Content check — College Physics (AP), Chapter 9: Statics and Torque

9-1 §9.2 source ↗

(a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges?

Answer

a) $46.8 N·m$

b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force’s application from the hinges. (Children don’t have a tougher time opening a door because they push lower than adults, they have a tougher time because they don’t push far enough from the hinges.)

9-3 §9.2 source ↗

Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.

Answer

23.3 N

9-5 §9.2 source ↗

Repeat the seesaw problem in Example 9.1 with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium.

Example 9.1 View it in the book ↗

Answer

Given:

\begin{matrix} m_{1} & = & 26.0 kg, m_{2} = 32.0 kg, m_{s} = 12.0 kg, \\ r_{1} & = & 1.60 m, r_{s} = 0.160 m, find (a) r_{2,} (b) F_{p} \end{matrix}

a) Since children are balancing:

\begin{matrix} net τ_{cw} = - net τ_{ccw} \\ \Rightarrow w_{1} r_{1} + m_{s} {gr}_{s} = w_{2} r_{2} \end{matrix}

So, solving for $r_{2}$ gives:

\begin{array}{l} r_{2} & = & \frac{w_{1} r_{1} + m_{s} {gr}_{s}}{w_{2}} = \frac{m_{1} {gr}_{1} + m_{s} {gr}_{s}}{m_{2} g} = \frac{m_{1} r_{1} + m_{s} r_{s}}{m_{2}} \\ = & \frac{(26.0 kg) (1.60 m) + (12.0 kg) (0.160 m)}{32.0 kg} \\ = & 1.36 m \end{array}

b) Since the children are not moving:

\begin{matrix} net F = 0 = F_{p} - w_{1} - w_{2} - w_{s} \\ \Rightarrow F_{p} = w_{1} + w_{2} + w_{s} \end{matrix}

So that

\begin{array}{l} F_{p} & = & (26.0 kg + 32.0 kg + 12.0 kg) (9.80 m / s^{2}) \\ = & 686 N \end{array}

9-6 §9.3 source ↗

Suppose a horse leans against a wall as in Figure 9.29, emulating a scene in the comedy movie Cat Ballou. Calculate the force exerted on the wall assuming that force is horizontal while using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 500 kg. Take the data to be accurate to three digits.

In part a, a horse is standing next to a wall with its legs crossed. A sleepy-looking rider is leaning against the wall. Part b is a drawing of the same horse from a rear view, but this time with no rider. The horse is crossing its rear legs, and its rump is leaning against the wall. The reaction of the wall F is acting on the horse at a height one point two meters above the ground. The weight of the horse is acting at its center of gravity near the base of the tail. The center of gravity is one point four meters above the ground. The line of action of weight is zero point three five meters away from the feet of the horse. — Figure 9.29 in the book ↗

Answer

F_{wall} =

1.43 \times 10^{3} N

9-8 §9.3 source ↗

(a) Calculate the magnitude and direction of the force on each foot of the horse in Figure 9.29 (two are on the ground), assuming the center of mass of the horse is midway between the feet. The total mass of the horse and rider is 500kg. (b) What is the minimum coefficient of friction between the hooves and ground? Note that the force exerted by the wall is horizontal.

Answer

a) $2.55 {×10}^{3} N, 16.3º to the left of vertical (i.e., toward the wall)$

b) 0.292

9-10 §9.3 source ↗

A 17.0-m-high and 11.0-m-long wall under construction and its bracing are shown in Figure 9.30. The wall is in stable equilibrium without the bracing but can pivot at its base. Calculate the force exerted by each of the 10 braces if a strong wind exerts a horizontal force of 650 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall.

