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## 2ND_Ed Test Bank Physics for The Life Sciences – Zinke-Allmang

2ND_Ed Test Bank Physics for The Life Sciences – Zinke-Allmang

**Sample**

CHAPTER 6—TORQUE AND EQUILIBRIUM

MULTIPLE CHOICE

1. A weigh-scale measures the weight of an object by balancing its weight against that of a fixed,

“reference” weight sliding along a scale until equilibrium is reached. If the weight to be

measured is suspended at twice its distance from the fulcrum, which of the following would

not restore equilibrium?

a. doubling the reference weight

b. doubling the distance of reference weight from fulcrum

c. doubling the reference weight and its distance from fulcrum

d. replacing the reference weight by two of half its mass, one at the original position

and the other 3 times further from the fulcrum

ANS: C

RAT: Sums of torques due to weights on each side of fulcrum are equal at equilibrium.

PTS: 1

REF: p. 131

BLM: Higher Order

2. A cyclist goes round a curve of radius 10 m at a speed of 10 km/hr. At what angle to the

horizontal must the cyclist lean to ensure that the net force of the ground on the person passes

through his centre of gravity?

a. 15

b. 20

c. 75

d. 85

ANS: D

RAT: The forces acting on the cyclist are a vertical normal force mg due to the weight of the

cyclist and bicycle, and a horizontal centrifugal force mv

2

/r due to motion in a circle. The

resultant of these must be directed from the point of contact on the ground to the centre of

mass of the person + bicycle system, at angle to the horizontal.

PTS: 1

REF: p. 131

BLM: Higher Order

3. A 20 kg child is sitting on a teeter-totter, 2 m from the pivot. What force is required, when

applied 0.7 m from the pivot, in order to lift the child?

a. 29 N

b. 57 N

c. 280 N

d. 560 N

ANS: D

RAT: This is a Class I lever, with load of 20g N and effort is the unknown quantity.

PTS: 1

REF: p. 131 | p. 141

BLM: Higher Order

6-2 Copyright © 2013 Nelson Education Limited

4. How much force must a man of shoulder-height 1.8 m and mass 70 kg exert on both hands in

order to do a push-up, assuming that his body’s centre of mass is 1.0 m from his feet?

a. 38.9 N

b. 77.8N

c. 381.1 N

d. 762.1 N

ANS: C

RAT: This is a Class II lever system, with load of 70g N, and effort is the unknown quantity.

PTS: 1

REF: p. 131 | p. 141-142

BLM: Higher Order

5. A woman is holding a 5.0kg suitcase at arm’s length, a distance of 0.4 m from her shoulder.

What is the torque on the shoulder joint if her arm is horizontal?

a. 1.0 Nm

b. 2.0 Nm

c. 19.6 Nm

d. 196.0 Nm

ANS: B

RAT: Torque is given by Equation 6.1, in which = 90

, and F = mg, with m = 5 kg.

PTS: 1

REF: p. 132

BLM: Higher Order

6. A man is holding a 5.0kg suitcase at arm’s length, a distance of 0.4 m from his shoulder.

What is the torque on the shoulder joint if his arm is 30

below the horizontal?

a. 0.9 N

b. 1.7 N

c. 17.0 N

d. 34.0 N

ANS: C

RAT: Apply Equation 6.1 with = 90

30

= 60

.

PTS: 1

REF: p. 132

BLM: Higher Order

Copyright © 2013 Nelson Education Limited 6-3

Figure 6.1

A uniform bar, with fulcrum (pivot) at its centre, with oblique forces acting at each end.

7. A uniform bar with pivot (fulcrum) at its centre is in equilibrium, as shown in Figure 6.1.

What is the condition for this equilibrium?

a. F2sin = F1cos

b. F1cos = F2cos

c. F1sin = F2sin

d. F1sin = F2cos

ANS: D

RAT: Torque = component of force perpendicular to bar multiplied by distance to pivot.

PTS: 1

REF: p. 134

BLM: Higher Order

6-4 Copyright © 2013 Nelson Education Limited

Figure 6.1

Two forces acting on a bar with fulcrum at the centre of the bar.

8. In Figure 6.1, two forces F1 and F2 act at the ends of a bar supported at its centre. The

magnitude of F2 is half that of F1, and = 45

. What is the value of angle for which the bar

is in equilibrium?

a. 16

b. 21

c. 30

d. 45

ANS: B

RAT: Balance the torques, given by Equations 6.3 or 6.4, using Equation 6.7.

