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Serway and Vuille - College Physics 11/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 33 / 66

Due : Monday, January 28, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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  • Instructions

    College Physics, 11th edition, by Raymond A. Serway and Chris Vuille, published by Cengage Learning helps students master physical concepts, improve their problem-solving skills, and enrich their understanding of the world around them. Serway/Vuille provides a consistent problem-solving strategy and an unparalleled array of worked examples to help students develop a true understanding of physics. The WebAssign component for this title engages students with immediate feedback, tutorials, videos, and an interactive eBook.

    Question 1 is an Active Example which reframes one of the examples from the Topic narrative as an interactive lesson. A capstone question provides an extension of the in-text Example to test student understanding.

    Question 2 is a Training Tutorial authored by Chris Vuille to assist students in understanding how to apply certain key concepts in a practical context.

    Questions 3 and 4 are Pre-Lecture Explorations which use interactive HTML objects along with conceptual and analytical questions to help promote a robust physical intuition.

    Question 5 is a problem with an optional Master It tutorial that walks students through a series of progressive steps aligned with the text’s problem-solving strategy.

    Question 6 is a complete Master It tutorial. Instructors have the option of assigning either an optional or full tutorial version of those problems that have Master It functionality.

    Questions 7 and 8 exemplify content new to the 11th edition with hint button functionality and detailed solutions that exactly match the version of the problem a student receives.

    Question 9 is an Interactive Video Vignette (IVV) question. IVVs encourage students to address their alternate conceptions outside of the classroom. They include online video analysis and interactive individual tutorials to address learning difficulties identified by PER (Physics Education Research).

    Question 10 includes a Watch It video solution to one algorithmic variation of the problem.

    Question 11 is a Conceptual Question (CQ). For the 11th edition 125 new Conceptual Questions were created to be more systematic and serve as a diagnostic for conceptual misalignment irrespective of mathematical ability.

    Also available to adopters is an additional collection of over 1,600 questions that feature feedback and tutorials, organized to match the table of contents of Serway and Vuille College Physics, 11th edition. This collection can be added from "Free Additional Content" when creating your WebAssign course at no additional cost.

    Questions 12 and 13 are from this exclusive additional collection. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

    The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

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1. 4/4 points  |  Previous Answers SerCP11 8.AE.002. My Notes
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EXAMPLE 8.2 The Swinging Door
(a) Top view of a door being pushed by a 300-N force. (b) The components of the 300-N force.
Goal Apply the more general definition of torque.

Problem   (a)   A man applies a force of F = 3.00 102 N at an angle of 60.0° to the door of Figure (a), 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.   (b)   Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?

Strategy Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum of these torques must be zero, a condition that can be used to find the wedge force.
SOLUTION
(a) Compute the torque due to the applied force exerted at 60.0°.
Substitute into the general torque equation.
τF = rFsin θ = (2.00 m)(3.00 ✕ 102 N) sin 60.0°
 = (2.00 m)(2.60 ✕ 102 N) = 5.20 102 N · m
(b) Calculate the force exerted by the wedge on the other side of the door.
Set the sum of the torques equal to zero.
τhinge + τwedge + τF = 0
The hinge force provides no torque because it acts at the axis (r = 0). The wedge force acts at an angle of −90.0°, opposite the upward 260 N component.
0 + Fwedge(1.50 m) sin (−90.0°) + 5.20 ✕ 102 N · m = 0
Fwedge = 347 N
LEARN MORE
Remarks Notice that the angle from the position vector to the wedge force is −90°. This is because, starting at the position vector, it's necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure (b) illustrates the fact that the component of the force perpendicular to the lever arm causes the torque.

Question To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob?
     Correct: Your answer is correct.