A seventeen meter high wall is standing on the ground with ten braces to support it. At the base of the figure a brown colored ground is visible. Only one brace is visible from a side. A brace makes an angle of thirty five degree with the wall. The point of contact of the brace is eight point five meters high. You have to find the force exerted by this brace on the wall to support. — Figure 9.30 in the book ↗

Answer

F_{B} =

2.12 \times 10^{4} N

9-12 §9.3 source ↗

Suppose the weight of the drawbridge in Figure 9.32 is supported entirely by its hinges and the opposite shore, so that its cables are slack. The mass of the bridge is 2500 kg. (a) What fraction of the weight is supported by the opposite shore if the point of support is directly beneath the cable attachments? (b) What is the direction and magnitude of the force the hinges exert on the bridge under these circumstances?

A small drawbridge is shown. There is one vertical and one horizontal wooden plank. The left end of the horizontal plank is attached to the vertical plank near its middle. At the point of contact, a hinge is shown. A wire is tied to the right end of the horizontal end, is passed over the top of the vertical plank and is connected to a pulley. The angle made by the wire with the horizontal plank is forty degrees. The reaction F at the hinge is inclined at an angle theta. — Figure 9.32 in the book ↗

Answer

a) 0.167, or about one-sixth of the weight is supported by the opposite shore.

b) $F = 2 . 0 \times 10^{4} N$ , straight up.

9-14 §9.3 source ↗

A sandwich board advertising sign is constructed as shown in Figure 9.33. The sign’s mass is 8.00 kg. (a) Calculate the tension in the chain assuming no friction between the legs and the sidewalk. (b) What force is exerted by each side on the hinge?

A sandwich board advertising sign is in form of a triangle. The base of the triangle is one point one zero meters. The other two sides are connected with a hinge at the top. A horizontal chain is connected to the two legs at zero point five zero meters below the hinge. The height of the hinge above the base is one point three zero meters. The centers of the gravity of the two legs are shown at their midpoints. The figure is labeled at uniform board with c g at the center. — Figure 9.33 in the book ↗

Answer

a) 21.6 N

b) 21.6 N

9-16 §9.3 source ↗

An athlete is attempting to perform splits. From the information given in Figure 9.34, calculate the magnitude and direction of the force exerted on each foot by the floor.

A gymnast with two pompoms in her hands is shown. One of the hand is horizontal toward left and the other is vertical. The gymnast is attempting to perform a full split. The span of her legs is one point eight meters, and the distance of one foot from the center of gravity is zero point nine meters. The weight of the girl is labeled as seven hundred newtons. The vertical distance of one foot from the center of gravity is zero point three zero meter. — Figure 9.34 in the book ↗

Answer

350 N directly upwards

9-19 §9.5 source ↗

What is the mechanical advantage of a nail puller—similar to the one shown in Figure 9.21 —where you exert a force

45 cm

from the pivot and the nail is

1.8 cm

on the other side? What minimum force must you exert to apply a force of

1250 N

to the nail?

There is a nail in a wooden plank. A nail puller is being used to pull the nail out of the plank. A hand is applying force F sub I downward on the handle of the nail puller. The top of the nail exerts a force F sub N downward on the puller. At the point where the nail puller touches the plank, the reaction of the surface force N is applied. At the top of the figure, a free body diagram is shown. — Figure 9.21 in the book ↗

Answer

25

50 N

9-21 §9.5 source ↗

a) What is the mechanical advantage of a wheelbarrow, such as the one in Figure 9.22, if the center of gravity of the wheelbarrow and its load has a perpendicular lever arm of 5.50 cm, while the hands have a perpendicular lever arm of 1.02 m? (b) What upward force should you exert to support the wheelbarrow and its load if their combined mass is 55.0 kg? (c) What force does the wheel exert on the ground?

