Binder 1

Chapter 4. Translational Equilibrium and Friction Physics, 6th Ed.

21

B A

W

B

400

Bx

By

W

B A

300 600

W

B A

600

W B

A

300 600

Chapter 4. Translational Equilibrium and Friction.

Note: For all of the problems at the end of this chapter, the rigid booms or struts are considered to be of negligible weight. All forces are considered to be concurrent forces. Free-body Diagrams 4-1. Draw a free-body diagram for the arrangements shown in Fig. 3-18. Isolate a point where

the important forces are acting, and represent each force as a vector. Determine the

reference angle and label components.

(a) Free-body Diagram (b) Free-body with rotation of axes to simplify work.

4-2. Study each force acting at the end of the light strut in Fig. 3-19. Draw the appropriate free-

body diagram.

There is no particular advantage to rotating axes.

Components should also be labeled on diagram.

Solution of Equilibrium Problems: 4-3. Three identical bricks are strung together with cords and hung from a scale that reads a total

of 24 N. What is the tension in the cord that supports the lowest brick? What is the tension

in the cord between the middle brick and the top brick?

Each brick must weight 8 N. The lowest cord supports only one brick,

whereas the middle cord supports two bricks. Ans. 8 N, 16 N.


22

W

B A

600

4-4. A single chain supports a pulley whose weight is 40 N. Two identical 80-N weights are

then connected with a cord that passes over the pulley. What is the tension in the

supporting chain? What is the tension in each cord?

Each cord supports 80 N, but chain supports everything.

T = 2(80 N) + 40 N = 200 N. T = 200 N

*4-5. If the weight of the block in Fig. 4-18a is 80 N, what are the tensions in ropes A and B?

By - W = 0; B sin 400 – 80 N = 0; B = 124.4 N

Bx – A = 0; B cos 400 = A; A = (124.4 N) cos 400

A = 95.3 N; B = 124 N.

*4-6. If rope B in Fig. 4-18a will break for tensions greater than 200 lb, what is the maximum

weight W that can be supported?

ΣFy = 0; By – W = 0; W = B sin 400; B = 200 N

W = (200 N) sin 400; W = 129 lb

*4-7. If W = 600 N in Fig. 18b, what is the force exerted by the rope on the end of the boom A in

Fig. 18b? What is the tension in rope B?

ΣFx = 0; A – Wx = 0; A = Wx = W cos 600

A = (600 N) cos 600 = 300 N

ΣFy = 0; B – Wy = 0; B = Wy = W sin 600

B = (600 N) sin 600 = 520 N

A = 300 N; B = 520 N

Wy

Wx

80 N 80 N

40 N

By

Bx

B 400

A

W

Bx

B 400

A

W


23

W

B = 800 N A

600

300

W

F N

*4-8. If the rope B in Fig. 18a will break if its tension exceeds 400 N, what is the maximum

weight W? ΣFy = By - W = 0; By = W

B sin 400 = 400 N ; B = 622 N ΣFx = 0

Bx – A = 0; B cos 400 = A; A = (622 N) cos 400 A = 477 N.

*4-9. What is the maximum weight W for Fig. 18b if the rope can sustain a maximum tension of

only 800 N? (Set B = 800 N).

Draw diagram, then rotate x-y axes as shown to right.

ΣFy = 0; 800 N – W Sin 600 = 0; W = 924 N.

The compression in the boom is A = 924 Cos 600 A = 462 N.

*4-10. A 70-N block rests on a 300 inclined plane. Determine the normal force and find the friction

force that keeps the block from sliding. (Rotate axes as shown.)

ΣFx = N – Wx = 0; N = Wx = (70 N) cos 300; N = 60.6 N

ΣFx = F – Wy = 0; F = Wy = (70 N) sin 300; F = 35.0 N

*4-11. A wire is stretched between two poles 10 m apart. A sign is attached to the midpoint of the

line causing it to sag vertically a distance of 50 cm. If the tension in each line segment is

2000 N, what is the weight of the sign? (h = 0.50 m)

tan φ = (0.5/5) or φ = 5.710 ; 2(2000 N) sin φ = W

W = 4000 sin 5.71; W = 398 N.

*4-12. An 80-N traffic light is supported at the midpoint of a 30-m length of cable between to

poles. Find the tension in each cable segment if the cable sags a vertical distance of 1 m.

h = 1 m; Tan φ = (1/15); φ = 3.810

T sin φ + T sin φ = 80 N; 2T sin 3.810 = 80 N

15 m

5 m

W = ?

h φ φ 2000 N 2000 N

5 m

15 m

W = 80 N

h φ φ T T

Bx

B 400

A

W

By


24

Solution to 4-12 (Cont.): T = =80

3816010

N2

Nsin .

; T = 601 N

*4-13. The ends of three 8-ft studs are nailed together forming a tripod with an apex that is 6ft

above the ground. What is the compression in each of these studs if a 100-lb weight is hung

from the apex?

Three upward components Fy hold up the 100 lb weight:

3 Fy = 100 lb; Fy = 33.3 lb sin φ = (6/8); φ = 48.90

F sin 48.90 = 33.3 lb; F = =333 44 4. . lb

sin 48.9 lb0 F = 44.4 lb, compression

*4-14. A 20-N picture is hung from a nail as in Fig. 4-20, so that the supporting cords make an

angle of 600. What is the tension of each cord segment?

According to Newton’s third law, the force of frame on nail (20 N)

is the same as the force of the nail on the rope (20 N , up).

ΣFy = 0; 20 N = Ty + Ty; 2Ty = 20 N; Ty = 10 N

Ty = T sin 600; So T sin 600 = 10 N, and T = 11.5 N.

Friction 4-15. A horizontal force of 40 N will just start an empty 600-N sled moving across packed snow.

After motion is begun, only 10 N is needed to keep motion at constant speed. Find the

coefficients of static and kinetic friction.

µ µs k= = = =40 10 N600 N

0.0667 N600 N

0.0167 µs = 0.0667; µk = 0.016

4-16. Suppose 200-N of supplies are added the sled in Problem 4-13. What new force is needed

to drag the sled at constant speed?

N= 200 N + 600 N = 800 N; Fk = µkN = (0.0167)(800 N); Fk = 13.3 N

φ

F Fy

h

600 600

T T

20 N


25

4-17. Assume surfaces where µs = 0.7 and µk = 0.4. What horizontal force is needed to just start

a 50-N block moving along a wooden floor. What force will move it at constant speed?

Fs = µsN = (0.7)(50 N) = 35 N ; Fk = µsN = (0.4)(50 N) = 20 N

4-18. A dockworker finds that a horizontal force of 60 lb is needed to drag a 150-lb crate across

the deck at constant speed. What is the coefficient of kinetic friction?

µ k =FN

; µ k = =60 lb

150 lb 0.400 µk = 0.400

4-19. The dockworker in Problem 4-16 finds that a smaller crate of similar material can be

dragged at constant speed with a horizontal force of only 40 lb. What is the weight of this

crate?

Fk = µsN = (0.4)W = 40 lb; W = (40 lb/0.4) = 100 lb; W = 100 lb.

4-20. A steel block weighing 240 N rests on level steel beam. What horizontal force will move

the block at constant speed if the coefficient of kinetic friction is 0.12?

Fk = µsN = (0.12)(240 N) ; Fk = 28.8 N.

4-21. A 60-N toolbox is dragged horizontally at constant speed by a rope making an angle of 350

with the floor. The tension in the rope is 40 N. Determine the magnitude of the friction

force and the normal force.

ΣFx = T cos 350 – Fk = 0; Fk = (40 N) cos 350 = 32.8 N

ΣFy = N + Ty – W = 0; N = W – Ty = 60 N – T sin 350

N = 60 N – (40 N) sin 350; N = 37.1 N Fk = 32.8 N

4-22. What is the coefficient of kinetic friction for the example in Problem 4-19?

µ k = =FN

32 8. ; N37.1 N

µk = 0.884

F N

T 350

W


26

4-23. The coefficient of static friction for wood on wood is 0.7. What is the maximum angle for

an inclined wooden plane if a wooden block is to remain at rest on the plane?

Maximum angle occurs when tan θ = µs; µs = tan θ = 0.7; θ = 35.00

4-24. A roof is sloped at an angle of 400. What is the maximum coefficient of static friction

between the sole of the shoe and the roof to prevent slipping?

Tan θ = µk; µk = Tan 400 =0.839; µk = 0.839

*4-25. A 200 N sled is pushed along a horizontal surface at constant speed with a 50-N force that

makes an angle of 280 below the horizontal. What is the coefficient of kinetic friction?

ΣFx = T cos 280 – Fk = 0; Fk = (50 N) cos 280 = 44.1 N

ΣFy = N - Ty – W = 0; N = W + Ty = 200 N + T sin 280

N = 200 N + (50 N) sin 350; N = 223 N

µ k = =FN

441. N223 N

µk = 0.198

*4-26. What is the normal force on the block in Fig. 4-21? What is the component of the weight

acting down the plane?

ΣFy = N - W cos 430 = 0; N = (60N) cod 430 = 43.9 N

Wx = (60 N) sin 350; Wx = 40.9 N

*4-27. What push P directed up the plane will cause the block in Fig. 4-21 to move up the plane

with constant speed? [From Problem 4-23: N = 43.9 N and Wx = 40.9 N]

Fk = µkN = (0.3)(43.9 N); Fk = 13.2 N down plane.

ΣFx = P - Fk – Wx = 0; P = Fk + Wx; P = 13.2 N + 40.9 N; P = 54.1 N

P

Fk

N

280

W

W

F

N P

430


27

*4-28. If the block in Fig. 4-21 is released, it will overcome static friction and slide rapidly down

the plane. What push P directed up the incline will retard the downward motion until the

block moves at constant speed? (Note that F is up the plane now.)

Magnitudes of F , Wx, and N are same as Prob. 4-25.

ΣFx = P +Fk – Wx = 0; P = Wx - Fk; P = 40.9 N - 13.2 N

P = 27.7 N directed UP the inclined plane

Challlenge Problems

*4-29. Determine the tension in rope A and the compression B in the strut for Fig. 4-22.

