PRACTICE PROBLEMS

INDYNAMICS

SUBMITTED BY:JOSEPH F. LEGASPI

SUBMITTED TO:ENGR. JORGE YASAY

1) A 90 kg woman stands in an elevator. Find the force which the floor of the elevator exerts on the woman a) when the elevator has an upward acceleration of 2 m/sec2; b) when the elevator is rising at constant speed; c) when the elevator has a downward acceleration of 2 m/sec2.

Solution: We have a simple application of Newton's 2nd Law of motion. We need only draw the vector force diagram for all forces acting on the woman. There are only two forces present, these are:

Fearth on woman = W = mg .

Ffloor on woman = N .

We select a CS with 'y' chosen upwards. The basic equation involved with the problem is the 2nd law:

Fnet = m a .

a). ay 0; Fy = N - mg = m ay N = m(g + a) = (90)(9.8 + 2) = 1062 N.

b). a 0; Fy = N - mg = m ay N = m(g + a) = (90)(9.8 - 2) = 702 N.

c) a = 0 (equilibrium) Fy = N - mg = 0 N = mg = (90)(9.8) = 882 N.

2) A Boeing 707 jet aircraft has a takeoff mass of 1.2 x 105 kg. Each of its four engines has a net thrust of 75 kN. Calculate the acceleration and the length of the runway needed to become airborne if the takeoff speed is 73 m/s. (Neglect any frictional forces and air-resistance)

Solution: Neglecting air-resistance & frictional forces then the net force acting is simply 4 times the force produced by each engine. From the 2nd law we then have:

Fnet = m ax Fx = (4)(75 x 103) = (1.2 x 105) ax .

Hence the acceleration is: a x = (300 kN)/(120 k-kg) = 2.5 m/sec2

The equations of motion for constant acceleration (x-direction) are:

x(t) = (1/2) ax t2 + vox t + xo ; vx(t) = ax t + vox .

The initial conditions are: x(t=0) = xo = 0; vx (t=0) = vox = 0. Thus the specific equations of motion are:

x(t) = (1/2)(2.5) t2 ; vx (t) = (2.5) t .05-2

If liftoff occurs at a time t', then we have: vx (t') = 73 = (2.5)t' t' = 29.2 sec.

The position of the plane at liftoff (= runway length) is: x(t') = (1/2)(2.5)(29.2)2 = 1065.8 m .

3) Two carts are in contact on a horizontal frictionless surface. Cart 'A' has a mass of 3 kg, and cart 'B' a mass of 2 kg. (a) Find the force F on 'A' needed to give the two carts an acceleration of 0.8 m/sec2 . What is then the force exerted on 'B' by 'A'? (b) If the carts are given an equal acceleration in the opposite direction by pushing on B, find the force exerted on 'A' by 'B'. Explain why the force exerted on 'B' by 'A' in part (a) is not equal to the force exerted by 'A' on 'B' in part (b).

Solution: In handling forces we should always be able to identify both the agent and object for each force. If we do so, then the 'reaction' force can be identified simply by reversing the 'agent' and 'object'.

Consider 'A' by itself. We draw all forces acting on 'A', and select a CS as shown. Newton's 2nd law for object 'A' then becomes:

Fx = F - F B on A = mA a x (1) ; Fy = NA - WA = 0

The 'y' equation can be solved for NA ( = 29.4 N) which is numerically equal to the weight of 'A'. The 'x' equation cannot be solved since we have two 'unknowns' F and F B on A.

This is to be expected. Since we have two objects involved in the problem, then we will usually have to consider both before a solution is possible.

Consider 'B' by itself. We draw all forces acting on 'B', and select a CS as shown. Newton's 2nd law for object 'B' then becomes:

Fx = FA on B = mB a x (2)

Fy = NB - WB = 0

Again we can immediately determine that NB = WB = 19.6 N. If we count up the 'unknown' quantities in equations (1) & (2) we find we have 3: F, FA on B, FB on A. The 3rd 'missing relation' is, of course Newton's 3rd Law which tells us that: FA on B = - FB on A.

From equation (2) we have: FA on B = mB ax = (2)(.8) = 1.6 N .