PRACTICE IT
Use the worked example above to help you solve this problem.
(a) A man applies a force of F = 3.00 102 N at an angle of 60.0° to a door, x = 2.50 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.
Correct: Your answer is correct. seenKey

650

 N · m

(b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?
Correct: Your answer is correct. seenKey

433

 N
EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!
A man ties one end of a strong rope 8.62 m long to the bumper of his truck, 0.587 m from the ground, and the other end to a vertical tree trunk at a height of 3.15 m. He uses the truck to create a tension of 8.42 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.
Correct: Your answer is correct. seenKey

2530

 N · m
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Tutorial 8.2: Applying the Rotational Second Law The rotational analog of Newton's second law is
τ
 = Iα,
where all torques
τ,
 moments of inertia
I,
 and angular accelerations α, are computed around a fixed axis. This equation states that the angular acceleration of an extended rigid object is proportional to the net torque acting on it. The moment of inertia
I
 is the rotational analog of translational mass.

For a collection of point particles, the moment of inertia is given by
I = 
mr2.
 For rigid objects of uniform composition, this table in the textbook lists moments of inertia about different axes.

Recall that the magnitude of torque generated by a force of magnitude
F
 is
τ = rF sin θ,
where, as before,
r
 is the length of the position vector,
F
 is the magnitude of the applied force, and θ is the angle between the vectors
r
 and
F.
 By convention, when an applied force causes an object to rotate counterclockwise, the torque on the object is positive; when the force causes the object to rotate clockwise, the torque on the object is negative.

The Guided Problem guide you through the following Problem-Solving Strategy.
•   Read the problem carefully at least once.
•   Draw a picture of the system, identify the object of interest, and indicate forces with arrows.
•   Label each force in the picture.
•   Draw a free-body diagram of the object of interest.
•   Apply the rotational analog of Newton's second law.
•   Solve for the desired unknown quantity and substitute the numbers.
Guided Problem
A small sphere of mass
m = 2.20 kg
 is attached to the end of a very light, rigid rod of length
L = 1.30 m
 as shown in the figure. The mass and weight of the rod are negligible, and the sphere is small enough to be treated as a point particle. The left end of the rod is attached to the top of the pivot but free to rotate without friction around an axis perpendicular to the plane of the figure, and the sphere is held so the rod is horizontal and at rest. At the instant the sphere is released, determine its angular acceleration about an axis perpendicular to the plane of the figure and passing through the pivot point.
Part 1 of 6
Read the problem carefully at least once.
Be sure to notice which quantities are known and which must be found. The known quantities are the sphere's mass and the length of the rod. The rod's mass and weight are negligible.

The unknown quantity to be determined is the angular acceleration of the sphere at the instant it is dropped.

Identify all of the forces acting on the rod and the sphere shown in the figure.
     Correct: Your answer is correct.
Correct.
Part 2 of 6
Draw a picture of the system, identify the object of interest, and indicate forces with arrows.
Together, the rod and the sphere form a system that is the object of interest. The figure below shows a picture of this system.
Label each force in the picture.
Each force in this figure is labeled in a way that brings to mind the associated physical quantity. The normal force
n
 is exerted by the pivot on the rod and
Fg
 is the gravity force acting on the sphere.
Part 3 of 6
Draw a free-body diagram of the object of interest.
Which of the following free-body diagrams for the sphere-rod system is correct?
     Correct: Your answer is correct.
Correct.
Part 4 of 6
Draw a free-body diagram of the object of interest. (cont.)
The correct FBD that describes all of the forces acting on the sphere-rod system is shown below.
Apply the rotational analog of Newton's second law.
Applying the rotational analog of Newton's second law means taking
τ
 = Iα,
 determining the axis of rotation, and writing out each term in the equation.

Consider the FBD above for the sphere-rod system. Which of the following is the correct substitution for the net torque
τ
 acting on the system? Remember that the rod is assumed to be massless. All torques are computed about an axis perpendicular to the screen and passing through the left end of the rod (the pivot). Positive torques tend to produce a counterclockwise rotation about this axis and negative torques tend to produce a clockwise rotation.
     Correct: Your answer is correct.
Correct.
Part 5 of 6
Apply the rotational analog of Newton's second law. (cont.)
The normal force
n
 produces no torque about the pivot because the normal force acts directly on the pivot so that
r = 0.
 The gravity force acting on the sphere exerts a torque of
τFg = −rF sin θ.
 The distance from the pivot to the sphere's center where the gravity force acts is
r = L = 1.30 m,
 the magnitude of the force
F = Fg,
 and
θ = 90°
 so that
τFg = −LFg.
 This torque is negative because the gravity force acting on the sphere tends to make the system rotate clockwise around the pivot. The net torque acting on the system is then
τ
 = −LFg.

