

4 4 4

1 3 1( ) cos30

4 4 8

C C

C

A x

v v v

v v v

v v v v

0 0 03 3 3

( ) sin 304 4 8A yv v v v 00.250A vv 60°

00.866B vv 30°

00.433C vv 30°


Position of mass center.

7 in.AG BG

(5 lb) (2 lb) 2.5 3.5 7 2 in.AG BG BG AG AG AG

(a) Linear and angular momentum

0 2

5 lb(10.5 ft/s) (1.6304 lb s)

32.2 ft/sAL m v i i

(1.630 lb s) L i

0 (2 in.) × (1.6304 lb s)

= (3.266 in lb s)

(0.27174 ft lb s)

G AGA m

H v j i

k

k

(0.272 ft lb s)G H k

(b) Velocities of A and B after 180° rotation

Conservation of linear momentum:

0

5 5 2(10.5)

A A A B B

A B

m v m v m v

v vg g g

5 2 52.5A Bv v (1)

Conservation of angular momentum about :G

0A A A A A B B Br m v r m v r m v

5 5 2(2 in.) (10.5) (2 in.) (5 in.)

g A Bv vg g

Multiplying by g and dividing by 2:

5 5 52.5A Bv v (2)

Add Eqs. (1) and (2): 7 105 15.00 ft/sB Bv v

From Eq. (1): 5 2(15) 52.5 4.50 ft/sA Av v

(4.50 ft/s) ; (15.00 ft/s)A B v i v i


Initial conditions.

Location of G:

1

32

3

B A B A

B

A

AG BG AG GB l

m m m m m

mAG l l

mm

BG l lm

Linear momentum:

0 0 03L m v v

Angular momentum about G:

0

2 2

( )

1 1 2 2(2 kg) (1 kg)

3 3 3 3

2

3

G A A B BH GA m GB m

l lw l lw

l w

v v

k k

k

Kinetic energy: Using Eq. (14.29),

2 2 2 2 20 0 0

2 22

0

2 2 20 0

1 1 1 1 1

2 2 2 2 2

1 1 1 1 2(3) (2) (1)

2 2 3 2 3

3 1

2 3

i i A A B Bi

T mv m v mv m v m v

v lw lw

T v l w

Conservation of linear momentum:

0

03 (2)(5 ) (1)(7.2 4.6 ) 7.2 5.4A A B Bm m m

v v v

v j i j i j


(a) 0 (2.4 m/s) (1.8 m/s) v i j

Conservation of angular momentum about O:

0 0(1.89 ) ( ) (2.56 ) (7.48 )G A A B Am m m j v H i v i v

Substituting for 0 ,v 0( ) , ,G A BH v v and masses:

2

2

2(1.89 ) 3(2.4 1.8 ) (2.56 ) 2(5 ) (7.48 ) (7.2 4.6 )

32

13.608 25.6 34.4083

l w

l w

j i j k i j i i j

k k k k

2 224.80 7.20

3l w l w (1)


2 2 2 2 20 0

3 1 1 1:

2 3 2 2A A B BT T v l w m v m v

2 2 2 2 2 2 23 1 1 1[(2.4) (1.8) ] (2)(5) (1)[(7.2) (4.6) ]

2 3 2 2l w

2 2 2 2113.5 25 36.5 144.0

3l w l w (2)

Dividing Eq. (2) by Eq. (1), member by member:

144.020.0 rad/s

7.20w

(c) Original rate of spin = 20.0 rad/s

Substituting for w into Eq. (1):

2 2(20.0) 7.20 0.360 0.600 ml l l

(b) Length of cord = 600 mm




The motion of a cam is defined by the relation 3 29 15 ,t t t where is expressed in radians and t inseconds. Determine the angular coordinate, the angular velocity, and the angular acceleration of the cam when(a) 0,t (b) 3 s.t


The assembly shown consists of the straight rod ABC, whichpasses through and is welded to the rectangular plate DEFH.The assembly rotates about the axis AC with a constantangular velocity of 9 rad/s. Knowing that the motion whenviewed from C is counterclockwise, determine the velocityand acceleration of corner F.