Then equation (1) gives: F - FB on A = mA a x or F = FB on A + mA a x = 1.6 + (3)(.8) = 4 N.

05-3

Note that this answer is obtainable by taking a 'systems' approach. If we treat the 2 objects as comprising a 'system' of mass 5 kg, then the only force acting in the x-direction is F, and we have:

(on system) Fx = M a x = (5)(.8) = 4 N.

(b) We now reverse the problem by applying a force F' on B. In this case there are 2 horizontal forces on B, and only 1 on A. Hence, the 2nd law for object A is now:

Fx = F'B on A = mA a x = (3)(.8) = 2.4 N.

The 2nd law applied to 'B' then gives:

Fx = F' - F'A on B = mB a x .

Solving for F' we have: F' = F'A on B + mB a x = 2.4 + (2)(.8) = 4 N.

Thus while the applied force must be the same since the system acquires the same magnitude acceleration, the forces between the objects are different in the two cases.

4) The three bodies A, B, and C in the figure have masses of 6, 3, and 2 kg respectively. If they are given a vertical acceleration upwards of 2 m/sec2 , find the tensions in the ropes.

05-4

Solution: If the only quantity desired was the tension T1, then we could take a 'systems' approach and immediately calculate T1. As a system the only 'external' forces acting are T1 and the pull of the earth on the system (the total weight). Hence:

Fy = T1 - W = M a y T1 = M(g + a y) = (11)(11.8) = 129.8 N.

In this approach the other tensions do not appear since they are 'internal' to the system. Since we are asked to calculate these other tensions, then we must make them 'external'.

We do this by treating each object separately. We draw the forces acting on each of the 3 objects:

We choose the same CS for all 3 so that the accelerations of all 3 will be the same. Applying Newton's 2nd law to each gives us the following set of equations:

For A: F y = T1 - T2 - = a y (1)

For B: F y = T2 - T3 - = a y (2)

For A: F y = T3 - = a y (3)

Known quantities have been 'boxed'. We recognize that we have 3 equations in 3 unknowns, and hence are guaranteed solutions for the 3 tensions. Equation (3) can be solved immediately for T3 (ans. 23.6 N). This is then inserted into equation (2). This permits determination of T2 (ans. 59 N). In turn this is substituted into equation (1) yielding T1 (ans. 129.8 N).

Note the relationship of the 3 equations for the individual objects above and the equation for the 'system' which we introduced first. If we add the 3 equations above, we find that the tensions T1 and T2 cancel giving us:

T1 - (mA + mB + mC) g = (mA + mB + mC) a y .

This is the 2nd law for the system.

5) An Atwood's machine consists of masses of 2 and 2.5 kg suspended from a massless, frictionless pulley. When the system is released from rest, how long does it take for the larger mass to descend 1.6 m? What is the tension in the cord connecting the two masses during the descent?

05-5

Solution: We note that since an answer for the tension in the connecting rope is requested, the 'systems' approach is insufficient to solve the complete problem. However, we will take this approach to solve the first part of the problem.

Fig. 1 illustrates the 'system' which has a total mass of 4.5 kg. As a system the only 'external' forces present are the weight of A, and the weight of B, (as shown). The system accelerates in a 'counterclockwise sense', and the 2nd law gives us:

Fccw = WA - WB = M a x .

Hence, the acceleration of the system is: a x = {(2.5)(9.8) - (2)(9.8)}/(2 + 2.5) = 1.09 m/sec2 .

The equation of motion for 'A' would be: x(t) = (1/2) a x t2 . Thus, if it descends 1.6 m in a time t', we have:

mA g mA

mC g

mB g mB

mC

x(t') = 1.6 = (1/2)(1.09) t'2 t' = 1.71 seconds.

As noted above, we cannot find the tension since the tension is 'internal' to the system, and hence doesn't appear in the 2nd law. To have the tension appear we must consider the two masses individually.

The force diagram for A & B are drawn & coordinate systems are chosen. (Note: 'x' is chosen upward for B so that the acceleration of B will be the same as that for A.) Applying Newton's Law to each object, we have:

For A: Fx = WA - T = mA a x (1)

For B: Fx = T - WB = mB a x (2)

We have a set of 2 equations in 2 unknowns (a & T). Solving, we find:

a x = 1.09 m/sec2 ; T = 21.8 N .