A wheelbarrow is shown in which the input force F sub I is shown as a vector in vertically upward direction below the handle of wheelbarrow. The weight of the wheelbarrow is downward at the center of gravity. The normal reaction of the ground is acting at the wheel in upward direction. The perpendicular distance between the normal reaction and the input force F sub I is labeled as R sub I and the distance between output force F sub O and normal reaction is labeled as R sub O. In figure b, a man is holding a shovel in his hands. One hand is at one end of the handle and the other hand is holding the shovel at the middle. The center of gravity of the shovel is at its flat end. The weight of the shovel is acting at the center of gravity. The input force is acting at the hand in the middle in upward direction and the end of the shovel is acting as pivot. A free body diagram is also shown at the right side of the figure. — Figure 9.22 in the book ↗

Answer

a) $MA = 18 . 5$

b) $F_{i} = 29.1 N$

c) 510 N downward

9-23 §9.5 source ↗

What force does the nail puller in Exercise 9.19 exert on the supporting surface? The nail puller has a mass of 2.10 kg.

Answer

1 . 3 \times 10^{3} N

9-25 §9.5 source ↗

Repeat Exercise 9.24 for the pulley shown in Figure 9.24(c), assuming you pull straight up on the rope. The pulley system’s mass is

7.00 kg

.

In figure a, a rope over two pulleys is shown. One pulley is fixed at the roof and the other is hanging through the rope. A weight is hanging from the second pulley. The tensions T are shown at the two parts of hanging pulley and at the free end of the rope. The mechanical advantage of the system is two. In figure b, a set of three pulleys is shown. A pulley is fixed at the roof with another pulley below it. The third pulley is hanging through the rope with a weight hanging at it. The tensions on the rope are shown as vectors on the rope and at the end of the rope. In figure c, set of three pulleys is shown. One of the pulleys is fixed at the roof. Two connected pulleys are hanging through a rope over the first pulley. The directions of the tensions are marked on the ropes and at the end of the rope. — Figure 9.24 in the book ↗

Answer

a) $T = 299 N$

b) 897 N upward

9-26 §9.6 source ↗

Verify that the force in the elbow joint in Example 9.4 is 407 N, as stated in the text.

Example 9.4 Muscles Exert Bigger Forces Than You Might Think in the book ↗

Answer

\begin{array}{l} F_{B} & = & 470 N; r_{1} = 4.00 cm; w_{a} = 2.50 kg; r_{2} = 16.0 cm; w_{b} = 4.00 kg; r_{3} = 38.0 cm \\ F_{E} & = & w_{a} \frac{r_{2}}{r_{1}} - 1 + w_{b} \frac{r_{3}}{r_{1}} - 1 \\ = & 2.50 kg 9.80 m / s^{2} \frac{16.0 cm}{4.0 cm} - 1 \\ + 4.00 kg 9.80 m / s^{2} \frac{38.0 cm}{4.00 cm} - 1 \\ = & 407 N \end{array}

9-28 §9.6 source ↗

The upper leg muscle (quadriceps) exerts a force of 1250 N, which is carried by a tendon over the kneecap (the patella) at the angles shown in Figure 9.36. Find the direction and magnitude of the force exerted by the kneecap on the upper leg bone (the femur).

The figure shows a side view of the bones of a knee and the quadriceps muscle. The upper bone is inclined at fifty five degrees to the horizontal and the tension exerted by the quadriceps muscle is one thousand two hundred and fifty newtons. The tendon from the knee cap to the lower bone is inclined at seventy five degrees below the horizontal. The force in this direction is the same as that provided by the quadriceps. — Figure 9.36 in the book ↗

Answer

\begin{matrix} 1.1 \times 10^{3} N \\ θ = 190 º ccw from positive x axis \end{matrix}

9-30 §9.6 source ↗

A person working at a drafting board may hold her head as shown in Figure 9.38, requiring muscle action to support the head. The three major acting forces are shown. Calculate the direction and magnitude of the force supplied by the upper vertebrae

F_{V}

to hold the head stationary, assuming that this force acts along a line through the center of mass as do the weight and muscle force.

The head of a person working at a drafting board in relaxed position is shown. The inclination of the head is theta to the horizontal and the center of gravity is near the top of the head. The weight of the head is fifty newtons and is acting downward at the center of gravity. Three major forces are shown. The force exerted along the neck is sixty newtons. — Figure 9.38 in the book ↗

Answer

F_{V} =

97 N, θ =

59º

9-32 §9.6 source ↗

Even when the head is held erect, as in Figure 9.39, its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles at the back of the neck should therefore exert a force to keep the head erect. That is why your head falls forward when you fall asleep in the class. (a) Calculate the force exerted by these muscles using the information in the figure. (b) What is the force exerted by the pivot on the head?