PTS: 1

REF: p. 134-136

BLM: Higher Order

9. The drive chain on a bicycle applies a torque of 0.85 Nm to the wheel of the bicycle. What

force must be applied tangentially at the rim of the wheel, of diameter 0.7 m, to stop it

accelerating?

a. 1.21 N, radially along the wheel

b. 1.21 N, tangentially to the wheel

c. 2.43 N, radially along the wheel

d. 2.43 N, tangentially to the wheel

ANS: D

RAT: If there is no acceleration of the wheel, then the torque due to the drive chain must be

equal and opposite to that applied to the rim of the wheel. Only the force component

perpendicular to the radius vector from the pivot to the point of application contributes to this

latter torque.

PTS: 1

REF: p. 134 | p. 138

BLM: Higher Order

Copyright © 2013 Nelson Education Limited 6-5

10. A car travels around a corner at high speed. Why is it more likely to roll when there is a

weight strapped to the roof?

a. There is more area for a side-wind to exert force.

b. The centrifugal force causes a sideways force on the car.

c. The centrifugal force is larger than the weight of the car.

d. The centre of gravity is higher off the ground, causing the centrifugal force to exert

larger torque on the car.

ANS: D

RAT: The net force on the car no longer passes through its centre of mass. The centrifugal

force in acts horizontally, and since the added mass is located above the normal centre of mass

of the car, thereby raising its centre of gravity and increasing the leverage.

PTS: 1

REF: p. 135

BLM: Remember

11. A truck is carrying a refrigerator that is 2.0 m tall and 0.8 m wide. The centre of mass of the

refrigerator is at its geometric centre. It is facing sideways and a short strip on the bed of the

truck keeps the refrigerator from sliding. What is the maximum acceleration that the truck can

have before the refrigerator begins to tip over?

a. 2.0 m/s2

b. 3.9 m/s2

c. 4.9 m/s2

d. 9.8 m/s2

ANS: B

RAT: Force acting at the centre of mass has a vertical component due to gravity and a

horizontal component due to the acceleration of the truck. In order to have stability, the line of

the resultant force through the centre of mass must pass on the near side of the line of contact

of the bottom of the fridge with the floor.

PTS: 1

REF: p. 136

BLM: Higher Order

6-6 Copyright © 2013 Nelson Education Limited

12. A refrigerator 2.0 m tall and 0.8 m wide has its centre of mass at its geometric centre. You are

attempting to slide it along the floor by pushing horizontally on the side of the refrigerator.

The coefficient of static friction between the floor and the refrigerator is 0.4. Depending on

where you push, the refrigerator may start to tip over before it starts to slide along the floor.

What is the highest distance above the floor that you can push the refrigerator so that it won’t

tip before it begins to slide?

a. 0.4 m

b. 1.0 m

c. 1.2 m

d. 2.0 m

ANS: B

RAT: The force applied to move the fridge must be at least 0.4 times its weight. At the point

of tipping, with this force applied, the torque about the far edge of the fridge where it touches

the floor, due to pushing at height h must balance the torque due to weight of refrigerator

pointing down from its centre of mass.

PTS: 1

REF: p. 136 | p. 138

BLM: Higher Order

Copyright © 2013 Nelson Education Limited 6-7

Figure 6.2

Force diagram with masses at each end of a pivoted bar of length (L+1) m.

13. In Figure 6.2, the normal force N is 20 N and masses mA and mB, where mB = 2mA, hang in

equilibrium at the ends of a uniform bar of mass mbar = mA/2 and length (L+1) m, exerting

loads wA and wB, respectively. What are the values of mbar and L that give equilibrium?

a. mbar = 0.29 kg, L = 2/7 m

b. mbar = 0.29 kg, L = 5/7 m

c. mbar = 2.86 kg, L = 2/7 m

d. mbar = 2.86 kg, L = 5/7 m

ANS: B

RAT: Write down the vertical force and torque balance equations, as in Equations 6.11.

Substitute for mB and mbar in terms of mA using the given relations, and note that wA = gmA.

This gives two linear equations in two unknowns mA and L, from which mbar = mA/2 can be

found.

PTS: 1

REF: p. 138

BLM: Higher Order

14. A person of mass 70 kg stands 0.5 m from one end of a 2 m long uniform plank (of negligible

mass) that is being held in equilibrium by two vertical ropes attached to the ends of the plank.

What are the tensions in the rope nearer to the person?

a. 23 N

b. 47 N

c. 229 N

d. 457 N

ANS: D

RAT: Set up torque and vertical force equations from Equations 6.11 in the two unknown

tensions. Solve this linear system for the two tensions: the rope closer to the person will have

the larger tension.