The circular plate shown is initially at rest. Knowingthat 200 mmr and that the plate has a constant angularacceleration of 20.3 rad/s , determine the magnitudeof the total acceleration of Point B when (a) 0,t (b) 2 s,t (c) 4 s.t


The system shown is held at rest by the brake-and-drum system.After the brake is partially released at t 0, it is observed that thecylinder moves 16 ft in 5 s. Assuming uniformly acceleratedmotion, determine (a) the angular acceleration of the drum, (b) theangular velocity of the drum at t 4 s.


The motion of rod AB is guided by pins attached at A and B, whichslide in the slots shown. At the instant shown, 40 and the pinat B moves upward to the left with a constant velocity of 6 in./s.Determine (a) the angular velocity of the rod, (b) the velocity ofthe pin at end A.


Arm AB rotates with an angular velocity of 42 rpm clockwise.Determine the required angular velocity of gear A for which(a) the angular velocity of gear B is 20 rpm counterclockwise,(b) the motion of gear B is a curvilinear translation.


Knowing that crank AB has a constant angular velocity of160 rpm counterclockwise, determine the angular velocityof rod BD and the velocity of collar D when (a) 0, (b) o90 .


In the position shown, bar AB has an angular velocityof 4 rad/s clockwise. Determine the angular velocityof bars BD and DE.




3 2

2

2

9 15

3 18 15

3( 6 5)

3( 5)( 1)

6 18

t t t

d

dt

t t

t t

t t

d

dtt

(a) 0:t

0

15.00 rad/s

218.00 rad/s

(b) 3 s:t

3 23 9(3) 15(3) 9.00 rad

23(3) 18(3) 15 12.00 rad/s

6(3) 18 0


(350 mm) (200 mm) (200 mm) 450 mm

350 200 200 1(7 4 4 )

450 91

(9 rad/s) (7 4 4 )9

(7 rad/s) (4 rad/s) (4 rad/s) 0

AC

AC

AC AC

AC

AC

i j k

i j k i j k

i j k

i j k

Corner F: /

/

( 175 mm) (100 mm)

(0.175 m) (0.100 m)

7 4 4

0.175 0 0.100

0.4 ( 0.7 0.7) 0.7

F B

F F B

r i ki + k

v ri j k

i j k

(0.4 m/s) (1.4 m/s) (0.7 m/s)F v i j k

/ /

0

( )

0

7 4 4

0.4 1.4 0.7

(2.8 5.6) ( 1.6 4.9) ( 9.8 1.6)

F F B F B

F

F F

=a r r

va v

i j k

i j k

2 2 2(8.4 m/s ) (3.3 m/s ) (11.4 m/s )F a i j k


Uniformly accelerated motion.

0

2 2 2

2 2 2 2 2 2 4 4 2 2 2 4

2 4 1/2

2

2 4 1/2

4 1/2

0

(1 )

(1 )

0.2 m, 0.3 rad/s

(0.2)(0.3)(1 (0.3) )

0.06(1 0.09 )

t n

B t n

B

B

t t t

a r a r r t

a a a r r t r t

a r t

r

a t

t

(a) 0:t 0.06(1 0)Ba 20.0600 m/sBa

(b) 2 s:t 4 1/20.06(1 0.09 2 )Ba 20.0937 m/sBa

(c) 4 s:t 4 1/20.06(1 0.09 4 )Ba 20.294 m/sBa


Block A: 20

2

1

21

16 ft 0 (5 s)2

s t at

a

21.28 ft/sa 21.28 ft/sa

Drum:

(a) a r

2

2

(1.28 ft/s ) (0.75 ft)

1.707 rad/s

21.707 rad/s

(b) Uniformly accelerated motion. 0 0 when 0t

0 t

When 4 s:t 20 (1.707 rad/s )(4 s)

6.83 rad/s 6.83 rad/s


/A B A B v v v

[ ] [6 in./sAv /15 ] [ A Bv 40 ]

Law of sines.

/ 6 in./s

sin 55 sin 75 sin 50A BA vv

(b) 6.42 in./sA v

/ 7.566 in./sA B v 40

/ ( ) 20 in.