ΣFy = 0; By – 400 N = 0; B = =400 462 N Nsin600

ΣFx = 0; Bx – A = 0; A = B cos 600

A = (462 N) cos 600; A = 231 N and B = 462 N

*4-30. If the breaking strength of cable A in Fig. 4-23 is 200 N, what is the maximum weight that

can be supported by this apparatus?

ΣFy = 0; Ay – W = 0; W = (200 N) sin 400 = 129 N

The maximum weight that can be supported is 129 N.

*4-31. What is the minimum push P parallel to a 370 inclined plane if a 90-N wagon is to be

rolled up the plane at constant speed. Ignore friction.

ΣFx = 0; P - Wx = 0; P = (90 N) sin 370

P = 54.2 N

W

F

N P

430

B

400 N

A 600 By

A

W

200 N

B 400 Ay

N P

370

W = 90 N


28

340 N

A B

W

4-32. A horizontal force of only 8 lb moves a cake of ice slides with constant speed across a floor

(µk = 0.1). What is the weight of the ice?

Fk = µkN = (0.3) W; Fk = 8 lb; (0.1)W = 8 lb; W = 80 lb.

*4-33. Find the tension in ropes A and B for the arrangement shown in Fig. 4-24a.

ΣFx = B – Wx = 0; B = Wx = (340 N) cos 300; B = 294 N

ΣFy = A – Wx = 0; A = Wy = (340 N) sin 300; A = 170 N

A = 170 N; B = 294 N

*4-34. Find the tension in ropes A and B in Fig. 4-24b.

ΣFy = By – 160 N = 0; By = 160 N ; B sin 500 = 294 N

B =160

500

Nsin

; B = 209 N

ΣFx = A – Bx = 0; A = Bx = (209 N) cos 500; A = 134 N

*4-35. A cable is stretched horizontally across the top of two vertical poles 20 m apart. A 250-N

sign suspended from the midpoint causes the rope to sag a vertical distance of 1.2 m.

What is the tension in each cable segment?.

h = 1.2 m; tan . ; .φ φ= =1210

6 840

2Tsin 6.840 = 250 N; T = 1050 N

*4-36. Assume the cable in Problem 4-31 has a breaking strength of 1200 N. What is the

maximum weight that can be supported at the midpoint?

2Tsin 6.840 = 250 N; 2(1200 N) sin 6.840 = W W = 289 N

Wy

Wy Wx 300

W = 160 N

B

A 500

W = 250 N

h φ φ T T

10 m 10 m


29

*4-37. Find the tension in the cable and the compression in the light boom for Fig. 4-25a.

ΣFy = Ay – 26 lb = 0; Ay = 26 lb ; A sin 370 = 26 lb

A =26 lb

sin;

370 A = 43.2 lb

ΣFx = B – Ax = 0; B = Ax = (43.2 lb) cos 370; B = 34.5 lb

*4-38. Find the tension in the cable and the compression in the light boom for Fig. 4-25b.

First recognize that φ = 900 - 420 = 480, Then W = 68 lb

ΣFy = By – 68 lb = 0; By = 68 lb ; B sin 480 = 68 lb

B =68 lb

sin;

480 A = 915 lb

ΣFx = Bx – A = 0; A = Bx = (91.5 lb) cos 480; B = 61.2 lb

*4-39. Determine the tension in the ropes A and B for Fig. 4-26a.

ΣFx = Bx – Ax = 0; B cos 300 = A cos 450; B = 0.816 A

ΣFy = A sin 450 – B sin 300 – 420 N = 0; 0.707 A – 0.5 B = 420 N

Substituting B = 0.816A: 0.707 A – (0.5)(0.816 A) = 420 N

Solving for A, we obtain: A = 1406 N; and B = 0.816A = 0.816(1406) or B = 1148 N

Thus the tensions are : A = 1410 N; B = 1150 N

*4-40. Find the forces in the light boards of Fig. 4-26b and state whether the boards are under

tension or compression. ( Note: θA = 900 - 300 = 600 )

ΣFx = Ax – Bx = 0; A cos 600 = B cos 450; A = 1.414 B

ΣFy = B sin 450 + A sin 600 – 46 lb = 0; 0.707 B + 0.866 A = 46 lb

Substituting A = 1.414B: 0.707 B + (0.866)(1.414 B) = 46 lb

Solving for B: B = 23.8 lb; and A = 1.414B = 01.414 (23.8 lb) or A = 33.7 lb

A = 33.7 lb, tension; B = 23.8 lb, compression

W = 26 lb

A

B 370

B

W 68 lb

A 480 By

420 N

A

B 300

450

W

46 lb

A B

600 450

W


30

Critical Thinking Questions 4-41. Study the structure drawn in Fig. 4-27 and analyze the forces acting at the point where the

rope is attached to the light poles. What is the direction of the forces acting ON the ends

of the poles? What is the direction of the forces exerted BY the poles at that point? Draw

the appropriate free-body diagram. Imagine that the poles are bolted together at their

upper ends, then visualize the forces ON that bolt and BY that bolt.

*4-42. Determine the forces acting ON the ends of the poles in Fig 3-27 if W = 500 N.


ΣFy = A sin 600 – B sin 300 – 500 N = 0; 0.866 A – 0.5 B = 500 N

Substituting B = 0.577 A: 0.866 A – (0.5)( 0.577 A) = 500 N

Solving for A, we obtain: A = 866 N; and B = 0.577 A = 0.577(866) or B = 500 N

Thus the forces are : A = 866 N; B = 500 N

Can you explain why B = W? Would this be true for any weight W?

Try another value, for example W = 800 N and solve again for B.

W

A

B 300

600

Forces ON Bolt at Ends (Action Forces):

The force W is exerted ON the bolt BY the weight. The force B is exerted ON bolt BY right pole. The force A is exerted ON bolt BY the middle pole. To understand these directions, imagine that the poles snap, then what would be the resulting motion.

Wr Ar

Br

300 600

Forces BY Bolt at Ends (Reaction Forces):

The force Wr is exerted BY the bolt ON the weight. The force Br is exerted ON bolt BY right pole. The force Ar is exerted BY bolt ON the middle pole. Do not confuse action forces with the reaction forces.

W

A

B 300

600


31

*4-43. A 2-N eraser is pressed against a vertical chalkboard with a horizontal push of 12 N. If

µs = 0.25, find the horizontal force required to start motion parallel to the floor? What if

you want to start its motion upward or downward? Find the vertical forces required to just

start motion up the board and then down the board? Ans. 3.00 N, up = 5 N, down = 1 N.

For horizontal motion, P = Fs = µsN

P = 0.25 (12 N); P = 3.00 N

For vertical motion, P – 2 N – Fk = 0

P = 2 N + 3 N; P = 5.00 N

For down motion: P + 2 N – Fs = 0; P = - 2 N + 3 N; P = 1.00 N

*4-44. It is determined experimentally that a 20-lb horizontal force will move a 60-lb lawn

mower at constant speed. The handle of the mower makes an angle of 400 with the

ground. What push along the handle will move the mower at constant speed? Is the

normal force equal to the weight of the mower? What is the normal force?

µ k = =20 0 333 lb60 lb

. ΣFy = N – Py - W= 0; W = 60 lb

N = P sin 400 + 60 lb; Fk = µkN = 0.333 N

ΣFy = Px - Fk = 0; P cos 400 – 0.333N = 0

P cos 400 – 0.333 (P sin 400 + 60 lb) = 0; 0.766 P = 0.214 P + 20 lb;

0.552 P = 20 lb; P = =200552

36 2 lb lb.

. ; P = 36.2 lb

The normal force is: N = (36.2 lb) sin 400 + 60 lb N = 83.3 lb

12 N P

2 N

F

N

2 N

F F P

P

Fk

N

400

W

12 N F

2 N

P

N


32

W = 70N

N F

400

*4-45. Suppose the lawn mower of Problem 4-40 is to be moved backward. What pull along the

handle is required to move with constant speed? What is the normal force in this case?

Discuss the differences between this example and the one in the previous problem.

µ k = =20 0 333 lb60 lb

. ΣFy = N + Py - W= 0; W = 60 lb

N = 60 lb - P sin 400; Fk = µkN = 0.333 N

ΣFy = Px - Fk = 0; P cos 400 – 0.333N = 0

P cos 400 – 0.333 (60 lb - P sin 400) = 0; 0.766 P - 20 lb + 0.214 P = 0;

0.980 P = 20 lb; P = =200 980

20 4 lb lb.

. ; P = 20.4 lb

The normal force is: N = 60 lb – (20.4 lb) sin 400 N = 46.9 lb

*4-46. A truck is removed from the mud by attaching a line between the truck and the tree. When

the angles are as shown in Fig. 4-28, a force of 40 lb is exerted at the midpoint of the line.

What force is exerted on the truck? φ = 200

T sin 200 + T sin 200 = 40 lb 2 T sin 200 = 40 lb

T = 58.5 lb

*4-47. Suppose a force of 900 N is required to remove the move the truck in Fig. 4-28. What

force is required at the midpoint of the line for the angles shown?.

2 T sin 200 = F; 2(900 N) sin 200 = F; F = 616 N

*4-48. A 70-N block of steel is at rest on a 400 incline. What is the static

friction force directed up the plane? Is this necessarily the

maximum force of static friction? What is the normal force?

F = (70 N) sin 400 = 45.0 N N = (70 N) cos 400 = 53.6 N

P Fk

N

400

W

F

h φ φ T T


33

*4-49. Determine the compression in the center strut B and the tension in the rope A for the

situation described by Fig. 4-29. Distinguish clearly the difference between the

compression force in the strut and the force indicated on your free-body diagram.


ΣFy = B sin 500 – A sin 200 – 500 N = 0; 0.766 B – 0.342 A = 500 N

Substituting B = 1.46 A: 0.766 (1.46 A) – (0.342 A) = 500 N

Solving for A, we obtain: A = 644 N; and B = 1.46 A = 1.46 (644) or B = 940 N

Thus the tensions are : A = 644 N; B = 940 N

*4-50. What horizontal push P is required to just prevent a 200 N block from slipping down a 600

inclined plane where µs = 0.4? Why does it take a lesser force if P acts parallel to the

plane? Is the friction force greater, less, or the same for these two cases?