PTS: 1

REF: p. 138

BLM: Higher Order

6-8 Copyright © 2013 Nelson Education Limited

15. A man with upper body mass of 40 kg bends forward at the hip at an angle of 60

from the

vertical, keeping his legs in a vertical position and his back straight. The effective pivot point

is the pelvic cavity, located 6 cm from the line of the back muscles used to maintain this

bending position. The centre of mass of the upper body is located 60 cm from the pelvic pivot.

What is the force exerted by the back muscles?

a. 40 N

b. 200 N

c. 392 N

d. 1960 N

ANS: D

RAT: Required only to balance torques, which are (6 cm) (force exerted by back muscles)

and (60 cm) (upper body weight)

PTS: 1

REF: p. 138

BLM: Higher Order

16. A 1-kg mass and a 0.7kg mass are attached to the two ends of a string that goes over a pulley

with a radius of 0.1 m. Because of friction, the pulley does not begin to rotate. What is the

magnitude of the frictional torque on the bearing of the pulley if the system is in static

equilibrium?

a. 0.03 Nm

b. 0.29 Nm

c. 0.58 Nm

d. 2.90 Nm

ANS: B

RAT: For equilibrium, the net torque must be zero. The components of the torque are due to

the two weights and to friction in the bearing which opposes the torque due to the difference

in the weights hung from the string

PTS: 1

REF: p. 138

BLM: Higher Order

17. A 75kg diver stands at the edge of a light 6m diving board, which is supported by two

pillars 2 m apart. Find the force exerted by the pillar further from the diver.

a. 735 N downwards

b. 735 N upwards

c. 1470 N downwards

d. 1470 N upwards

ANS: C

RAT: Select the inner pillar as pivot, then balance torques due to weight of diver and force

exerted by the pillar further from the diver.

PTS: 1

REF: p. 138

BLM: Higher Order

Copyright © 2013 Nelson Education Limited 6-9

18. A 75kg diver stands at the edge of a light 6m diving board, which is supported by two

pillars 2 m apart. Find the force exerted by the pillar nearer to the diver.

a. 735 N downwards

b. 735 N upwards

c. 1102 N upwards

d. 2205 N upwards

ANS: D

RAT: Select the outer pillar (further from the diver) as pivot, then balance torques due to

weight of diver and force exerted by the pillar nearer to the diver.

PTS: 1

REF: p. 138

BLM: Higher Order

19. A car drives over a weigh-scale. When the two front wheels are on the scale, the weight

measured is 6000 N. When the rear two wheels are on the scale, the weight measured is 7500

N. How far from the front wheels is the car’s centre of mass, if the wheels are 3.5 m apart?

a. 0.7 m

b. 1.56 m

c. 1.94 m

d. 2.8 m

ANS: C

RAT: Balance torques due to front and rear wheels about the unknown position of centre of

mass, and solve for this unknown.

PTS: 1

REF: p. 138

BLM: Higher Order

20. A 6.0 kg mass is held in the hand, with the upper arm vertical and the forearm bent 30

below

the horizontal. The pivot is at the elbow, and the bicep muscle acts in a vertical line 5 cm in

front of the pivot. The forearm is 40 cm long, and has a mass of 2.5 kg, and may be assumed

to have its centre of mass in the middle. How much force must a bicep muscle exert to

maintain this configuration?

a. 58 N

b. 116 N

c. 284.2 N

d. 568.4 N

ANS: D

RAT: The force F due to the bicep acts between the pivot and the load (Class III lever). It

exerts a torque of F x (5 cm) about the pivot. This must balance the sum of the torques due to

the ball of weight = 6g N located at (40 cos 30

) cm from the pivot, and to the weight of the

forearm (2.5g N) located at (20 cos 30

) cm from the pivot.

PTS: 1

REF: p. 138 | p. 141

BLM: Higher Order

6-10 Copyright © 2013 Nelson Education Limited

TRUE/FALSE

1. If the net force on an object is zero, the net torque on the object must be zero.

ANS: F

RAT: If, for example, two equal and opposite forces act along different lines through the

body, there will be a net torque, even though the net force is zero.

PTS: 1

REF: p. 131

BLM: Remember

2. When two uniform spherical balls of equal mass but different diameters undergo a head-on

collision, the rotation speeds of the balls change.

ANS: F

RAT: The impulsive force acting during collision acts along a line passing through the centres

of mass of both balls, and hence cannot impart torque to either ball. Therefore, their rotation is

unaffected by the collision.