7.566 in./s (20 in.)A B AB

AB

v AB AB

(a) 0.3783 rad/sAB 0.378 rad/sAB


Arm AB: 42 rpm 1.4 rad/sAB

(15)(1.4 ) 21 in./sB AB ABr v

(a) Gear B:2

20 rpm in./s3B

/2

(9) 6 in./s3D B B Br

v

/D B D B v v v

21 6 27 in./s

Gear A: DA

A

v

r

274.5 rad/s

6A

135.0 rpmA

(b) Gear B: /0, 0B D B v

21 in./sD B v v

Gear A:21

3.5 rad/s6A

105.0 rpmA


16160 rpm rad/s

3AB

16( ) (3) 50.265 in./s

3B ABv AB

(a) 0.

B Bvv D Dvv

6sin , 36.87

10

/ ( )D B BDBD v 10 BD

/ Draw vector diagram.D B D B v v v

/50.265

62.831 in./scos cos

BD B

vv

/ 62.8316.2831 rad/s

10D B

BD

v

BD

60.0 rpmBD

tan 50.265 tan 37.7 in./sD Bv v

37.7 in./sD v

(b) 90 .

B Bvv D Dvv

Bar is in translation.BD 0BD

D Bv v 50.3 in./sD v


Bar AB: 1 0.4tan 26.56

0.80.8

0.8944 mcos

( ) (0.8944 m)(4 m/s)B AB

AB

AB

v

3.578 m/sB v 26.56

Bar DE: 1 0.4tan 38.66

0.50.5

0.6403 mcos

( )D DE

DE

v DE

(0.6403 m)D DEv 38.66

Bar BD:

/D B D B v v v

[ Dv ] [ B v /] [ D B v ]

Law of sines.

/ 3.578 m/s

sin 63.44 sin 65.22 sin 51.34D BD vv

4.099 m/s

(0.6403 m) 4.099 m/sD

DE

v

6.4 rad/sDE

/ 4.160 m/s

(0.8 m) 4.16 m/sD B

BD

v

v

5.2 rad/sBD




The spool of tape shown and its frame assembly are pulled upward ata speed 750Av mm/s. Knowing that the 80-mm-radius spool has anangular velocity of 15 rad/s clockwise and that at the instant shown thetotal thickness of the tape on the spool is 20 mm, determine (a) theinstantaneous center of rotation of the spool, (b) the velocities ofPoints B and D.


An overhead door is guided by wheels at A and B that roll inhorizontal and vertical tracks. Knowing that when 40 thevelocity of wheel B is 1.5 ft/s upward, determine (a) the angularvelocity of the door, (b) the velocity of end D of the door.


Rod AB can slide freely along the floor and the inclined plane. Knowing that 20 , 50 , l 0.6 m, and3 m/s,Av determine (a) the angular velocity of the rod, (b) the velocity of end B.


A 3-in.-radius drum is rigidly attached to a 5-in.-radius drum asshown. One of the drums rolls without sliding on the surface shown,and a cord is wound around the other drum. Knowing that at theinstant shown end D of the cord has a velocity of 8 in./s and anacceleration of 230 in./s , both directed to the left, determine theaccelerations of Points A, B, and C of the drums.


In the two-cylinder air compressor shown, the connecting rodsBD and BE are each 190 mm long and crank AB rotates aboutthe fixed Point A with a constant angular velocity of 1500 rpmclockwise. Determine the acceleration of each piston when 0.


Arm AB has a constant angular velocity of 16 rad/s counter-clockwise. At the instant when 0, determine theacceleration (a) of collar D, (b) of the midpoint G of bar BD.


Knowing that at the instant shown rod AB has a constantangular velocity of 6 rad/s clockwise, determine theacceleration of Point D.


Two rotating rods are connected by slider block P. The rodattached at A rotates with a constant angular velocity .A Forthe given data, determine for the position shown (a) theangular velocity of the rod attached at B, (b) the relativevelocity of slider block P with respect to the rod on which itslides.

b 300 mm, 10 rad/s.A




750 mm/sA v

15 rad/s

75050 mm

15Av

x

(a) The instantaneous center lies 50 mm to the right of the axle.