(a) ΣFy = N – Wy– Py = 0; Wy = (200 N) cos 600 = 100 N

Py = P sin 600 = 0.866 P; N = 100 N + 0.866 P

F = µN = 0.4(100 N + 0.866 P); F = 40 N + 0.346 P

ΣFx = Px – Wx + F = 0; P cos 600 - (200 N) sin 600 + (40 N + 0.346 P) = 0

0.5 P –173.2 N + 40 N + 0.346 P = 0 Solving for P gives: P = 157 N

(b) If P were parallel to the plane, the normal force would be LESS, and therefore the

friction force would be reduced. Since the friction force is directed UP the plane, it is

actually helping to prevent slipping. You might think at first that the push P (to stop

downward slipping) would then need to be GREATER than before, due to the lesser

friction force. However, only half of the push is effective when exerted horizontally.

If the force P were directed up the incline, a force of only 133 N is required. You

should verify this value by reworking the problem.

W A

B

200 500

x

W 600

600

F N

P 600


34

*4-51. Find the tension in each cord of Fig. 4-30 if the suspended weight is 476 N.

Consider the knot at the bottom first since more information is given at that point.

Cy + Cy = 476 N; 2C sin 600 = 476 N

C = =476 275 N

2sin60 N0

ΣFy = A sin 300 - (275 N) sin 600 = 0

A = 476 N; ΣFx = A cos 300 – C cos 600 – B = 0; 476 cos 300 – 275 cos 600 – B = 0

B = 412 N – 137 N = 275 N; Thus: A = 476 N, B = 275 N, C = 275 N

*4-52. Find the force required to pull a 40-N sled horizontally at constant speed by exerting a pull

along a pole that makes a 300 angle with the ground (µk = 0.4). Now find the force

required if you push along the pole at the same angle. What is the major factor that

changes in these cases?

(a) ΣFy = N + Py - W= 0; W = 40 N

N = 40 N - P sin 300; Fk = µkN

ΣFx = P cos 300 - µkN = 0; P cos 400- 0.4(40 N - P sin 300) =0;

0.866 P – 16 N + 0.200 P = 0; P = 15.0 N

(b) ΣFy = N - Py - W= 0; N = 40 N + P sin 300; Fk = µkN

ΣFx = P cos 300 - µkN = 0; P cos 400- 0.4(40 N + P sin 300) =0;

0.866 P – 16 N - 0.200 P = 0; P = 24.0 N Normal force is greater!

476 N

A C C

600 600 B

C 275 N

600 300

300

P

Fk

N

300

W

P Fk

N

W


35

**4-53. Two weights are hung over two frictionless pulleys as shown in Fig. 4-31. What weight

W will cause the 300-lb block to just start moving to the right? Assume µs = 0.3. Note:

The pulleys merely change the direction of the applied forces.

ΣFy = N + (40 lb) sin 450 + W sin 300 – 300 lb = 0

N = 300 lb – 28.3 lb – 0.5 W; F = µsN

ΣFx = W cos 300 - µsN – (40 lb) cos 450 = 0

0.866 W – 0.3(272 lb – 0.5 W) – 28.3 lb = 0; W = 108 lb

**4-54. Find the maximum weight than can be hung at point O in Fig. 4-32 without upsetting the

equilibrium. Assume that µs = 0.3 between the block and table.

We first find F max for the block

F = µsN = 0.3 (200 lb) = 60 lb

Now set A = F = 60 lb and solve for W:

ΣFx = B cos 200 – A = 0; B cos 200 = 60 lb; B = 63.9 lb

ΣFy = B sin 200 – W = 0; W = B sin 200 = (63.9 lb) sin 200; W = 21.8 lb

F

40 lb N

W 450 300

300 lb

W

200 F

B A A

Chapter 6 Uniform Acceleration Physics, 6th Edition

47

Chapter 6. Uniform Acceleration

Problems:

Speed and Velocity

6-1. A car travels a distance of 86 km at an average speed of 8 m/s. How many hours were

required for the trip?

s vt= 86,000 m 1 h10,750 s 8 m/s 3600 s

t = =

t = 2.99 h

6-2. Sound travels at an average speed of 340 m/s. Lightning from a distant thundercloud is

seen almost immediately. If the sound of thunder reaches the ear 3 s later, how far away is

the storm?

t st

= = =20 m

340 m / s 0.0588 s t = 58.8 ms

6-3. A small rocket leaves its pad and travels a distance of 40 m vertically upward before

returning to the earth five seconds after it was launched. What was the average velocity for

the trip?

v st

= = =40 80 m + 40 m

5 s m

5 s v = 16.0 m/s

6-4. A car travels along a U-shaped curve for a distance of 400 m in 30 s. It’s final location,

however is only 40 m from the starting position. What is the average speed and what is the

magnitude of the average velocity?

Average speed: v st

= =400 m30 s

v = 13.3 m/s

Average velocity: v = =Dt

m30 s40 v = 1.33 m/s, E

s = 400 m

D = 40 m


48

6-5. A woman walks for 4 min directly north with a average velocity of 6 km/h; then she

walks eastward at 4 km/h for 10 min. What is her average speed for the trip?

t1 = 4 min = 0.0667 h; t2 = 10 min = 0.167 h

s1 = v1t1 = (6 km/h)(0.0667 h) = 0.400 km

s1 = v2t2 = (4 km/h)(0.167 h) = 0.667 km

v s st t

=++

=1 2

1 2

0.4 km + 0.667 km0.0667 h + 0.167 h

v = 4.57 km/h

6-6. What is the average velocity for the entire trip described in Problem 6-5?

D = =( . ; tan ..

0 667 0 40 667

km) + (0.400 km) km km

2 2 θ D = 0.778 km, 31.00

v = =0 778 3 33. . km

0.0667 h + 0.167 h km / h v = 3.33 km/h, 31.00

6-7. A car travels at an average speed of 60 mi/h for 3 h and 20 min. What was the distance?

t = 3 h + 0.333 h = 3.33 h; s = vt = (60 mi/h)(3.33 h); s = 200 mi

6.8 How long will it take to travel 400 km if the average speed is 90 km/h?

t st

= =400 km

90 km / h t = 4.44 h

*6-9. A marble rolls up an inclined ramp a distance of 5 m, then stops and returns to a point 5

m below its starting point. The entire trip took only 2 s. What was the average speed

and what was the average velocity? (s1 = 5 m, s2 = -10 m)

speed = 5 m + 10 m2 s

v = 7.50 m/s

velocity = Dt=

5 m - 10 m2 s

v = – 2.5 m/s, down plane.

D

s2 s1

θ 6 km/h, 4 min

4 km/h, 10 min

D θ

s1

s2

C B

A E


49

Uniform Acceleration

6-10. The tip of a robot arm is moving to the right at 8 m/s. Four seconds later, it is moving to

the left at 2 m/s. What is the change in velocity and what is the acceleration.

∆v = vf - vo = (–2 m/s) – (8 m/s) ∆v = –10 m/s

a vt

= =−∆ 10 m / s

4 s a = –2.50 m/s2

6-11. An arrow accelerates from zero to 40 m/s in the 0.5 s it is in contact with the bow string.

What is the average acceleration?

av v

tf o=−

=40 m / s - 0

0.5 s a = 80.0 m/s2

6-12. A car traveling initially at 50 km/h accelerates at a rate of 4 m/s2 for 3 s. What is the

final speed?

vo = 50 km/h = 13.9 m/s; vf = vo + at

vf = (13.9 m/s) + (4 m/s2)(3 s) = 25.9 m/s; vf = 25.9 m/s

6-13. A truck traveling at 60 mi/h brakes to a stop in 180 ft. What was the average acceleration

and stopping time?

vo = 60 mi/h = 88.0 ft/s 2as = vf2 – vo

2

2 2 20 (88.0 ft/s)2 2(180 ft)

f ov va

s− −

= = a = – 21.5 ft/s2

0

0

2 2(180 ft);2 88.0 ft/s + 0

f

f

v v xx t tv v

+ = = = + t = 4.09 s


50

6-14. An arresting device on a carrier deck stops an airplane in 1.5 s. The average acceleration

was 49 m/s2. What was the stopping distance? What was the initial speed?

vf = vo + at; 0 = vo + (– 49 m/s2)(1.5 s); vo = 73.5 m/s

s = vf t - ½at2 ; s = (0)(1.5 s) – ½(-49 m/s2)(1.5 s)2; s = 55.1 m

6-15. In a braking test, a car traveling at 60 km/h is stopped in a time of 3 s. What was the

acceleration and stopping distance? ( vo = 60 km/h = 16.7 m/s)

vf = vo + at; (0) = (16.7 m/s) + a (3 s); a = – 5.56 m/s2

( )0 16.6 m/s + 0 3 s2 2

fv vs t

+ = =

; s = 25.0 m

6-16. A bullet leaves a 28-in. rifle barrel at 2700 ft/s. What was its acceleration and time in the

barrel? (s = 28 in. = 2.33 ft)

2as = vo2 - vf

2 ; av v

sf=−

=−2

02

2(2700 ft / s) 0

2(2.33 ft)

2

; a = 1.56 x 106 m/s2

sv v

t sv v

f

f

=+

+=0

022 2 2 33; t = ft)

0 + 2700 ft / s( . ; t = 1.73 ms

6-17. The ball in Fig. 6-13 is given an initial velocity of 16 m/s at the bottom of an inclined

plane. Two seconds later it is still moving up the plane, but with a velocity of only 4

m/s. What is the acceleration?

vf = vo + at; av v

tf=−

=0 4 m / s - (16 m / s)2 s

; a = -6.00 m/s2

6-18. For Problem 6-17, what is the maximum displacement from the bottom and what is the

velocity 4 s after leaving the bottom? (Maximum displacement occurs when vf = 0)

2as = vo2 - vf

2; sv v

af=−

=−2

02

20 (16 m / s)

2(-6 m / s )