PTS: 1

REF: p. 132

BLM: Higher Order

3. All forces acting on a body contribute to the torque on the body.

ANS: F

RAT: Those forces acting along a line through the pivot do not contribute to the torque.

PTS: 1

REF: p. 132 | p. 134

BLM: Remember

4. The magnitude of the torque exerted by a force of magnitude F is equal to the perpendicular

distance from the axis of rotation multiplied by F. This is equivalent to saying it is equal

to the distance from the axis of rotation r multiplied by the magnitude of the perpendicular

component of the force, .

ANS: T

RAT: The torque is given by Equation 6.1, which may be written as either r(Fsin) or

(rsin)F, equivalent to rF or rF, respectively.

PTS: 1

REF: p. 134 | p. 134-135

BLM: Remember

Copyright © 2013 Nelson Education Limited 6-11

5. If a uniform rectangular box resting on the ground is pushed, depending on the height at

which it is pushed, it may either slide or tip.

ANS: T

RAT: The force applied to move the box must exceed the frictional force proportional to its

weight. At the point of tipping, with this force applied, the torque about the far edge of the

fridge where it touches the floor, due to pushing at height h must balance the torque due to

weight of refrigerator pointing down from its centre of mass.

PTS: 1

REF: p. 136 | p. 138

BLM: Higher Order

6. For equilibrium of an extended body, the vector sum of all forces must be zero.

ANS: T

RAT: This is one condition, the other being the vector sum of torques is zero.

PTS: 1

REF: p. 138

BLM: Remember

7. When two (circular) curling rocks collide, they will not begin to rotate if they collide head-on.

ANS: T

RAT: Head-on collision means that the impulsive force of contact is directed towards the

centres of mass of the two rocks.

PTS: 1

REF: p. 138

BLM: Remember

8. If the torque on a body is zero, then the vector sum of all forces is zero.

ANS: F

RAT: Torque depends on where on the body the forces are applied, and balancing torques

does not necessarily imply balancing forces.

PTS: 1

REF: p. 138

BLM: Remember

9. A uniform thickness equilateral triangular plate is placed on a pivot passing through the centre

of area of the triangle. The triangle is in equilibrium.

ANS: T

RAT: The symmetry of the equilateral triangle implies that the centre of area lies at the

common intersection of the perpendiculars from each vertex to the corresponding opposite

sides. Since the plate is uniform this is also the centre of mass of the triangle.

PTS: 1

REF: p. 138

BLM: Higher Order

6-12 Copyright © 2013 Nelson Education Limited

10. When lifting a weight by the hand, the force exerted by the muscle in the forearm is less than

the weight being lifted.

ANS: F

RAT: This is a Class III lever system, with the effort closer to the fulcrum than the load.

Therefore, to balance torque, the effort must exceed the load.

PTS: 1

REF: p. 142

BLM: Remember

ESSAY

1. Explain why it is safer to lift a heavy weight by first bringing it closer to your feet before

lifting.

ANS:

The torque exerted on your body is greater the further away the weight is located from the

fulcrum, which in this case are your feet. This torque can cause damage to the spine if the

compression force on it, caused by the lifting, is not directed along the length of the (straight)

spine. Attempting to lift a weight located far from the body requires bending the spine as well

as a large torque due to forces acting perpendicular to the spine.

RAT: Torque is proportional to distance from the fulcrum.

PTS: 1

REF: p. 131

BLM: Higher Order

2. A hockey player swings his stick to hit a stationary puck. Describe the torques exerted on the

player by this action.

ANS:

Use Newton’s Third Law: The impulsive force of the stick on the ball exerts an equal and

opposite impulsive force on the stick, giving rise to an impulsive torque on the stick which is

communicated to the player. Due to the low coefficient of friction of ice, this will impart

rotation to the player.

RAT: Torque is force times distance of pivot from line of action of force.

PTS: 1

REF: p. 132

BLM: Higher Order

Copyright © 2013 Nelson Education Limited 6-13

3. A uniform bar has different masses attached at each end, and rests on ice. It is pushed in the

middle at right angles to its length. Assuming no friction with the ice, describe the subsequent

motion of this object.

ANS:

The centre of mass of the object will lie to the side of the greater mass, and hence pushing it

will exert a torque on the object, causing it to rotate about its centre of mass. As we are

neglecting friction with the ice, the centre of mass will have constant velocity and rotation

speed will be constant.

RAT: The initial push imparts linear and angular momentum to object. After this, both linear

and angular momentum is conserved.