80 20 50 50 mmCB

(b) ( ) (50)(15) 750 mm/sBv CB

750 mm/sB v

80 50 130 mmCD

( ) (130)(15) 1950 mm/sDv CD

1.950 m/sD v


Locate instantaneous center at intersection of lines drawn perpendicularto Av and .Bv

(a) Angular velocity.

( )

1.5 ft/s (3.214 ft)

0.4667 rad/s

Bv BC

0.467 rad/s

(b) Velocity of D:

In CDE: 1 6.427tan 59.2

3.836.427

7.482 ftsin

( )

(7.482 ft)(0.4667 rad/s)

3.49 ft/s

D

CD

v CD

3.49 ft/sD v 59.2


Locate the instantaneous center at intersection of lines draw perpendicular to Av and .Bv

Law of sines.

sin[90 ( )] sin(90 )

sin

cos( ) cos

sincos( )

sincos

sin

AC BC

l

AC BC

l

AC l

BC l

Angular velocity:cos( )

( )sin

sin

cos( )

A

A

v AC l

v

l

Velocity of B:cos sin

( )sin cos( )

cos

cos( )

AB

B A

vv BC l

l

v v

Data: 20 , 50 , 0.6 m, 3 m/sAl v

(a)sin 3 m/s sin 50

cos( ) 0.6 m cos(50 20 )Av

l

4.423 rad/s 4.42 rad/s

(b)cos

cos( )

cos 20(3 m/s)

cos(50 20 )

B Av v

3.2552 m/sBv 3.26 m/sB v 50°


Velocity analysis. v v 8 in./sD A

Instantaneous center is at Point B. ( ) , 8 (5 3)Av AB

4 rad/s

Acceleration analysis. [B Baa ] for no slipping.

2[30 in./sA a ] [( )A na ]

[G Gaa ]

/ /( ) ( )B A B A t B A n a a a a

[ Ba ] [30 ] [( )A na ] [(5 3) 2] [5 3) ]

Components : 20 30 2 15 rad/s

/( ) ( )/B G B G t nB G a a a a

[ Ba ] [ Ga ] [5 2] [5 ]

Components : 20 5 5 75 in./sG Ga a

: 2 2(5)(4) 80 in./sBa 280.0 in./sB a

/ /( ) ( )A G A G t A G n a a a a

[75 ] [3 2] [3 ]

[75 ] [45 ] [48 ]

2[30 in./s 2] [48 in./s ]

256.6 in./sA a 58.0

/ /( ) ( )C G C G t C G n a a a a

[75 ] [5 2] [5 ]

[75 ] [75 ] [80 ]

2[155 in./s 2] [75 in./s ]

2172.2 in./sC a 25.8


Crank AB.

0 0 1500 rpm 157.08 rad/sA A AB v a , 0AB

/ 0 [0.05B A B A AB v v v 45°] 7.854 m/s 45°

/ /( ) ( )B A B A t B A n a a a a

0 [0.05 AB 45°] 2[0.05 AB 45°]

2[(0.05)(157.08) 45°] 21233.7 m/s 45°

Rod BD. D Dvv 45° BD BD

/D B B D v v v

Dv 45 [7.854 45 ] [0.19 BD 45 ]

Components 45 : 0 7.854 0.19 41.337 rad/sBD BD

D Daa 45

/ /( ) ( )D B D B t D B n a a a a

[ Da 45 ] [1233.7 45 ] [0.19 BD 245 ] [0.19 BD 45 ]

Components 45 : 2 21233.7 (0.19)(41.337) 1558.4 m/sDa

21558 m/sD a 45°

Rod BE.0.05

sin , 15.258 , 45 29.7420.19

E Evv 45°


Since Ev is parallel to ,Bv 0.BE

E Eaa 45° 2/( ) 0.19 0B E n BEa

/( ) ( )E/B t E B taa /( )E B B E t a a a

Draw vector addition diagram.

2

45

15.258

tan

1233.7 tan

336.52 m/s

E Ba a

2337 m/sE a 45°


Velocity. ( )

(3 in.)(16 rad/s)B ABv AB

48 in./sB v

( )

48 in./s (8 in.)B BD

BD

v BC

6 rad/sBD

Acceleration.