An erect head is shown. The weight of the head is fifty newtons. The center of gravity of the head lies in front of its support. The perpendicular distance between the support and the weight of the head is two point five centimeters. Between these forces, there is a point where a vertical force vector is shown. This force is marked as F sub J. At the back of the head, five point zero centimeters behind the support point, is a downward vector labeled F sub m. — Figure 9.39 in the book ↗

Answer

(a) 25 N downward

(b) 75 N upward

9-33 §9.6 source ↗

A 75-kg man stands on his toes by exerting an upward force through the Achilles tendon, as in Figure 9.40. (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown—that force is representative of forces in the ankle joint.

A foot of a person is shown. The ankle is slightly above the ground. There is a force in F-A on the back part of ankle, which is in upward direction. The weight of the leg is downward. The normal reaction is acting at the front foot in upward direction. The perpendicular distance between the normal reaction and the force F-A is sixteen centimeters. There is a point between these two forces where a force F-P is shown, which acts as fulcrum of the simplified lever system. — Figure 9.40 in the book ↗

Answer

(a) $F_{A} =$ $2.21 \times 10^{3} N$ upward

(b) $F_{B} =$ $2.94 \times 10^{3} N$ downward

9-35 §9.6 source ↗

Unlike most of the other muscles in our bodies, the masseter muscle in the jaw, as illustrated in Figure 9.42, is attached relatively far from the joint, enabling large forces to be exerted by the back teeth. (a) Using the information in the figure, calculate the force exerted by the lower teeth on the bullet. (b) Calculate the force on the joint.

The masseter muscles of a jaw of a man are shown. The force F sub M is equal to two hundred newtons and is acting on the muscle in upward direction and the force F sub J is acting to the left end of the muscle downward. The span of the muscle at upper part is five centimeters. At the joint of jaw, the reaction force is downward. — Figure 9.42 in the book ↗

Answer

(a) $F_{teeth on bullet} =$ $1.2 \times 10^{2} N$ upward

(b) $F_{J} = 84 N$ downward

9-37 §9.6 source ↗

(a) What force should the woman in Figure 9.43 exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.75 cm, and she exerts force on the floor at a horizontal distance of 20.0 cm from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) How much work does she do if her center of mass rises 0.240 m? (d) What is her useful power output if she does 25 pushups in one minute?

A woman doing pushups is shown. The weight W of her body is acting at the middle point of the length of her body. Her palms are on the ground. The distance between the palm and the feet is one point five meters. The distance between the center of gravity and the feet is zero point nine meters. The normal reaction on her hands is acting upward. — Figure 9.43 in the book ↗

Answer

(a) 147 N downward

(b) 1680 N, 3.4 times her weight

(c) 118 J

(d) 49.0 W

9-39 §9.6 source ↗

Unreasonable Results

Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?

Answer

a) ${\bar{x}}_{2} = 2.33 m$

b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board.

c) The position of the first child must be shortened, i.e. brought closer to the pivot.

AP Test Prep 9.1 §9.2 source ↗

Which of the following is not an example of an object undergoing a torque?

a.
A car is rounding a bend at a constant speed.
b.
A merry-go-round increases from rest to a constant rotational speed.
c.
A pendulum swings back and forth.
d.
A bowling ball rolls down a bowling alley.

Answer

A. A car is rounding a bend at a constant speed.

AP Test Prep 9.3 §9.3 source ↗

Using the concept of torque, explain why a traffic cone placed on its base is in stable equilibrium, while a traffic cone placed on its tip is in unstable equilibrium.