Consider the FBD and the discussion above of the sphere-rod system. Which of the following is the correct substitution for the moment of inertia
I
 about the chosen axis?
     Correct: Your answer is correct.
Correct.
Part 6 of 6
Apply the rotational analog of Newton's second law. (cont.)
Collecting these results, we have
τ
 = −FgL
 and the sphere's moment of inertia
I
 about the chosen axis is
I = mL2,
 where
Fg = mg,
L = 1.30 m,
 and
m = 2.20 kg.

Solve for the desired unknown quantity and substitute the numbers.
The rotational analog of Newton's second law is
τ
 = Iα,
where
τ
 = −FgL
 and
I = mL2.
 In these expressions,
Fg = mg,
L = 1.30 m,
 and
m = 2.20 kg
 so that
−FgL = (mL2)α.
The unknown quantity to be determined is the angular acceleration α of the sphere at the instant it is dropped. We solve for this angular acceleration to obtain
α = − 
Fg
mL
 = − 
mg
mL
 = − 
g
L
 
 = − 
9.80 m/s2
1.30 m
 = −7.54 rad/s2.
The negative angular acceleration indicates that when the sphere is dropped, it begins to rotate clockwise around the pivot point.
You have now completed the Guided Problem.
Additional Problems

Question 8.2a:
The figure shows a block of mass
m = 2.80 kg
 hanging from a thin rope wrapped around a solid, disk-shaped pulley. The pulley has mass
M = 8.70 kg,
 radius
R = 0.127 m,
 and is free to rotate on a frictionless, horizontal axle. If the block is released from rest, determine the magnitude of its downward acceleration.
Incorrect: Your answer is incorrect. seenKey

3.84
Incorrect. See m/s2

Question 8.2b:
A simple yo-yo is made by wrapping a thin string of negligible mass around the perimeter of a solid disk of radius
R = 0.120 m
 and mass
M = 2.49 kg.
 The free end of the string is held stationary by a person's hand, as in the figure, and the yo-yo is released from rest. Determine the magnitude of the disk's downward acceleration.
Incorrect: Your answer is incorrect. seenKey

6.53
Incorrect. See m/s2
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4/7
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Prelecture Exploration: Calculating the Center of Mass
Two classmates, Madhur and Jayden, are interested in understanding the center of mass. Although they know that most objects are three-dimensional, they think that by examining one-dimensional examples, they can learn about how to extend the process of finding the "imaginary" point at which a point mass would represent the combined object to all three dimensions. They learned in class that the center of mass is the weighted average position of the mass of a system along the line connecting their centers.

Madhur and Jayden use the simulation to help them explore the concept of the center of mass. In this simulation, they can use the sliders to adjust the positions and masses of the two objects. The coordinate xCM of the center of mass of the system along the line between the objects is shown for each configuration possibility.

Click here to open the simulation in a new window.
Part 1 of 6 - Location of the Center of Mass When m2 Changes
Jayden sets m1 = 1 kg, x1 = 25 cm, and x2 = 150 cm. Now he changes m2 and observes how the location of the center of mass of the system changes. What can he say happens?
     Correct: Your answer is correct.
Jayden is correct. Imagine the two masses to be connected by a massless rod and balanced on a fulcrum. If m2 increases, to return the system to balance, the fulcrum would need to shift to the right. The point at which the system is balanced is the center of mass.
Part 2 of 6 - Location of the Center of Mass When m1 = m2
Now Madhur sets m1 = 1 kg, m2 = 1 kg, x1 = 25 cm, and x2 = 150 cm. How does she describe the location of the center of mass to Jayden?
     Correct: Your answer is correct.
Madhur is correct. When the two masses are equal, the center of mass is directly between the two, or
25 cm + 
(150 − 25) cm
2
 = 87.5 cm
.
Part 3 of 6 - Location of the Center of Mass When the Location of m2 Changes
Jayden sets m1 = 4 kg, m2 = 2 kg, x1 = 25 cm, and x2 = 150 cm. Now he increases the position of m2 by moving the slider and observing how the location of the center of mass of the system changes. How does he describe what happens?
    