Rod AB: 2 2 2( ) (3 in.)(16 rad/s) 768 in./sB ABAB a

Plane motion Trans. with B Rotation about B

(a) / / /( ) ( )D B D B B D B t D B na a a a a a

D B a a ( )BD 2( )BD

2768 in./sD a (10 in.) 2(10 in.)(6 rad/s)

2768 in./sD a (10 in.) 2360 in./s

Vector diagram.

2 24(10 in.) 216 in./s ; 27 rad/s

5 α

2

3768 288 (10 in.)

53

768 288 (10)(27) 768 288 162 1218 in./s5

Da

21218 in./sD a


(b) / / /( ) ( )G B G B B G B t G B n a a a a a a

( )G Ba BG a 2( )BG

2768 in./s (5 in.)(27 rad/s)G a 2(5 in.)(6 rad/s )

2 2768 in./s 135 in/sG a 2180 in./s

components:

2 2

4 3( ) 135 180

5 5

108 in./s 108 in./s

( ) 0

G y

G y

a

a

components:3 4

( ) 768 135 1805 5

768 81 144

G xa

2993 in./s 2993 in./sG a


Velocity analysis. 6 rad/sAB

( )

(90)(6)

540 mm/s

B ABAB

v

B Bvv , D Dvv

The instantaneous center of bar BDE lies at .

Then 0 and 540 mm/sBD D Bv v

5403 rad/s

180D

CDv

CD

.Acceleration analysis 0AB

2 2( ) [(90)(6)B ABAB a 2] 3240 mm/s

[( )D CDCD a 2] [( ) CDCD ] [180 CD 2] [(180)(3) ]

[180 CD 2] [1620 mm/s ]

/ [90D B BDa ] [225 BD ] 2 [225 BD 2] [90 BD ]

[90 BD ] [225 BD

/ Resolve into components.D B D B a a a

: 1620 3240 225 ,BD 27.2 rad/sBD

: 2180 0 (90)(7.2), 3.6 rad/sCD

[3240D a ] [(90)(7.2) ] [(225)(7.2) ]

[648 2] [1620 mm/s ]

21745 mm/s 68.2° 21.745 m/sD a 68.2°


Dimensions:

Law of sines.

300 mm

sin 20 sin120 sin 40

159.63 mm

404.19 mm

AP BP

AP

BP

10 rad/sAD

Velocities.

Note: P Point of AD coinciding with P.

( )

(159.63 mm)(10 rad/s)P ADAP

v

1596.3 mm/s 30°

/P P P AD v v v

[ Pv 70 ] [1596.3 /30 ] [ P ADv 60 ]

(a) (1596.3)/cos 40°

2083.8 mm/s

pv

2083.8 mm/s

404.19 mm5.155 rad/s

pBP

v

BP

5.16 rad/sBD

(b) / (1596.3) tan 40 1339.5 mm/sP ADv / 1.339 m/sP AD v 60°




The sleeve BC is welded to an arm that rotates about A with aconstant angular velocity In the position shown rod DF isbeing moved to the left at a constant speed 16 in./su relativeto the sleeve. For the given angular velocity , determine theacceleration (a) of Point D, (b) of the point of rod DF thatcoincides with Point E.

(3 rad/s) . j


A chain is looped around two gears of radius 40 mm that canrotate freely with respect to the 320-mm arm AB. The chainmoves about arm AB in a clockwise direction at the constantrate of 80 mm/s relative to the arm. Knowing that in theposition shown arm AB rotates clockwise about A at theconstant rate 0.75 rad/s, determine the acceleration ofeach of the chain links indicated.

Links 1 and 2.


A conveyor system is fitted with vertical panels, and a 300-mm rod AB ofmass 2.5 kg is lodged between two panels as shown. Knowing that theacceleration of the system is 21.5 m/s to the left, determine (a) the forceexerted on the rod at C, (b) the reaction at B.


A 20-kg cabinet is mounted on casters that allow it to movefreely ( 0) on the floor. If a 100-N force is applied as shown,determine (a) the acceleration of the cabinet, (b) the range ofvalues of h for which the cabinet will not tip.


A completely filled barrel and its contents have acombined weight of 200 lb. A cylinder C is connectedto the barrel at a height h 22 in. as shown. Knowing

0.40s and 0.35,k determine the maximumweight of C so the barrel will not tip.