Answer

Both objects are in equilibrium. However, they will respond differently if a force is applied to their sides. If the cone placed on its base is displaced to the side, its center of gravity will remain over its base and it will return to its original position. When the traffic cone placed on its tip is displaced to the side, its center of gravity will drift from its base, causing a torque that will accelerate it to the ground.

AP Test Prep 9.5 §9.4 source ↗

A group of students is stacking a set of identical books, each one overhanging the one below it by 1 inch. They would like to estimate how many books they could place on top of each other before the stack tipped. What information below would they need to know to make this calculation?

I. The mass of each book

II. The width of each book

III. The depth of each book

a.
I only
b.
I and II only
c.
I and III only
d.
II only
e.
I, II, and III

Three rectangles with thick blue borders on the bottom, left, and right side and a think black slightly indented line of the right indicate three books. There are two short black vertical lines going down on the left and right of the bottom book with a double-headed black arrow and the label Width below. On the left side of the bottom book are two short horizontal lines with a black vertical double-headed short arrow labeled Depth. The three books are stacked with the second book offset to the right and the third offset even more to the right on the second book. The edge of the second book is marked with a double-headed arrow marked 1” between a vertical line above the left edge of the bottom book and a vertical line above the left edge of the second book. — Figure 9.45 in the book ↗

Answer

D. II only

AP Test Prep 9.7 §9.4 source ↗

A bridge is supported by two piers located 20 meters apart. Both the left and right piers provide an upward force on the bridge, labeled F_L and F_R respectively.

(a) If a 1000 kg car comes to rest at a point 5 meters from the left pier, how much force will the bridge provide to the left and right piers?

(b) How will F_L and F_R change as the car drives to the right side of the bridge?

Answer

(a) F_L = 7350 N, F_R = 2450 N

(b) As the car moves to the right side of the bridge, F_L will decrease and F_R will increase. (At exactly halfway across the bridge, F_L and F_R will both be 4900 N.)

AP Test Prep 9.9 §9.5 source ↗

As a young student, you likely learned that simple machines are capable of increasing the ability to lift and move objects. Now, as an educated AP Physics student, you are aware that this capability is governed by the relationship between force and torque.

In the space below, explain why torque is integral to the increase in force created by a simple machine. You may use an example or diagram to assist in your explanation. Be sure to cite the mechanical advantage in your explanation as well.

Answer

The student should mention that the guiding principle behind simple machines is the second condition of equilibrium. Though the torque leaving a machine must be equivalent to torque entering a machine, the same requirement does not exist for forces. As a result, by decreasing the lever arm to the existing force, the size of the existing force will be increased. The mechanical advantage will be equivalent to the ratio of the forces exiting and entering the machine.

AP Test Prep 9.11 §9.6 source ↗

When you use your hand to raise a 20 lb dumbbell in a curling motion, the force on your bicep muscle is not equal to 20 lb.

(a) Compare the size of the force placed on your bicep muscle to the force of the 20 lb dumbbell lifted by your hand. Using the concept of torque, which force is greater and explain why the two forces are not identical.

(b) Does the force placed on your bicep muscle change as you curl the weight closer toward your body? (In other words, is the force on your muscle different when your forearm is 90° to your upper arm than when it is 45° to your upper arm?) Explain your answer using torque.

Answer

(a) The force placed on your bicep muscle will be greater than the force placed on the dumbbell. The bicep muscle is closer to your elbow than the downward force placed on your hand from the dumbbell. Because the elbow is the pivot point of the system, this results in a decreased lever arm for the bicep. As a result, the force on the bicep must be greater than that placed on the dumbbell. (How much greater? The ratio between the bicep and dumbbell forces is equal to the inverted ratio of their distances from the elbow. If the dumbbell is ten times further from the elbow than the bicep, the force on the bicep will be 200 pounds!)

(b) The force placed on your bicep muscle will decrease. As the forearm lifts the dumbbell, it will get closer to the elbow. As a result, the torque placed on the arm from the weight will decrease and the countering torque created by the bicep muscle will do so as well.