Part 4 of 6 - Center of Mass within an Object
Madhur and Jayden consider a system made of two spherical masses. How can Madhur describe the circumstances that could allow the center of mass of this two-mass system to be located inside one of the masses?
     Correct: Your answer is correct.
Madhur is correct. It is not necessarily true that under this condition the center of mass will lie within one of the bodies, but it is possible. In the simulation the friends can try, for example, m1 = 1 kg, m2 = 8 kg, x1 = 25 cm, and x2 = 150 cm. The center of mass of the Earth-Moon system is, in fact, approximately 1700 km below Earth's surface.
Part 5 of 6 - Center of Mass within an Object
The friends can think of the center of mass as the average location of a system's mass. Jayden writes an expression for the x-component of the center of mass of a two-mass system. Which expression is correct?
     Correct: Your answer is correct.
Jayden is correct. This is the weighted average of the location of the masses of the two particles.
Part 6 of 6 - Analyze
Now the friends try a homework problem.

An object is placed at the position x1 = 52 cm and a second mass that is 5/2 times as large is placed at x2 = 245 cm. Find the location of the center of mass of the system.
Incorrect: Your answer is incorrect. seenKey

1.9
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m

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Total
3/16
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Prelecture Exploration: Energy Conservation for Rolling Objects
Darcel and Chandra are sitting in the park after physics class, and they notice some children rolling various objects down a slight incline in the sidewalk. Darcel is curious how the masses and shapes of the objects affect their motion, and Chandra says that it might be fun to think about the energy of motion associated with rolling objects. The friends decide to use the simulation to explore the motion and energies of various objects rolling down an incline.

They can click on an object (solid sphere, spherical shell, hoop, or cylinder) to position it at the top of the incline and click "roll" to release it. The translational and rotational speeds of the object and the time interval the object is on the incline are shown, together with a graph of the percentages of the different energy types as the object rolls down the incline. The "roll" button turns into a "pause" button while the object is in motion to allow them to examine different parameters at various points. For the object–Earth system in the simulation, the zero configuration of gravitational potential energy is defined to occur when the object is at the bottom of the incline. For the simulation, air resistance and any rolling friction effects are neglected, and the masses of all the objects are equal.

Click here to open the simulation in a new window.
Part 1 of 11 - Comparison of Similar Objects of Different Sizes
Darcel clicks on the large solid sphere, then clicks "roll." He pays attention to the time the sphere takes to roll down the incline. Then he clicks on the small solid sphere and repeats. Which of Darcell's statements about his observations of the time it takes for the two objects to reach the bottom is correct?
     Correct: Your answer is correct.
Darcel is correct. He observed that both the large and small sphere take the same amount of time to reach the bottom.
Part 2 of 11 - Comparison of Similar Objects of Different Sizes
Now Chandra clicks on the large hollow sphere, and clicks "roll." She pays attention to the time the sphere takes to roll down the incline. Then she clicks on the small hollow sphere and repeats. Which statement about her observations of the times is correct?
     Correct: Your answer is correct.
Chandra is correct. Both the large and small sphere take the same amount of time to reach the bottom.
Part 3 of 11 - Comparison of Similar Objects with Different Mass Distributions
Now Darcel compares the times it takes for the large solid sphere and the large hollow sphere to roll down the incline. What can he say about his observations of the time it take for the two objects to reach the bottom?
     Correct: Your answer is correct.
Darcel is correct. The time for the solid sphere to reach the bottom is 4.00 s, while for the hollow sphere, the time is 4.36 s.
Part 4 of 11 - Speeds of Objects at the Bottom of the Incline
The speed of the object at the bottom of the incline is equal to the translational speed once the object reaches the bottom. Which of Chandra's statements is supported by her observations?
    