2

2 ; s = +21.3 m

vf = vo + at = 16 m/s = (-6 m/s2)(4 s); vf = - 8.00 m/s, down plane


51

6-19. A monorail train traveling at 80 km/h must be stopped in a distance of 40 m. What average

acceleration is required and what is the stopping time? ( vo = 80 km/h = 22.2 m/s)

2as = vo2 - vf

2; av v

sf=−

=−2

02

20 (22.2 m / s)

2(40 m)

2

; a = -6.17 m/s2

sv v

t sv v

f

f

=+

+=0

022 2 40; t = m)

22.2 m / s + 0( ; t = 3.60 m/s

Gravity and Free-Falling Bodies

6-20. A ball is dropped from rest and falls for 5 s. What are its position and velocity?

s = vot + ½at2; s = (0)(5 s) + ½(-9.8 m/s2)(5 s)2 ; s = -122.5 m

vf = vo + at = 0 + (-9.8 m/s2)(5 s); v = -49.0 m/s

6-21. A rock is dropped from rest. When will its displacement be 18 m below the point of

release? What is its velocity at that time?

s = vot + ½at2; (-18 m) = (0)(t) + ½(-9.8 m/s2)t2 ; t = 1.92 s

vf = vo + at = 0 + (-9.8 m/s2)(1.92 s); vf = -18.8 m/s

6-22. A woman drops a weight from the top of a bridge while a friend below measures the time

to strike the water below. What is the height of the bridge if the time is 3 s?

s = vot + ½at2 = (0) + ½(-9.8 m/s2)(3 s)2; s = -44.1 m

6-23. A brick is given an initial downward velocity of 6 m/s. What is its final velocity after

falling a distance of 40 m?

2as = vo2 - vf

2 ; v v asf = + = + −02 2 40(-6 m / s) 2(-9.8 m / s m)2 2 )( ;

v = ±28.6 m/s; Since velocity is downward, v = - 28.6 m/s


52

6-24. A projectile is thrown vertically upward and returns to its starting position in 5 s. What

was its initial velocity and how high did it rise?

s = vot + ½at2; 0 = vo(5 s) + ½(-9.8 m/s2)(5 s)2 ; vo = 24.5 m/s

It rises until vf = 0; 2as = vo2 - vf

2 ; s = −0 ()

24.5 m / s)2(-9.8 m / s

2

2 ; s = 30.6 m

6-25. An arrow is shot vertically upward with an initial velocity of 80 ft/s. What is its

maximum height? (At maximum height, vf = 0; a = g = -32 ft/s2)

2as = vo2 - vf

2; sv v

af=−

=2

02

20 - (80 ft / s)2(-32 ft / s

2

2 ); s = 100 ft

6-26. In Problem 6-25, what are the position and velocity of the arrow after 2 s and after 6 s?

s = vot + ½at2 = (80 ft/s)(2 s) + ½(-32 ft/s2)(2 s)2 ; s = 96 ft

vf = vo + at = (80 ft/s) + (-32 ft/s2)(2 s); vf = 16 ft/s

s = vot + ½at2 = (80 ft/s)(6 s) + ½(-32 ft/s2)(6 s)2 ; s = -96 ft

vf = vo + at = (80 ft/s) + (-32 ft/s2)(6 s); vf = -112 ft/s

6-27. A hammer is thrown vertically upward to the top of a roof 16 m high. What minimum

initial velocity was required?

2as = vo2 - vf

2 ; v v asf02 2 16= − = −(0) 2(-9.8 m / s m)2 2 )( ; vo = 17.7 m/s

Horizontal Projection

6-28. A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, how far

will it have traveled horizontally and how far has it fallen vertically?

x = vox t = (20 m/s)(2.5 s) ; x = 50.0 m

y = voy + ½gt2 = (0)(2.5 s) + ½(-9.8 m/s2)(0.25 s)2 y = -0.306 m


53

0

6-29. An airplane traveling at 70 m/s drops a box of supplies. What horizontal distance will the

box travel before striking the ground 340 m below?

First we find the time to fall: y = voy t + ½gt2 t yg

= =−

−2 2

9 8(

.340 m) m / s2

t = 8.33 s ; x = vox t = (70 m/s)(8.33 s) ; x = 583 m

6-30. At a lumber mill, logs are discharged horizontally at 15 m/s from a greased chute that is

20 m above a mill pond. How far do the logs travel horizontally?

y = ½gt2; t yg

= =−

−2 2

9 8(.

20 m) m / s2 ; t = 2.02 s

x = vox t = (15 m/s)(8.33 s) ; x = 30.3 m

6-31. A steel ball rolls off the edge of a table top 4 ft above the floor. If it strikes the floor 5 ft

from the base of the table, what was its initial horizontal speed?

First find time to drop 4 ft: t yg

= =−

−2 2

32( 4 ft)

ft / s2 ; t = 0.500 s

x = vox t ; v xtx0

505

= = ft s.

; vox = 10.0 ft/s

6-32. A bullet leaves the barrel of a weapon with an initial horizontal velocity of 400 m/s. Find

the horizontal and vertical displacements after 3 s.

x = vox t = (400 m/s)(3 s) ; x = 1200 m

y = voy + ½gt2 = (0)(3 s) + ½(-9.8 m/s2)(3 s)2 y = -44.1 m

6-33. A projectile has an initial horizontal velocity of 40 m/s at the edge of a roof top. Find

the horizontal and vertical components of its velocity after 3 s.

vx = vox = 40 m/s vy = voy t + gt = 0 + (-9.8 m/s2)(3s); vy = -29.4 m/s


54

The More General Problem of Trajectories 6-34. A stone is given an initial velocity of 20 m/s at an angle of 580. What are its horizontal

and vertical displacements after 3 s?

vox = (20 m/s) cos 580 = 10.6 m/s; voy = (20 m/s) sin 580 = 17.0 m/s

x = voxt = (10.6 m/s)(3 s); x = 31.8 m

y = voyt + ½gt2 = (17.0 m/s)(3 s) +½(-9.8 m/s2)(3 s)2; y = 6.78 m

6-35. A baseball leaves the bat with a velocity of 30 m/s at an angle of 300. What are the

horizontal and vertical components of its velocity after 3 s?


vx = vox = 26.0 m/s ; vx = 26.0 m/s

vy = voy + gt = (15 m/s) + (-9.8 m/s2)(3 s) ; vy = -14.4 m/s

6-36. For the baseball in Problem 6-33, what is the maximum height and what is the range?

ymax occurs when vy = 0, or when: vy = voy + gt = 0 and t = - voy/g

tvg

toy=−

=−−

=30 309 8

1530sin

.; .

m / s s ; Now we find ymax using this time.

ymax = voyt + ½gt2 = (15 m/s)(1.53 s) + ½(-9.8 m/s2)(1.53 s)2; ymax = 11.5 m

The range will be reached when the time is t’ = 2(1.53 s) or t’ = 3.06 s, thus

R = voxt’= (30 m/s) cos 300 (3.06 s); R = 79.5 m

6-37. An arrow leaves the bow with an initial velocity of 120 ft/s at an angle of 370 with the

horizontal. What are the horizontal and vertical components of is displacement two

seconds later?

vox = (120 ft/s) cos 370 = 104 ft/s; voy = (120 ft/s) sin 300 = 60.0 ft/s


55

6-37. (Cont.) The components of the initial velocity are: vox = 104 ft/s; voy = 60.0 ft/s

x = voxt = (104 ft/s)(2 s); x = 208 ft

y = voyt + ½gt2 = (60.0 m/s)(2 s) +½(-32 ft/s2)(2 s)2; y = 56.0 ft

*6-38. In Problem 6-37, what are the magnitude and direction of arrow’s velocity after 2 s?

vx = vox = 104 ft/s ; vx = 104 ft/s

vy = voy + gt = (60 m/s) + (-32 ft/s2)(2 s) ; vy = -4.00 ft/s

*6-39. A golf ball in Fig. 6-14 leaves the tee with a velocity of 40 m/s at 650. If it lands on a

green located 10 m higher than the tee, what was the time of flight, and what was the

horizontal distance to the tee?


y = voyt + ½gt2: 10 ft = (36.25 m/s) t + ½(-9.8 m/s2)t2

Solving quadratic (4.9t2 – 36.25t + 10 = 0) yields: t1 = 0.287 s and t2 = 7.11 s

The first time is for y = +10 m on the way up, the second is y = +10 m on the way down.

Thus, the time from tee to green was: t = 7.11 s

Horizontal distance to tee: x = voxt = (16.9 m/s)(7.11 s); x = 120 m

*6-40. A projectile leaves the ground with a velocity of 35 m/s at an angle of 320. What is the

maximum height attained.


ymax occurs when vy = 0, or when: vy = voy + gt = 0 and t = - voy/g

tvg

toy=−

=−

−=

18559 8

1890.

.; .

m / s s2 ; Now we find ymax using this time.

ymax = voyt + ½gt2 = (18.55 m/s)(1.89 s) + ½(-9.8 m/s2)(1.89 s)2; ymax = 17.5 m


56

*6-41. The projectile in Problem 6-40 rises and falls, striking a billboard at a point 8 m above

the ground. What was the time of flight and how far did it travel horizontally.


y = voyt + ½gt2: 8 m = (18.55 m/s) t + ½(-9.8 m/s2)t2

Solving quadratic (4.9t2 – 18.55t + 8 = 0) yields: t1 = 0.497 s and t2 = 3.36 s

The first time is for y = +8 m on the way up, the second is y = +8 m on the way down.

Thus, the time from tee to green was: t = 3.29 s

Horizontal distance to tee: x = voxt = (29.7 m/s)(3.29 s); x = 97.7 m

Challenge Problems

6-42. A rocket travels in space at 60 m/s before it is given a sudden acceleration. It’s velocity

increases to 140 m/s in 8 s, what was its average acceleration and how far did it travel in

this time?

av v

tf=−

=0 (140 m / s) - (60 m / s)8 s

; a = 10 m/s2

sv v

tf=+

=FHG IKJ0

2140 8 m / s + 60 m / s

2 sb g; t = 800 s

6-43. A railroad car starts from rest and coasts freely down an incline. With an average

acceleration of 4 ft/s2, what will be the velocity after 5 s? What distance does it travel?

vf = vo + at = 0 + (4 ft/s2)(5 s); vf = 20 ft/s

s = vot + ½at2 = 0 + ½(4 ft/s2)(5 s)2; s = 50 ft


57

*6-44. An object is projected horizontally at 20 m/s. At the same time, another object located

12 m down range is dropped from rest. When will they collide and how far are they

located below the release point?