The 180-mm disk is at rest when it is placed in contact with a beltmoving at a constant speed. Neglecting the weight of the link AB andknowing that the coefficient of kinetic friction between the disk and thebelt is 0.40, determine the angular acceleration of the disk whileslipping occurs.


The 6-lb disk A has a radius 3Ar in. and an initial angular velocity

0 375 rpm clockwise. The 15-lb disk B has a radius 5Br in. and isat rest. A force P of magnitude 2.5 lb is then applied to bring the disksinto contact. Knowing that 0.25k between the disks and neglectingbearing friction, determine (a) the angular acceleration of each disk,(b) the final angular velocity of each disk.


A 3-kg sprocket wheel has a centroidal radius of gyration of 70 mm andis suspended from a chain as shown. Determine the acceleration ofPoints A and B of the chain, knowing that 14 NAT and 18 N.BT




(a) Point D. / / /

2

(16 in./s) ; 0

(5 in.) (12 in.)

(3 rad/s) [ (5 in.) (12 in.) ]

(36 in./s)

(3 rad/s) (36 in./s)

(108 in./s )

D F D BC F BC

D

D D

AD

AD

v v k a

j k

vj j ki

a vj i

k

/

2

/2 2

2

(3 rad/s) (16 in./s)

(96 in./s )

(108 in./s ) 0 (96 in./s )

C D F

D D D F C

a vj k

ia a a a

k i

2 2(96 in./s ) (108 in./s )D a i k

(b) Point P of DF that coincides with E.

/ / /

/

2

/

(16 in./s)k; 0

(5 in.) ; (3 rad/s) ( 5 in./s) 0

0

2

2(3 rad/s) (16 in./s)

(96 in./s )

P F D BC P F

P

D E

C E F

P P P F C

AE AE

v v aj v j j

a va v

j ki

a a a a

20 0 (96 in./s )P a i 2(96 in./s )P a i


Let the arm AB be a rotating frame of reference. 0.75 rad/s (0.75 rad/s) : k

Link 1: 1 1/

2 21 1

2 22

1/

/

(40 mm) , (80 mm/s)

(0.75) ( 40) (22.5 mm/s)

80160 mm/s (160 mm/s )

40

2 (2)( 0.75 ) (80 )

(120 mm/s)

AB

AB

P AB

u

u

r i v j

a r i

a i

v k ji

1 1 1/ 1/

2

2

(302.5 mm/s )

AB AB

a a a v

i 21 303 mm/sa

Link 2: 2

2/2

2 22

2 2

2/

(160 mm) (40 mm)

(4 in./s)

(0.75) (160 40 )

(90 mm/s ) (22.5 mm/s )

0

AB

AB

u

r i jv i

a ri ji j

a

2/

2

22 2/ 2/

2 2

2 22

2

2 (2)( 0.75 ) (80 )

(120 mm/s )

2

90 22.5 120

(90 mm/s ) (142.5 mm/s )

(90) (142.5)

168.5 mm/s

AB

AB AB

v k i

j

a a a vi j j

i j

a

142.5tan , 57.7

90 2

2 168.5 mm/sa 57.7


Geometry200 mm

212.84 mmcos 20

(0.15 mm)sin 20° 51.3 mm

d

b

Weight 2(2.5 kg)9.81 m/s 24.525 NW mg

eff( ) : (212.84 mm) (51.3 mm) (140.95 mm)B BM M C W ma

0.241 0.6622

0.241(24.525 N) 0.6622(2.5 kg)( )

C W ma

C a

5.911 N 1.656C a (1)

The given acceleration is 21.5 m/sa

(a) Force at C.