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A 440-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 140-N rod as indicated in the figure below. The left end of the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical. (Assume the cable is connected to the very end of the 6.00-m-long rod, and that there are 2.00 m separating the wall from the sign.)
(a) Find the (magnitude of the) tension T in the cable.
Correct: Your answer is correct. seenKey

420

 N

(b) Find the horizontal and vertical components of the force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions. Indicate the direction with the sign of your answer.)
horizontal component N
vertical component N

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11/11
 
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Tutorial Exercise
A 147-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.850 rev/s in 2.00 s?
Step 1
The diagram below shows the view of the merry-go-round from above.
The moment of inertia of a uniform, solid disk of mass M = 147 kg and radius r = 1.50 m is
I = 
1
2
Mr2
 = 
1
2
147 Correct: Your answer is correct. seenKey

147

kg
1.5 Correct: Your answer is correct. seenKey

1.5

m
2
 
 = 165 Correct: Your answer is correct. seenKey

165

kg · m2.
Step 2
The desired angular acceleration α for the merry-go-round, where ω is the final angular speed, ω0 the initial angular speed, and Δt the time interval from rest, is given by
α = 
ω − ω0
Δt
 
 = 
.85 Correct: Your answer is correct. seenKey

0.85

rev/s
 − 0
2 Correct: Your answer is correct. seenKey

2

s
 = 
.425 Correct: Your answer is correct. seenKey

0.425

rev/s2
2π rad
1 rev
 = 2.67 Correct: Your answer is correct. seenKey

2.67

rad/s2.
Step 3
We take counterclockwise torques (and angular accelerations) as positive, and we use the rotational form of Newton's second law,
Στ = Fr = Iα,
 in order to find the tension F in the rope required to produce this angular acceleration. Solving for F gives
F = 
Iα
r
 
 = 
165 Correct: Your answer is correct. seenKey

165

kg · m2
2.67 Correct: Your answer is correct. seenKey

2.67

rad/s2
1.5 Correct: Your answer is correct. seenKey

1.5

m
 = 294 Correct: Your answer is correct. seenKey

294

 N.
SUMMARY
Using the moment of inertia equation for a uniform circular disk of given mass and radius, we calculated the moment of inertia for the merry-go-round. Then we calculated the angular acceleration from the given final angular speed, knowing the initial angular speed and the interval of time before the final speed is attained. Finally, we used the rotational form of Newton's second law to find the tension in the rope required to produce the desired angular acceleration.
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Spectators watch a bicycle stunt rider travel off the end of a 52.5° ramp, rise to the top of his trajectory and, at that instant, suddenly push his bike horizontally away from him so that he falls vertically straight down, reaching the ground 0.525 s later. How far (in m) from the rider does the bicycle land if the rider has a mass
mrider = 74.0 kg
 and the bike has a mass
mbike = 12.5 kg?
 Neglect air resistance and assume the ground is level.
Incorrect: Your answer is incorrect. seenKey

14.3

 m


Solution or Explanation
First, use kinematics to find the initial speed of the rider–bicycle system. The system is launched at θ0 = 52.5° and rises from
y0 = 0,
 returning to
y = 0
 in a total time of
t = 2(0.525 s) = 1.05 s.
 Solve for its initial speed v0 to find the following.
y = y0 + (v0 sin(θ0))t − 
1
2
gt2
v0 = 
1
2
gt
sin(θ0)
 = 
1
2
(9.80 m/s2)(1.05 s)
sin(52.5°)
 = 6.49 m/s
The rider–bicycle system's horizontal velocity component is then
vx = v0 cos(θ0) = (6.49 m/s)cos(52.5°) = 3.95 m/s.
Set
x = 0
 at the rider's horizontal position after pushing off the bike at
t = 0.
 At
t = 0.525 s,
 the center of mass is located at
xcm = vxt = (3.95 m/s)(0.525 s) = 2.07 m.
From the definition of center of mass, with
xrider = 0,
 we have the following.
xcm = 
mi xi
mi
 = 
mriderxrider + mbikexbike
mrider + mbike
 = 
mbike
mrider + mbike
xbike
Solve for the final position of the bike, xbike, to find the following.
xbike = 
mrider + mbike
mbike
xcm = 
74.0 kg + 12.5 kg
12.5 kg
2.07 m
 = 14.3 m