A: vox = 20 m/s, voy = 0; B: vox = voy = 0

Ball B will have fallen the distance y at the same time t as ball A. Thus,

x = voxt and (20 m/s)t = 12 m; t = 0.600 s

y = ½at2 = ½(-9.8 m/s2)(0.6 s)2 ; y = -1.76 m

6-45. A truck moving at an initial velocity of 30 m/s is brought to a stop in 10 s. What was the

acceleration of the car and what was the stopping distance?

av v

tf=−

=0 0 - 30 m / s10 s

; a = -3.00 m/s2

sv v

tf=+

=FHG IKJ0

230 10 m / s + 0

2 sb g; s = 150 m

6-46. A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its

position and velocity after 2s, after 4 s, and after 8 s?

Apply s = vot + ½at2 and vf = vo + at for time of 2, 4, and 8 s:

(a) s = (23 m/s)(2 s) + ½(-9.8 m/s2)(2 s)2 ; s = 26.4 m

vf = (23 m/s) + (-9.8 m/s2)(2 s) ; vf = 3.40 m/s

(b) s = (23 m/s)(4 s) + ½(-9.8 m/s2)(4 s)2 ; s = 13.6 m

vf = (23 m/s) + (-9.8 m/s2)(4 s) ; vf = -16.2 m/s

(c) s = (23 m/s)(8 s) + ½(-9.8 m/s2)(8 s)2 ; s = -130 m

vf = (23 m/s) + (-9.8 m/s2)(8 s) ; vf = -55.4 m/s

y

B A

12 m


58

6-47. A stone is thrown vertically downward from the top of a bridge. Four seconds later it

strikes the water below. If the final velocity was 60 m/s. What was the initial velocity of

the stone and how high was the bridge?

vf = vo + at; v0 = vf – at = (-60 m/s) - (-9.8 m/s)(4 s); vo = -20.8 m/s

s = vot + ½at2 = (-20.8 m/s)(4 s) + ½(-9.8 m/s)(4 s)2; s = 162 m

6-48. A ball is thrown vertically upward with an initial velocity of 80 ft/s. What are its

position and velocity after (a) 1 s; (b) 3 s; and (c) 6 s

Apply s = vot + ½at2 and vf = vo + at for time of 2, 4, and 8 s:

(a) s = (80 ft/s)(1 s) + ½(-32 ft/s2)(1 s)2 ; s = 64.0 ft

vf = (80 ft/s) + (-32 ft/s2)(2 s) ; vf = 16.0 ft/s

(b) s = (80 ft/s)(3 s) + ½(-32 ft/s2)(3 s)2 ; s = 96.0 ft

vf = (80 ft/s) + (-32 ft/s2)(3 s) ; vf = -16.0 ft/s

(c) s = (80 ft/s)(6 s) + ½(-32 ft/s2)(6 s)2 ; s = 64.0 ft

vf = (80 ft/s) + (-32 ft/s2)(6 s) ; vf = -96.0 ft/s

6-49. An aircraft flying horizontally at 500 mi/h releases a package. Four seconds later, the

package strikes the ground below. What was the altitude of the plane?

y = ½gt2 = ½(-32 ft/s2)(4 s)2; y = -256 ft

*6-50. In Problem 6-49, what was the horizontal range of the package and what are the

components of its final velocity?

vo = 500 mi/h = 733 ft/s; vx = vox = 733 ft/s; voy = 0; t = 4 s

x = vxt = (733 ft/s)(4 s); x = 2930 ft

vy = voy + at = 0 + (-32 ft/s)(4 s); vy = -128 ft/s; vx = 733 m/s


59

*6-51. A putting green is located 240 ft horizontally and 64 ft vertically from the tee. What

must be the magnitude and direction of the initial velocity if a ball is to strike the green

at this location after a time of 4 s?

x = voxt; 240 ft = vox (4 s); vox = 60 m/s

s = vot + ½at2; 64 ft = voy(4 s) + ½(-32 ft/s2)(4 s)2; voy = 80 ft/s

v v vx y= + = +2 2 60( (80 ft / s) ft / s)2 2 ; tanθ =80 ft / s60 ft / s

v = 100 ft/s, θ = 53.10

Critical Thinking Questions 6-52. A long strip of pavement is marked off in 100-m intervals. Students use stopwatches to

record the times a car passes each mark. The following data is listed:

Distance, m 0 10 m 20 m 30 m 40 m 50 m

Time, s 0 2.1 s 4.3 s 6.4 s 8.4 s 10.5 s

Plot a graph with distance along the y-axis and time along the x-axis. What is the

significance of the slope of this curve? What is the average speed of the car? At what

instant in time is the distance equal to 34 m? What is the acceleration of the car?

Data taken directly from the graph (not drawn): Ans. Slope is v, 4.76 m/s, 7.14 s, 0.

6-53. An astronaut tests gravity on the moon by dropping a tool from a height of 5 m. The

following data are recorded electronically.

Height, m 5.00 m 4.00 m 3.00 m 2.00 m 1.00 m 0 m

Time, s 0 1.11 s 1.56 s 1.92 s 2.21 s 2.47 s


60

6-53. (Cont.) Plot the graph of this data. Is it a straight line? What is the average speed for the

entire fall? What is the acceleration? How would you compare this with gravity on earth?

Data taken directly from the graph (not drawn): Ans. Slope is v, 4.76 m/s, 7.14 s, 0.

*6-54. A car is traveling initially North at 20 m/s. After traveling a distance of 6 m, the car

passes point A where it's velocity is still northward but is reduced to 5 m/s. (a) What

are the magnitude and direction of the acceleration of the car? (b) What time was

required? (c) If the acceleration is held constant, what will be the velocity of the car

when it returns to point A?

(a) vo = 20 m/s, vf = 5 m/s, x = 6 m

2as = vo2 - vf

2;2 2 2 2

0 (5 m/s) (20 m/s)2 2(6 m)

fv va

s− −

= = ; a = -31.2 m/s2

(b) 0

0

2 2(6 m);2 20 m/s + 5 m/s

f

f

v v ss t tv v

+ = = = + ; t = 0.480 s

(c) Starts at A with vo = + 5 m/s then returns to A with zero net displacement (s = 0):

2as = vo2 - vf

2; 0 = (5 m/s)2 – vf2; v f = = ±(5 m / s) m / s2 5 ; vf = - 5 m/s

*6-55. A ball moving up an incline is initially located 6 m from the bottom of an incline and has

a velocity of 4 m/s. Five seconds later, it is located 3 m from the bottom. Assuming

constant acceleration, what was the average velocity? What is the meaning of a negative

average velocity? What is the average acceleration and final velocity?

vo = + 4 m/s; s = -3 m; t = 5 s Find vavg

s = vavg t; v =−3 m5 s

; vavg = -0.600 m/s

Negative average velocity means that the velocity was down the plane most of the time.

x = 6 m x = 0

A v = 5 m/s v = 20 m/s

4 m/s 6 m

3 m

s = 0


61

*6-55. (Cont.) s = vot + ½at2; -3 m = (4 m/s)(5 s) + ½a (5 s)2; a = -1.84 m/s2

vf = vo + at = 4 m/s + (-1.84 m/s2)(5 s); vf = -5.20 m/s

*6-56. The acceleration due to gravity on an distant planet is determined to be one-fourth its

value on the earth. Does this mean that a ball dropped from a height of 4 m above this

planet will strike the ground in one-fourth the time? What are the times required on the

planet and on earth?

The distance as a function of time is given by: s = ½at2 so that

one-fourth the acceleration should result in twice the drop time.

t sge

e

= =2 2(4 m)

9.8 m / s2 te = 0.904 s t sgp

p

= =2 2(4 m)

2.45 m / s2 tp = 1.81 s

*6-57. Consider the two balls A and B shown in Fig. 6-15. Ball A has a constant acceleration of

4 m/s2 directed to the right, and ball B has a constant acceleration of 2 m/s2 directed to

the left. Ball A is initially traveling to the left at 2 m/s, while ball B is traveling to the left

initially at 5 m/s. Find the time t at which the balls collide. Also, assuming x = 0 at the

initial position of ball A, what is their common displacement when they collide?

Equations of displacement for A and B:

s = so + vot + ½at2 (watch signs)

For A: sA = 0 + (-2 m/s)t + ½(+4 m/s2) t2

For B: sB = 18 m + (-5 m/s)t + ½(-2 m/s2) t2; Next simplify and set sA = sB

- 2t + 2t2 = 18 – 5t - t2 → 3t2 + 3t – 18 = 0 → t1 = - 3 s, t2 = +2 s

Accept t = +3 s as meaningful answer, then substitute to find either sA or sB:

sA = -2(2 s) + 2(2 s)2; x = + 4 m

v = - 5 m/s2

v = - 2 m/s

+

aa = +4 m/s2

x = 18 m x = 0

A B

ab = -2 m/s2


62

*6-58. Initially, a truck with a velocity of 40 ft/s is located distance of 500 ft to the right of a

car. If the car begins at rest and accelerates at 10 ft/s2, when will it overtake the truck?

How far is the point from the initial position of the car?

Equations of displacement for car and truck:

s = so + vot + ½at2 (watch signs)

For car: sC = 0 + ½(+10 ft/s2) t2 ; Truck: sT = 500 ft + (40 ft/s)t + 0;

Set sC = sT 5t2 = 500 + 40t or t2 – 8t –100 = 0; t1 = -6.77 s; t2 = +14.8 s

Solve for either distance: sC = ½(10 ft/s2)(14.8 s)2; s = 1092 ft

*6-59. A ball is dropped from rest at the top of a 100-m tall building. At the same instant a

second ball is thrown upward from the base of the building with an initial velocity of 50

m/s. When will the two balls collide and at what distance above the street?