Note that the equation for the system is obtained if equations (1) & (2) above are added together.

6) A 5 kg block 'A' on a rough plane (angle of inclination = 30o) is connected to a massless string which passes over a frictionless pulley at the top of the plane, and is connected to a 4 kg block 'B' which hangs vertically. If the coefficient of kinetic friction between 'A' and the plane is 0.25, find the acceleration of the system when released, and the tension in the string.

05-6

Solution: The figure is shown. Since all forces present are constants, then we know that the net force must be constant and hence we have a constant acceleration problem. The only question is, if it accelerates, in which direction does it go?

We will assume that 'A' slides down the plane. If this assumption is wrong, then our answer will indicate this. When we solve for 'a', if a > 0, then we guessed right! If a < 0, then 'A' slides up the plane.

We draw the force diagram for 'A' choosing a CS with 'x' drawn down the plane. We apply Newton's 2nd Law (in component form) to obtain:

For A: Fx = - sin 30 + T + fk = a x (1)

Fy = N - cos 30 = 0 (2)

Since we have kinetic friction present then we have a 3rd relation for 'A': fk = N (3)

Known quantities are 'boxed.' We see we have a total of 4 unknowns (a, T, fk, N). While we can solve for some (e.g., N = 42.4 N, fk = 10.6 N) solution for all 4 is impossible at this stage. Hence, we consider object 'B', drawing the force diagram & choosing a CS with 'x' downwards.

Note, we choose 'x' down here so that the numerical value of the acceleration of 'B' will be identical to that for 'A'. (If we had chosen 'x' up, then aB = - aA ).

For B: Fx = - T + = a x (4)

We now have 4 equations in 4 unknowns and can proceed with numerical solution. We add equ (4) and equ (1) to eliminate T, and we have:

- mA g sin 30 + f + mB g = (mA + mB) a x or

- (5)(9.8)(.5) + 10.6 + (4)(9.8) = (9) a x a x = + 2.81 m/sec2.

The positive answer indicates that we guessed wrong! 'A' slides up the plane, not down. Note that the acceleration up the plane is not 2.81 m/sec2! If 'A' slides up the plane, then the frictional force on 'A' must be down the plane. This means that we must change the sign of 'f' in equation (1). Then when we add to equ. (4) we have:

- mA g sin 30 - f + mB g = (mA + mB) a x or

- (5)(9.8)(.5) - 10.6 + (4)(9.8) = (9) a x a x = + 0.455 m/sec2.

Substituting this back into equation (4) gives: T = 37.4 N .05-7

7) The driver of a 500 kg car, heading directly for a RR crossing 300 m away, applies the brakes in a panic stop. The car is initially moving at 40 m/sec and the brakes are capable of producing a force of 1200 N. (a) How fast will the car be moving when it reaches the crossing? (b) Will the driver escape collision with a freight train that at the instant the brakes are applied is blocking the road, and still requires 11 seconds to clear the crossing?

mA g m

mA g

mB g m

Solution: We have a 1-d motion problem with constant acceleration. To find the acceleration we apply Newton's 2nd law to the car. The force diagram is drawn, and CS chosen. We also include the initial conditions in the figure since the 2nd part of the problem will utilize the equations of motion of the car.

Fx = - F = m a x -1200 = (500) a x

Hence, the car's acceleration (x-direction) is: a x = - 2.4 m/sec2 . Note that while we are not asked, we can calculate the normal force N (4900 N), and assuming that the 1200 N force is due to kinetic friction between the tires and the road, we could also find the coefficient of kinetic friction (0.245).

Inserting initial conditions & acceleration into the general equations of motion for constant acceleration, we arrive at the specific equations for the car:

x(t) = (1/2)(-2.4) t2 + (40) t ; v x (t) = - 2.4 t + 40 .

To determine the time to reach the crossing, we solve x(t=t') = 300 m.

x(t') = 300 = - 1.2 t'2 + 40 t' .

This is a quadratic equation & yields two solutions for t': t'1 = 11.4 sec; t'2 = 21.9 sec. (Can you explain the meaning of the second time?)

v(t=11.4s) = -(2.4)(11.4) + (40) = 12.6 m/sec.