PTS: 1

REF: p. 132

BLM: Higher Order

4. The brakes on a car are applied by brake shoes pressing down radially onto a circular brake

drum. Describe the motion of the car as the pressure on the brake shoes increases with time.

ANS:

The forward motion of the car is proportional to the rotation speed of the wheels. Applying

the brakes means the tangential friction force on the brake drum, which is proportional to the

radially applied force of the brake shoes, will exert an increasing torque on the wheels,

causing the car to decelerate at an increasing rate.

RAT: Slipping frictional force is proportional to applied normal force, and torque is

proportional to the tangential frictional force.

PTS: 1

REF: p. 132

BLM: Higher Order

5. Suppose you stir a cup of coffee, then place it at rest on an almost frictionless surface, such as

ice. Describe the subsequent motion of the cup of coffee.

ANS:

The rotating fluid (coffee) exerts a tangential force (“wall friction”) on the walls of the cup,

giving rise to a torque about the vertical axis of the cup through the centre of mass. This will

cause an accelerated rotation of the cup, until the cup is rotating at the same rate as the fluid

contents, when the wall friction ceases. Negligible momentum is assumed to be exchanged

with the surface on which the cup rests.

RAT: Friction forces give rise to torques, causing a change in rotational motion.

PTS: 1

REF: p. 132

BLM: Higher Order

6-14 Copyright © 2013 Nelson Education Limited

6. Explain why placing the heaviest items in a bookcase on the lowest shelf makes it more stable

against tipping over than if the heaviest books were on higher shelves.

ANS:

Putting the heaviest books nearest the floor ensures the centre of mass of the bookshelf is as

low as possible, and allows a greater angle of tipping along the edge in contact with the floor

before the centre of mass is located vertically above this edge, after which the torque it exerts

reverses, the torques all act in the same direction, and the bookcase tips over.

RAT: Balance of torques requires some to be clockwise and others counter-clockwise.

PTS: 1

REF: p. 136 | p. 138

BLM: Higher Order

7. If a uniform rectangular box resting on the ground is pushed, depending on the height at

which it is pushed it may either slide or tip. Discuss what conditions determine whether

sliding or tipping occurs.

ANS:

The force applied to move the box must exceed the static frictional force, determined by the

coefficient of friction. To prevent tipping when this force is applied, the torque about the far

edge of the box, where it touches the floor, due to pushing on its side at a certain height h

must balance the torque due to weight of box pointing down from its centre of mass. Clearly,

this depends on h relative to the height of the box. Since all forces are proportional to the

weight of the box, their directions are independent of the weight of the box, so whether

tipping or sliding occurs does not depend on the weight.

RAT: Find critical conditions for slipping and balancing of torques

PTS: 1

REF: p. 136 | p. 138

BLM: Higher Order

8. Why must the two pillars supporting a diving board not be too close together?

ANS:

Place the pivot at the rear pillar. Then the distance between the diver and the nearer pillar

must not be so large as to require a force acting upwards on this pillar, creating a

counteracting torque to balance that due to the diver, to be so great that it breaks the diving

board.

RAT: Balance torque about one of the pillars and due to the torque exerted by the diver and

by the force on the other pillar.

PTS: 1

REF: p. 138

BLM: Higher Order

Copyright © 2013 Nelson Education Limited 6-15

9. Explain why, when a car turns a corner, the car dips down on the outside of the curve.

ANS:

The frictional force between the road and the tires provides the centripetal acceleration, and

acts below the centre of mass of the car. For a car turning to the right, this torque is

counter-clockwise as seen by the driver. The normal forces on the left and right tires are

redistributed to produce a clockwise torque to counteract the torque produced by friction

between the tires and the road. This requires an increase in the normal force on the left side

and a decrease in the normal force on the right side. As a result, the car dips down on the left

side, which is on the outside of the curve.

RAT: The total torque about the axis of the car and through its centre of mass, must balance

to prevent the car from rolling over.

PTS: 1

REF: p. 138

BLM: Higher Order

10. Explain why, when resting a ladder against a wall, it should not be at a large angle from the

vertical.

ANS:

The coefficient of static friction for the ladder in contact with the floor sets an upper limit to

the angle at which ladder may make with the wall, in order to balance torques and forces.

RAT: Balance forces and torques with pivot at the point of contact of ladder with the wall (or

floor).

PTS: 1

REF: p. 138

BLM: Higher Order

6-16 Copyright © 2013 Nelson Education Limited

2ND_Ed Test Bank Physics for The Life Sciences – Zinke-Allmang

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