For A: sA = 100 m + v0At + ½gt2 = 100 m + 0 + ½(-9.8 m/s2) t2

For B: sB = 0 + (50 m/s)t + ½(-9.8 m/s2) t2 Set sA = sB

100 – 4.9 t2 = 50 t – 4.9 t2; 50 t = 100; t = 2.00 s

Solve for s: sA = 100 m – (4.9 m/s2)(2 s)2; s = 80.4 m

*6-60. A balloonist rising vertically with a velocity of 4 m/s releases a sandbag at the instant

when the balloon is 16 m above the ground. Compute the position and velocity of the

sandbag relative to the ground after 0.3 s and 2 s. How many seconds after its release

will it strike the ground?

The initial velocity of the bag is that of the balloon: voB = + 4 m/s

From ground: s = soB + voBt + ½gt2; s = 18 m + (4 m/s)t + ½(-9.8 m/s2)t2

s = 18 m + (4 m/s)(0.3 s) – (4.9 m/s2)(0.3 s)2 ; s = 16.8 m

v = 0

s = 0 s = 500 ft

v = 40 ft/s +

A

B s = 0

s = 100 m


63

*6-61. An arrow is shot upward with a velocity of 40 m/s. Three seconds later, another arrow is

shot upward with a velocity of 60 m/s. At what time and position will they meet?

Let t1 = t be time for first arrow, then t2 = t - 3 for second arrow.

s1 = (40 m/s)t1 + ½(-9.8 m/s2)t12 ; s1 = 40t – 4.9t2

s2 = (60 m/s)t2 + ½(-9.8 m/s2)t22 ; s2 = 60(t – 3) - 4.9(t – 3)2

s1 = s2; 40t – 4.9t2 = 60t – 180 – 4.9(t2 – 6t + 9)

The solution for t gives: t = 4.54 s

Now find position: s1 = s2 = (40 m/s)(4.54 s) – (4.9 m/s2)(4.54 s)2; s = 80.6 m

*6-62. Someone wishes to strike a target, whose horizontal range is 12 km. What must be the

velocity of an object projected at an angle of 350 if it is to strike the target. What is the

time of flight?

y = voyt + ½gt2 = 0; ( vo sin 350)t = (4.9 m/s2)t2 or t v t v= =0 574

4 901170

0.

.; .

R = voxt = 12 km; (vo cos 350)t = 12,000 m; tv

=14 649

0

, Set t = t

05744 9

14 6490

0

..

,vv

= ; From which vo = 354 m/s and t = 41.4 s

*6-63. A wild boar charges directly toward a hunter with a constant speed of 60 ft/s. At the

instant the boar is 100 yd away, the hunter fires an arrow at 300 with the ground. What

must be the velocity of the arrow if it is to strike its target?

y = 0 = (v0 sin 300)t + ½(-32 ft/s2)t2; Solve for t

t v v= =05 2

320 031250

0. ( ) . ; t = 0.03125 vo

s1 =( v0 cos 300) t = (0.866 vo)(0.03125 vo); s1 = 0.0271 vo2

s1 = s2 60 m/s

40 m/s

v = -60 ft/s

s1 = s2 s = 300 ft s = 0

vo

300


64

*6-63. (Cont.) s1 = 0.0271 vo2 ; t = 0.03125 vo

vB = - 60 ft/s; soB = 300 ft

s2 = soB + vBt = 300 ft + (-60 ft/s)t

s2 = 300 – 60 (0.03125 vo) = 300 – 1.875 vo Now, set s1 = s2 and solve for vo

0.0271 vo2 = 300 – 1.875 vo or vo

2 + 69.2 vo – 11,070 = 0

The quadratic solution gives: vo = 76.2 ft/s

s1 = s2 s = 300 ft s = 0

vo

300

Chapter 7 Newton’s Second Law Physics, 6th Edition

65

Chapter 7. Newton’s Second Law

Newton’s Second Law 7-1. A 4-kg mass is acted on by a resultant force of (a) 4 N, (b) 8 N, and (c) 12 N. What are the

resulting accelerations?

(a) a = =4N4 kg

1 m/s2 (b) a = =8N4 kg

2 m/s2 (c) a = =12N4 kg

3 m/s2

7-2. A constant force of 20 N acts on a mass of (a) 2 kg, (b) 4 kg, and (c) 6 kg. What are the

resulting accelerations?

(a) a = =20N2 kg

10 m/s2 (b) a = =20N4 kg

5 m/s2 (c) a = =20N6 kg

3.33 m/s2

7-3. A constant force of 60 lb acts on each of three objects, producing accelerations of 4, 8, and

12 N. What are the masses?

m = =60 lb

4 ft / s2 15 slugs m = =60 lb

8 ft / s2 7.5 slugs m = =60 lb

12 ft / s2 5 slugs

7-4. What resultant force is necessary to give a 4-kg hammer an acceleration of 6 m/s2?

F = ma = (4 kg)(6 m/s2); F = 24 N

7-5. It is determined that a resultant force of 60 N will give a wagon an acceleration of 10 m/s2.

What force is required to give the wagon an acceleration of only 2 m/s2?

m = =60 6 N

10 m / s slugs2 ; F = ma = (6 slugs)(2 m/s2); F = 12 N

7-6. A 1000-kg car moving north at 100 km/h brakes to a stop in 50 m. What are the magnitude

and direction of the force? Convert to SI units: 100 km/h = 27.8 m/s

22

0 27 82

2 22 2 2

as v v av v

saf o

f o= − =−

=−

=; ( ) ( .(50 m)

; m / s) 7.72 m / s2

2

F = ma = (1000 kg)(7.72 m/s2); F = 772 N, South.


66

The Relationship Between Weight and Mass

7-7. What is the weight of a 4.8 kg mailbox? What is the mass of a 40-N tank?

W = (4.8 kg)(9.8 m/s2) = 47.0 N ; m =40 N

9.8 m / s2 = 4.08 kg

7-8. What is the mass of a 60-lb child? What is the weight of a 7-slug man?

m =60 lb

32 ft / s2 = 1.88 slugs ; W = (7 slugs)(32 ft/s2) = 224 lb

7-9. A woman weighs 180 lb on earth. When she walks on the moon, she weighs only 30 lb.

What is the acceleration due to gravity on the moon and what is her mass on the moon? On

the Earth?

Her mass is the same on the moon as it is on the earth, so we first find the constant mass:

me = =180 5625 lb

32 ft / s slugs;2 . mm = me = 5.62 slugs ;

Wm = mmgm gm =30 lb

5.625 slugs; gm = 5.33 ft/s2

7-10. What is the weight of a 70-kg astronaut on the surface of the earth. Compare the resultant

force required to give him or her an acceleration of 4 m/s2 on the earth with the resultant

force required to give the same acceleration in space where gravity is negligible?

On earth: W = (70 kg)(9.8 m/s2) = 686 N ; FR = (70 kg)(4 m/s2) = 280 N

Anywhere: FR = 280 N The mass doesn’t change.

7-11. Find the mass and the weight of a body if a resultant force of 16 N will give it an

acceleration of 5 m/s2.

m =16

50 N

m / s2. = 3.20 kg ; W = (3.20 kg)(9.8 m/s2) = 31.4 N


67

7-12. Find the mass and weight of a body if a resultant force of 200 lb causes its speed to

increase from 20 ft/s to 60 ft/s in a time of 5 s.

a m= = =60 ft / s - 20 ft / s

5 s ft / s lb

8 ft / s2

28 200; = 25.0 slugs

W = mg = (25.0 slugs)(32 ft/s2); W = 800 lb

7-13. Find the mass and weight of a body if a resultant force of 400 N causes it to decrease its

velocity by 4 m/s in 3 s.

a vt

a= =−

= −∆ 4

3133 m / s

s m / s2; . ; m =

−−

400133

N m / s2.

; m = 300 kg

W = mg = (300 kg)(9.8 m/s2); W = 2940 N

Applications for Single-Body Problems:

7-14. What horizontal pull is required to drag a 6-kg sled with an acceleration of 4 m/s2 if a

friction force of 20 N opposes the motion?

P – 20 N = (6 kg)(4 m/s2); P = 44.0 N

7-15. A 2500-lb automobile is speeding at 55 mi/h. What resultant force is required to stop the

car in 200 ft on a level road. What must be the coefficient of kinetic friction?

We first find the mass and then the acceleration. (55 mi/h = 80.7 m/s)

m as v vf= = = −2500 2 2

02 lb

32 ft / s 78.1 slugs; Now recall that: 2

ft / s)2(200 ft)

and - 16.3 m / s2

2av v

saf=

−=

−=

202

20 7( ) (80. ;

F = ma = (78.1 slugs)(-16.3 ft/s2); F = -1270 lb

k1270 lb; ;2500 lbk kF Nµ µ= = µk = 0.508

6 kg 20 N P


68

7-16. A 10-kg mass is lifted upward by a light cable. What is the tension in the cable if the

acceleration is (a) zero, (b) 6 m/s2 upward, and (c) 6 m/s2 downward?

Note that up is positive and that W = (10 kg)(9.8 m/s2) = 98 N.

(a) T – 98 N = (10 kg)(0 m/s2) and T = 98 N

(b) T – 98 N = (10 kg)(6 m/s2) and T = 60 N + 98 N or T = 158 N

(c) T – 98 N = (10 kg)(-6 m/s2) and T = - 60 N + 98 N or T = 38.0 N

7-17. A 64-lb load hangs at the end of a rope. Find the acceleration of the load if the tension in

the cable is (a) 64 lb, (b) 40 lb, and (c) 96 lb.

(a) 2

64 lb; 64 lb 64 lb =32 ft/s

WT W a ag

− = −

; a = 0

(b) 2

64 lb; 40 lb 64 lb =32 ft/s

WT W a ag

− = −

; a = -12.0 ft/s2

(b) 2

64 lb; 96 lb 64 lb =32 ft/s

WT W a ag

− = −

; a = 16.0 ft/s2

7-18. An 800-kg elevator is lifted vertically by a strong rope. Find the acceleration of the

elevator if the rope tension is (a) 9000 N, (b) 7840 N, and (c) 2000 N.