From Eq. (1): 5.911 1.656(1.5);C 3.43 NC 20°

(b) Reaction at B.

eff( ) : sin 20 0y y yF F B W C

24.525 N (3.43 N)sin 20 23.35 NyB

eff( ) :x xF F cos 20xB C ma

2(3.43 N)cos 20 (2.5 kg)(1.5 m/s)xB

3.22 3.75 6.97 NxB 24.4 NB 73.4°


(a) Acceleration

eff( ) : 100 Nx xF F ma

100 N (20 kg)a

25.00 m/sa

(b) For tipping to impend : 0A

eff( ) : (100 N) (0.3 m) (0.9 m)B BM M h mg ma 2(100 N) (20 kg)(9.81 m/s )(0.3 m) (100 N)(0.9 m)

1.489 m

h

h

For tipping to impend : 0B

eff( ) : (100 N) (0.3 m) (0.9 m)A AM M h mg ma 2(100 N) (20 kg)(9.81 m/s )(0.3 m) (100 N)(0.9 m)

0.311 m

h

h

Cabinet will not tip: 0.311 m 1.489 mh# #


Kinematics: Assume that the barrel is sliding, but not tipping.

0 G a a a

Since the cord is inextensible, C aa

Kinetics: Draw the free body diagrams of the barrel and the cylinder.

The barrel is sliding. 0.35F kF N N

Assume that tipping is impending, so that the line of action of the reaction on the bottom of the barrel passesthrough Point B.

10 in.e

For the barrel. 0: 0y B BF N W N W

0: 4 (18)(0.35 ) 0GM Ne T N

(18)(0.35) 10 6.30.925

4 4 B Be

T N W W

: 0.35B Bx

W WF a T N a

g g

0.35 0.925 0.35 0.575B B

a T N

g W W

For the cylinder: :CC C

W aF a W T W

g g

0.925(2.1765)(200 lb)

1 0.5751

BC

WTW

a

g

435 lbCW


Belt: kF N

Disk:

eff( ) :x xF F cos 0ABN F

cosABF N (1)

eff( ) : sin 0y y k ABF F N F mg

sinAB kF mg N (2)

Eq. (2): tan

Eq. (1)kmg N

N

tan

tan

tan

k

kk

N mg N

mgN

kmg

F Nk

eff( ) :A AM M Fr I

21 tan22

tan

k

k

k

k

rF

Imgr

mr

g

r

Data: 0.18 m

60

0.40k

r

22(9.81 m/s ) 0.40

0.18 m tan 60 0.40

220.4 rad/s


While slipping occurs:

(a) Angular accelerations.

0.25(2.5 lb)

0.625 lb

k

k

F N

P

Disk A:

2eff

1( ) :

2A A A A A A AM M Fr I m r

2

1

21 6 lb 3

0.625 lb ft2 1232.2 rad/s

A A

A

F m r

226.833 rad/sA 226.8 rad/sA α

Disk B:

2eff

1( ) :

2B B B B B B B BM M Fr I m r

2

1

21 15 lb 5

0.625 lb ft2 1232.2 rad/s

B B B

B

F m r

26.44 rad/sB 26.44 rad/sB

(b) Final angular velocities. 02

( ) 375 rpm 39.27 rad/s60A

0( ) 0B


When disks stop sliding

:P P A A B Bv v (1)

0

0

[( ) ] ( )

(39.27 26.833 )(3 in.) (6.44 )(5 in.)

1.0454 s

( )

39.27 (26.833)(1.0454)

A B A B B

A A A

t t

t t

t

t

6011.22 rad/s

2A

107.1 rpmA ω

Eq. (1) (107.2 rpm)(3 in.) (5 in.)B 64.3 rpmB ω


2

2

3 2

2

(3 kg)(0.07 m)

14.7 10 kg m

0.08 m

(3 kg)(9.81 m/s )

29.43 N

I mk

r

W mg

W

eff( ) :y y A BF F T T W ma

29.43 N (3 kg)A BT T a

1( 29.43)

3 A Ba T T (1)

eff( ) : (0.08 m) (0.08 m)G G B AM M T T I

3 2( )(0.08 m) (14.7 10 kg m )B AT T

5.442 ( )B AT T (2)

14 N, 18 NA BT T

Eq. (1): 21(14 18 29.43) 0.8567 m/s

3a

Eq. (2): 25.442(18 14) 21.769 rad/s

( )

0.8567 (0.08)(21.769)

A A ta

a r

a

20.885 m/s 20.885 m/sA a

( )

0.8567 (0.08)(21.769)

B B ta

a r

a

22.60 m/s 22.60 m/sB a

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