Clearly the driver escapes injury since his car arrives at the crossing 0.4 seconds after the train has cleared the crossing.

8) In the figure assume you are given values for:

mA, mB, mC, , B, C .

You are told that 'A' accelerates downward. Apply Newton's 2nd Law to each object in the system in order to develop a set of equations which could be solved for the acceleration of the system. (Identify your 'knowns' & 'unknowns'.)

05-8

We must construct a force diagram for each of the objects in the problem.

For A: Force diagram & CS as shown:

Fx = - T1 = a x (1)

For B: Force diagram & CS as shown:

Fx = T1 - T2 - sin - fB = a x (2)

Fy = NB - cos = 0 (3)

fB = NB (4)

For C: Force diagram & CS as shown:

Fx = T2 - fC = a x (5)

Fy = NC - = 0 (6)

fC = NC (7)

Note that we have selected the 'x' direction for all three objects in the same 'sense'. This means that we need only one acceleration 'a' in our equations.

Known quantities are 'boxed'. The unknowns are: T1, a, T2, fB, NB, fC, NC.

Thus we have 7 equations in 7 unknowns and are guaranteed a numerical solution for all 7.

9)

mA g

mB g

mC g

m

m

m

C

B

mB g

12 - 1

A 200-lb block rests on a horizontal plane. Find the magnitude of the force P required to give the block an accelera-tion or 10 ft/s2 to the right. The coef-ficient of kinetic friction between the block and plane is k0.25.

SOLUTION:Resolve the equation of motion

for the block into two rectangular component equations.

Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns.

12 - 1

N

NF

g

Wm

k

25.0

ft

slb21.6

sft2.32

lb200

2

2

x

y

O

SOLUTION:Resolve the equation of motion for the block

into two rectangular component equations.

:maFx

lb1.62

sft10ftslb21.625.030cos 22

NP

:0 yF

0lb20030sin PN

Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns.

lb1.62lb20030sin25.030cos

lb20030sin

PP

PN

lb151P

10)

12 - 1

The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord.

SOLUTION:Write the kinematic relationships

for the dependent motions and accelerations of the blocks.

Write the equations of motion for the blocks and pulley.

Combine the kinematic relationships with the equations of motion to solve for the accelerations and cord tension.

12 - 1

Write equations of motion for blocks and pulley.

:AAx amF AaT kg1001

:BBy amF

B

B

BBB

aT

aT

amTgm

kg300-N2940

kg300sm81.9kg300

2

22

2

:0 CCy amF

02 12 TT

SOLUTION:Write the kinematic relationships for the

dependent motions and accelerations of the blocks.

ABAB aaxy21

21

x

y

O

12 - 1

N16802

N840kg100

sm20.4

sm40.8

12

1

221

2

TT

aT

aa

a

A

AB

A

Combine kinematic relationships with equations of motion to solve for accelerations and cord tension.

ABAB aaxy21

21

AaT kg1001

A

B

a

aT

21

2

kg300-N2940

kg300-N2940

0kg1002kg150N2940

02 12

AA aa

TT

x

y

O

11)

12 - 1

The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface. Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge.

SOLUTION:The block is constrained to slide

down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plus the acceleration of the block relative to the wedge.

Write the equations of motion for the wedge and block.

Solve for the accelerations.

12 - 1

SOLUTION:The block is constrained to slide down the

wedge. Therefore, their motions are dependent.

ABAB aaa

Write equations of motion for wedge and block.

x

y

:AAx amF

AA

AA

agWN

amN

1

1

5.0

30sin

:30cos ABABxBx aamamF

30sin30cos

30cos30sin

gaa

aagWW

AAB

ABABB

:30sin AByBy amamF

30sin30cos1 ABB agWWN

12 - 1

AA agWN 15.0

Solve for the accelerations.

30sinlb12lb302

30coslb12sft2.32

30sin2

30cos

30sin30cos2

30sin30cos

2

1

A

BA

BA

ABBAA

ABB

a

WW

gWa

agWWagW

agWWN

2sft07.5Aa

30sinsft2.3230cossft07.5

30sin30cos

22AB

AAB

a

gaa

2sft5.20ABa

12)

12 - 1

The bob of a 2-m pendulum describes an arc of a circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and accel-eration of the bob in that position.