Newton’s law for the problem is: T – mg = ma (up is positive)

(a) 9000 N – (800 kg)(9.8 m/s2) = (800 kg)a ; a = 1.45 m/s2

(a) 7840 N – (800 kg)(9.8 m/s2) = (800 kg)a ; a = 0

(a) 2000 N – (800 kg)(9.8 m/s2) = (800 kg)a ; a = -7.30 m/s2

7-19. A horizontal force of 100 N pulls an 8-kg cabinet across a level floor. Find the

acceleration of the cabinet if µk = 0.2.

F = µkN = µk mg F = 0.2(8 kg)(9.8 m/s2) = 15.7 Ν

100 N – F = ma; 100 N – 15.7 N = (8 kg) a; a = 10.5 m/s2

W = mg

+ T

10 kg

W

+ T

m = W/g

+

m mg

T

mg

F 100 N N


69

7-20. In Fig. 7-10, an unknown mass slides down the 300 inclined plane.

What is the acceleration in the absence of friction?

ΣFx = max; mg sin 300 = ma ; a = g sin 300

a = (9.8 m/s2) sin 300 = 4.90 m/s2, down the plane

7-21. Assume that µk = 0.2 in Fig 7-10. What is the acceleration?

Why did you not need to know the mass of the block?

ΣFx = max; mg sin 300 - µkN = ma ; N = mg cos 300

mg sin 300 - µk mg cos 300 = ma ; a = g sin 300 - µk g cos 300

a = (9.8 m/s2)(0.5) – 0.2(9.8 m/s2)(0.866); a = 3.20 m/s2, down the plane.

*7-22. Assume that m = I 0 kg and µk = 0. 3 in Fig. 7- 10. What push P directed up and along

the incline in Fig.7-10 will produce an acceleration of 4 m/s2 also up the incline?

F = µkN = µkmg cos 300; F = 0.3(10 kg)(9.8 m/s2)cos 300 = 25.5 N

ΣFx = ma; P – F – mg sin 300 = ma

P – 25.5 N – (10 kg)(9.8 m/s2)(0.5) = (10 kg)(4 m/s2)

P – 25.5 N – 49.0 N = 40 N; P = 114 N

*7-23. What force P down the incline in Fig. 7-10 is required to cause the acceleration DOWN

the plane to be 4 m/s2? Assume that in = IO kg and µk = 0. 3.

See Prob. 7-22: F is up the plane now. P is down plane (+).

ΣFx = ma; P - F + mg sin 300 = ma ; Still, F = 25.5 N

P - 25.5 N + (10 kg)(9.8 m/s2)(0.5) = (10 kg)(4 m/s2)

P - 25.5 N + 49.0 N = 40 N; P = 16.5 N

+

+

mg

N

300

300

F

mg

N

300

300

P

F

mg

N

300

300

F

300 P

mg

N

300


70

Applications for Multi-Body Problems

7-24. Assume zero friction in Fig. 7-11. What is the acceleration of the system? What is the

tension T in the connecting cord?

Resultant force = total mass x acceleration

80 N = (2 kg + 6 kg)a; a = 10 m/s2

To find T, apply F = ma to 6-kg block only: 80 N – T = (6 kg)(10 m/s2); Τ = 20 Ν

7-25. What force does block A exert on block B in Fig, 7-12?

ΣF = mTa; 45 N = (15 kg) a; a = 3 m/s2

Force ON B = mB a = (5 kg)(3 m/s2); F = 15 N

*7-26. What are the acceleration of the system and the tension in the

connecting cord for the arrangement shown in Fig. 7-13?

Assume zero friction and draw free-body diagrams.

For total system: m2g = (m1 + m2)a (m1g is balanced by N)

a m gm m

=+

=2

1 2

6( ) kg)(9.8 m / s4 kg + 6 kg

2

; a = 5.88 m/s2 Now, to find T, consider only m1

ΣF = m1a T = m1a = (4 kg)(5.88 m/s2); Τ = 23.5 Ν

*7-27. If the coefficient of kinetic friction between the table and the 4 kg block is 0.2 in

Fig. 7-13, what is the acceleration of the system. What is the tension in the cord?

ΣFy = 0; N = m1g; F = µkN = µkm1g

For total system: m2g - µkm1g = (m1 + m2)a

a m g m gm m

k=−+

=−2 1

1 2

6µ ( ) ( kg)(9.8 m / s 0.2)(4 kg)(9.8 m / s )4 kg + 6 kg

2 2

T 6 kg

2 kg 80 N

45 N 5 kg 10 kg

A B

N

m1 g

m2 g

T

T +a

Fk N

m1 g

m2 g

T

T +a


71

*7-27. (Cont.) a =−58.8 N 7.84 N

10 kg or a = 5.10 m/s2

To find T, consider only m2 and make down positive:

ΣFy = m2a ; m2g – T = m2a; T = m2g – m2a

Τ = (6 kg)(9.8 m/s2) – (6 kg)(5.10 m/s2); T = 28.2 N

*7-28. Assume that the masses m1 = 2 kg and m2 = 8 kg are connected by a cord

that passes over a light frictionless pulley as in Fig. 7-14. What is the

acceleration of the system and the tension in the cord?

Resultant force = total mass of system x acceleration

m2g – m1g = (m1 + m2)a a m g m gm m

=−+

2 1

1 2

a =(8 kg)(9.8 m / s ) - (2 kg)(9.8 m / s

kg + 8 kg

2 2 )2

a = 5.88 m/s2 Now look at m1 alone:

T - m1g = m1 a; T = m1(g + a) = (2 kg)(9.8 m/s2 – 5.88 m/s2); T = 31.4 N

*7-29. The system described in Fig. 7-15 starts from rest. What is the

acceleration assuming zero friction? (assume motion down plane)

ΣFx = mT a; m1g sin 320 – m2g = (m1 + m2) a

(10 kg)(9.8 m/s2)sin 320 – (2 kg)(9.8 m/s2) = (10 kg + 2 kg)a

a =519. N - 19.6 N

12 kg a = 2.69 m/s2

*7-30. What is the acceleration in Fig. 7-15 as the 10-kg block moves down the plane against

friction (µk = 0.2). Add friction force F up plane in figure for previous problem.

m1g sin 320 – m2g – F = (m1 + m2) a ; ΣFy = 0 ; N = m1g cos 320

Fk N

m1 g

m2 g

T

T +a

+a

m2g m1g

T T

T

320

m1g

N

320 m2g

T

+a


72

*7-30. (Cont.) m1g sin 320 – m2g – F = (m1 + m2) a ; F = µkN = µk m1g cos 320

m1g sin 320 – m2g – µk m1g cos 320 = (m1 + m2) a ; a = 1.31 m/s2

*7-31 What is the tension in the cord for Problem 7-30? Apply F = ma to mass m2 only:

T – m2g = m2 a; T = m2(g + a) = (2 kg)(9.8 m/s2 + 1.31 m/s2); T = 22.2 N

Challenge Problems

7-32. A 2000-lb elevator is lifted vertically with an acceleration of 8 ft/s2.

Find the minimum breaking strength of the cable pulling the elevator?

ΣFy = ma; m Wg

m= = =2000 lb32 ft / s

62.5 slugs2 ;

T – mg = ma; T = m(g + a); T = (62.5 slugs)(32 ft/s2 + 8 ft/s2); T = 2500 lb

7-33. A 200-lb worker stands on weighing scales in the elevator of Problem 7-32.

What is the reading of the scales as he is lifted at 8 m/s?

The scale reading will be equal to the normal force N on worker.

N – mg = ma; N = m(g + a); T = (62.5 slugs)(32 ft/s2 + 8 ft/s2); T = 2500 lb

7-34. A 8-kg load is accelerated upward with a cord whose breaking strength is 200 N. What is

the maximum acceleration?

Tmax – mg = ma a T mgm

=−

=max )2008

N - (8 kg)(9.8 m / s kg

2

a = 15.2 m/s2

+a T

2000 lb

+a

200 lb

N

8 kg +a

T = 200 N

mg


73

7-35. For rubber tires on a concrete road µk = 0.7. What is the horizontal stopping distance for a

1600-kg truck traveling at 20 m/s? The stopping distance is determined by the

acceleration from a resultant friction force F = µkN, where N = mg:

F = -µkmg = ma; a = -µkg = - (0.7)(9.8 m/s2); a = -6.86m/s2

Recall that: 2as = vo2 - vf

2; sv v

af=

−=

−202

20 (20 m / s)2(-6.86 m / s )

2

2 ; s = 29.2 km

*7-36. Suppose the 4 and 6-kg masses in Fig. 7-13 are switched so that the larger mass is on the

table. What would be the acceleration and tension in the cord neglecting friction?

For total system: m2g = (m1 + m2); m1 = 6 kg; m2 = 4 kg

a m gm m

=+

=2

1 2

4( ) kg)(9.8 m / s6 kg + 4 kg

2

; a = 3.92 m/s2

ΣF = m1a T = m1a = (6 kg)(5.88 m/s2); Τ = 23.5 Ν

*7-37. Consider two masses A and B connected by a cord and hung over a single pulley. If

mass A is twice that of mass B, what is the acceleration of the system?

mA = 2mB ; If the left mass B is m, the right mass A will be 2m.

2mg – mg = (2m + m)a mg = 3ma

a g= =

39 8. m / s

3

2

a = 3.27 m/s2

*7-38. A 5-kg mass rests on a 340 inclined plane where µk = 0.2. What push

up the incline, will cause the block to accelerate at 4 m/s2?

F = µkN = µkmg cos 340; F = 0.2(5 kg)(9.8 m/s2)cos 340 = 8.12 N

ΣFx = ma; P – F – mg sin 340 = ma

P – 8.12 N – (5 kg)(9.8 m/s2) sin 340 = (5 kg)(4 m/s2) P = 47.4 N

N

m1 g

m2 g

T

T +a

B A 2m m

+a

2mg mg

T T

F 340

P

mg

N

340

+


74

Forc

e, N

02468

101214

0 1 2 3 4 5 6 7 8 9Acceleration, m/s2

∆F

∆a

*7-39. A 96-lb block rests on a table where µk = 0.4. A cord tied to this block passes over a

light frictionless pulley. What weight must be attached to the free end if the system is to

accelerated at 4 ft/s2?

F = µkN = 0.2 (96 lb); F = 19.2 lb

( )22

96 lb + WW - 19.2 lb= 4 ft/s32 ft/s

W – 19.2 lb = 12 lb + 0.125 W; W = 35.7 lb

Critical Thinking Questions

7-40. In a laboratory experiment, the acceleration of a small car is measured by the separation of

spots burned at regular intervals in a paraffin-coated tape. Larger and larger weights are

transferred from the car to a hanger at the end of a tape that passes over a light frictionless

pulley. In this manner, the mass of the entire system is kept constant. Since the car moves

on a horizontal air track with negligible friction, the resultant force is equal to the weights

at the end of the tape. The following data are recorded:

Weight, N 2 4 6 8 10 12

Acceleration, m/s2 1.4 2.9 4.1 5.6 7.1 8.4

Plot a graph of weight (force) versus acceleration. What is the significance of the slope of

this curve? What is the mass?

The slope is the change in Force over

the change in acceleration, which is the

mass of the system. Thus, the mass is

found to be: m = 1.42 kg

F N

96 lb

W

T

T +a


75

7-41. In the above experiment, a student places a constant weight of 4 N at the free end of the

tape. Several runs are made, increasing the mass of the car each time by adding weights.

What happens to the acceleration as the mass of the system is increased? What should the

value of the product of mass and acceleration be for each run? Is it necessary to include

the mass of the constant 4 N weight in these experiments?

The acceleration increases with increasing mass. According to Newton’s second law, the

product of the total mass of the system and the acceleration must always be equal to the

resultant force of 4 N for each run. It is necessary to add the mass of the 4-N weight to

each of the runs because it is part of the total mass of the system.

7-42. An arrangement similar to that described by Fig. 7-13 is set up except that the masses are

replaced. What is the acceleration of the system if the suspended

mass is three times that of the mass on the table and µk = 0.3.

ΣFy = 0; N = mg; F = µkN = µkmg

For total system: 3mg - µkmg = (3m + m)a ; (3 - µk)mg = 4 ma

a gk=−

=( )3

4µ (3 - 0.3)(9.8 m / s )

4

2

a = 6.62 m/s2

7-43. Three masses, 2 kg, 4 kg, and 6 kg, are connected (in order) by strings and

hung from the ceiling with another string so that the largest mass is in the lowest

position. What is the tension in each cord? If they are then detached from the

ceiling, what must be the tension in the top string in order that the system

accelerate upward at 4 m/s2? In the latter case what are the tensions in the

strings that connect masses?

Fk N

mg

3mg

T

T +a

C

B

A

4 kg

6 kg

2 kg


76

7-43. (Cont.) The tension in each string is due only to the weights BELOW the string. Thus,

TC = (6 kg)(9.8 m/s2) = 58.8 N ; TB = (6 kg + 4 kg)(9.8 m/s2) = 98.0 N ;

TA = (6 kg + 4 kg + 2 kg)(9.8 m/s2) = 118 N

Now consider the upward acceleration of 4 m/s2.

ΣFy = 0; TA = (2 kg + 4 kg + 6 kg)(4 m/s2); TA = 48 N

TB = (4 kg + 6 kg)(4 m/s2) = 40 N ; TC = (6 kg)(4 m/s2) = 24 N

7-44. An 80-kg astronaut on a space walk pushes against a 200-kg solar panel that has become

dislodged from a spacecraft. The force causes the panel to accelerate at 2 m/s2. What

acceleration does the astronaut receive? Do they continue to accelerate after the push?

The force on the solar panel Fp is equal and opposite that on the astronaut Fa.

Fp = mpap; Fa = maaa ; Thus, mpap = - maaa ; solve for aa:

am amap p

a

= − = −(200 kg)(2 m / s )

80 kg

2

; a = - 5 m/s2

Acceleration exists only while a force is applied, once the force is removed, both astronaut

and solar panel move in opposite directions at the speeds obtained when contact is broken..

7-45. A 400-lb sled slides down a hill (µk = 0.2) inclined at an angle of 600. What is the normal

force on the sled? What is the force of kinetic friction? What is the resultant force down

the hill? What is the acceleration? Is it necessary to know the weight of the sled to

determine its acceleration?

ΣFy = 0; N – W cos 600 = 0; N = (400 lb)cos 600 = 200 lb ;

F = µkN = (0.2)(200 lb); F = 40 lb

ΣFx = W sin 600 – F = (400 lb)sin 600 – 40 lb; FR = 306 lb

a = + 4 m/s2

C

B

A

4 kg

6 kg

2 kg

W = mg = 400 lb

+a

N F

600

600


77

7-45 (Cont.) Since FR = ma; we note that: W sin 600 - µkW = (W/g)a; Thus, the weight

divides out and it is not necessary for determining the resultant acceleration.

*7-46. Three masses, m1 = 10 kg, m2 = 8 kg, and m3 = 6 kg, are connected as shown in Fig. 7-

16. Neglecting friction, what is the acceleration of the system? What are the tensions in

the cord on the left and in the cord on the right? Would the acceleration be the same if

the middle mass m2 were removed?

Total mass of system = (10 + 8 +6) = 24 kg

Resultant Force on system = m1g – m3g

The normal force N balances m2g; ΣF = mTa

m1g – m3g = (m1 + m2 +m3)a ; (10 kg)(9.8 m/s2) – (6 kg)(9.8 m/s2) = (24 kg) a

(24 kg)a = 98.0 N – 58.8 N; a = 1.63 m/s2 ; The acceleration is not affected by

m2.

To find TA apply ΣF = m1a to 10-kg mass: m1g – TA = m1a ; TA = m1g – m1a

TA = m1(g – a) = (10 kg)(9.8 m/s2 − 1.63 m/s2); TA = 81.7 N

Now apply to 6-kg mass: TB – m3g = m3a; TB = m3g + m3a

TB = (6 kg)(9.8 m/s2 + 1.63 m/s2) ; TB = 68.6 N

*7-47. Assume that µk = 0.3 between the mass m2 and the table in Fig. 7-16. The masses m2

and m3 are 8 and 6 kg, respectively. What mass m1 is

required to cause the system to accelerate to the

left at 2 m/s2? ( F = µkm2g acts to right. )

Apply ΣF = mTa to total system, left is positive.

N

10 kg 6 kg

+ +

TB TA

TB TA

m3g m1g

m2g 8 kg

N

6 kg

+ +

TB TA

TB TA

m3g m1g

m2g 8 kg

F


78

m1g – F – m3g = (m1 + m2 +m3)a ; F = µkm2g = 0.3(8 kg)(9.8 m/s2); F = 23.5 N

m1(9.8 m/s2) – 23.5 N - (6 kg)(9.8 m/s2) = (m1 + 14 kg)(2 m/s2)

9.8 m1 – 23.5 kg – 58.8 kg = 2m1 + 28 kg ; m1 = 14.1 kg

*7-48. A block of unknown mass is given a push up a 400 inclined plane and then released. It

continues to move up the plane (+) at an acceleration of –9 m/s2.

What is the coefficient of kinetic friction?

Since block is moving up plane, F is directed down plane.

F = µkN ; ΣFy = 0; N = mg cos 400; F = µkmg cos 400

ΣFx = ma; -F - mg sin 400 = ma; -µkmg cos 400 - mg sin 400 = ma

a = -µkg cos 400 - g sin 400 ; -9 m/s2 = -µk(9.8 m/s2) cos 400 - (9.8 m/s2) sin 400

Solving for µk we obtain: µk = 0.360

*7-49. Block A in Fig. 7-17 has a weight of 64 lb. What must be the weight of block B if Block

A moves up the plane with an acceleration of 6 ft/s2. Neglect friction.

ΣFx = ma: WB – WA sin 600 = (mA + mB) a

sin 60 A BB A

W WW W ag

+− =

;

2

2

6 ft/s 0.18832 ft/s

ag

= =

WB – (64 lb)(0.866) = 0.188(64 lb + WB)

WB – 55.4 lb = 12.0 lb + 0.188WB ; WB = 83.0 lb

*7-50. The mass of block B in Fig. 7-17 is 4 kg. What must be the mass of block A if it is to

move down the plane at an acceleration of 2 m/s2? Neglect friction.

ΣFx = ma: mAg sin 600 - mBg = (mA + mB) a

(9.8 m/s2)(0.866)mA – (4 kg)(9.8 m/s2) = mA(2 m/s2) + (4 kg)(2 m/s2)

+ F

F

mg

400 400

mBg

T

T 600

mAg

+a N

600

T

T

WB

600

WA = 64 lb

+a N

600


79

8.49 mA – 39.2 kg = 2 mA + 8 kg; mA = 7.28 kg.

*7-51. Assume that the masses A and B in Fig. 7-17 are 4 kg and 10 kg, respectively. The

coefficient of kinetic friction is 0.3. Find the acceleration if (a) the system is initially

moving up the plane, and (b) if the system is initially moving down the plane?

(a) With upward initial motion, F is down the plane.

F = µkN ; ΣFy = 0; N = mAg cos 600; F = µkmAg cos 600

Resultant force on entire system = total mass x acceleration

mBg – mAg sin 600 - µkmAg cos 600 = (mA + mB)a

(10 kg)(9.8 m/s2) – (4 kg)(9.8 m/s2)(0.866) – 0.3(10 kg)(9.8 m/s2)(0.5) = (14 kg)a

98 N – 33.9 N – 14.7 N = (14 kg)a; or a = 3.53 m/s2

(b) If initial motion is down the plane, then F is up the plane, but the resultant force is

still down the plane. The block will side until it stops and then goes the other way.

ΣFx = ma; mBg – mAg sin 600 + µkmAg cos 600 = (mA + mB)a

98 N – 33.9 N + 14.7 N = (14 kg)a

a = 5.63 m/s2 The greater acceleration results from

the fact that the friction force is increasing the resultant force

instead of decreasing it as was the case in part (a).

F

mBg

T

T 600

mAg

+a N

600

F

mBg

T

T 600

mAg

+a N

600

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