SOLUTION:Resolve the equation of motion

for the bob into tangential and normal components.

Solve the component equations for the normal and tangential accelerations.

Solve for the velocity in terms of the normal acceleration.

12 - 1

SOLUTION:Resolve the equation of motion for the bob

into tangential and normal components.

Solve the component equations for the normal and tangential accelerations.

:tt maF

30sin

30sin

ga

mamg

t

t

2sm9.4ta

:nn maF

30cos5.2

30cos5.2

ga

mamgmg

n

n

2sm03.16na

Solve for velocity in terms of normal acceleration.

22

sm03.16m2 nn avv

a

sm66.5v

12 - 1

SOLUTION:Resolve the equation of motion for the bob

into tangential and normal components.

Solve the component equations for the normal and tangential accelerations.

:tt maF

30sin

30sin

ga

mamg

t

t

2sm9.4ta

:nn maF

30cos5.2

30cos5.2

ga

mamgmg

n

n

2sm03.16na

Solve for velocity in terms of normal acceleration.

22

sm03.16m2 nn avv

a

sm66.5v

14) The maximum speed with which a 945-kg car makes a 180-degree turn is 10.0 m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction and the coefficient of friction acting upon the car.

m = 945 kg

v = 10.0 m/s

R = 25.0 m

Ffrict = ???

μ = ????

(μ - coefficient of friction)

The mass of the object can be used to determine the force of gravity acting in the downward direction. Use the equation

Fgrav = m * g

where g is 9.8 m/s/s. Knowing that there is no vertical acceleration of the car, it can be concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9261 N. This allows us to determine two of the three forces identified in the free-body diagram. Only the friction force remains unknown.

13)

12 - 1

SOLUTION:The car travels in a horizontal

circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface.

Resolve the equation of motion for the car into vertical and normal components.

:0 yF

cos

0cos

WR

WR

:nn maF

2sin

cos

sin

v

g

WW

ag

WR n

Solve for the vehicle speed.

18tanft400sft2.32

tan2

2 gv

hmi1.44sft7.64 v

Since the force of friction is the only horizontal force, it must be equal to the net force acting upon the object. So if the net force can be determined, then the friction force is known. To determine the net force, the mass and the kinematic information (speed and radius) must be substituted into the following equation:

Substituting the given values yields a net force of 3780 Newton. Thus, the force of friction is 3780 N.

Finally the coefficient of friction (μ) can be determined using the equation that relates the coefficient of friction to the force of friction and the normal force.

Substituting 3780 N for Ffrict and 9261 N for Fnorm yields a coefficient of friction of 0.408.

15) The coefficient of friction acting upon a 945-kg car is 0.850. The car is making a 180-degree turn around a curve with a radius of 35.0 m. Determine the maximum speed with which the car can make the turn.

m = 945 kg

μ = 0.85 (coefficient of friction)

R = 35.0 m

v = ???

(the minimum speed would be the speed achieved with the given friction coefficient)

The mass of the car can be used to determine the force of gravity acting in the downward direction. Use the equation

Fgrav = m * g

where g is 9.8 m/s/s. Knowing that there is no vertical acceleration of the car, it can be concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9261 N. Since the coefficient of friction (μ) is given, the force of friction can be determined using the following equation:

This allows us to determine all three forces identified in the free-body diagram.

The net force acting upon any object is the vector sum of all individual forces acting upon that object. So if all individual force values are known (as is the case here), the net force can be calculated. The vertical forces add to 0 N. Since the force of friction is the only horizontal force, it must be equal to the net force acting upon the object. Thus, Fnet = 7872 N.

Once the net force is determined, the acceleration can be quickly calculated using the following equation.

Fnet = m*a

Substituting the given values yields an acceleration of 8.33 m/s/s. Finally, the speed at which the car could travel around the turn can be calculated using the equation for centripetal acceleration:

Substituting the known values for a and R into this equation and solving algebraically yields a maximum speed of 17.1 m/s.

 

SOLUTION:

• Unknown

A 200-lb

12 - 24