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A 1.00 kg solid, uniform sphere rolls without slipping across a level surface, translating at 3.25 m/s. If the sphere's radius is 0.360 m, find the following.
(a)
the sphere's translational kinetic energy (in J)
J
(b)
the sphere's rotational kinetic energy (in J)
J

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Watch the video vignette below, and be sure to follow any instructions when prompted during the videos. Then answer the questions below about the concepts discussed in the video.
(a)
As you saw in the video, the two bullets in the experiment were fired from the same rifle. Assume the bullets also had the same mass and were fired from identically prepared cartridges. From this, what can we conclude about the initial momentum of the bullets just before they struck the blocks?
     Correct: Your answer is correct.
(b)
As you saw in the video, the two blocks had about the same mass, and both were initially at rest. Which of the following is true about the two bullet–block systems, just after the bullets collided with their respective blocks?
     Correct: Your answer is correct.
(c)
Which of the following statements provides correct reasoning for why both bullet–block systems reached the same maximum height?
     Correct: Your answer is correct.
(d)
Which of the following statements are true about the energy of the two bullet–block systems immediately after the collisions? (Select all that apply.)
Correct: Your answer is correct.

(e)
Imagine comparing the two blocks after the experiment. In which block would we expect the bullet to be embedded the greatest distance?
     Correct: Your answer is correct.


Solution or Explanation
The two bullets have the same initial momentum, directed upward. They have the same mass, and because they are fired from identically prepared cartridges from the same rifle, they have the same velocity before entering the block.
The two blocks both have the same mass and are both initially at rest before the collision.
Upon impact, conservation of momentum states the momentum of the bullet before impact will be equal to the momentum of the bullet–block system after impact. So both systems have the same upward momentum after impact. And because they have the same mass, they therefore have the same upward velocity after impact.
Because they have the same upward (center-of-mass) velocity after the collision, the two systems then travel to the same maximum height (by constant acceleration kinematics).
The two bullets have the same kinetic energy before the collision, because they have the same mass and speed. Conservation of energy states that both bullet–block systems have the same total energy after the collision.
However, the energy takes different forms after the collision in the two cases. They both do have the same translational kinetic energy after the collision (because they have the same momentum, mass, and center-of-mass speed). But in the first case, more energy has gone into deforming the block, and the bullet is embedded deeper. In the second case, less energy goes into deforming the block—the edge of the block "moves with" the bullet and the bullet does not get embedded as much. The remaining energy instead goes into the rotational kinetic energy of the system.
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10. –/3 points SerCP11 8.3.P.027. My Notes
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A uniform plank of length 2.00 m and mass 32.5 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 725–N person is d = 0.500 m from the left end.
magnitude of
T1
     		N
magnitude of
T2
     		N
magnitude of
T3
     		N

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A solid disk and a hoop are simultaneously released from rest at the top of an incline and roll down without slipping. Which object reaches the bottom first?
     Correct: Your answer is correct.
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A deli owner creates a lunch special display board by taking a uniform board which has a weight of 142 N, cutting it in half, hinging the halves together with a frictionless hinge, and setting it up as an inverted "V". Determine the minimum coefficient of static friction needed between the board and the ground in order for her to set the display board up with an angle of 30° between the two sides.
Two hinged boards of equal length are set on the ground with the hinge in the air making a triangle with one point at the top with an internal angle of 30 degrees.
Tutorial
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A uniform ladder stands on a rough floor and rests against a frictionless wall as shown in the figure.
On the xy coordinate plane there is a ladder leaning against a wall with a man standing near the top of it. The ladder reaches a distance d up on the wall. The man is a distance labeled b to the right of the base of the ladder. The base of the ladder has two arrows extending from it, an arrow pointing up labeled N_1 and an arrow pointing right labeled f_1. The man has an arrow pointing down from his feet labeled with the expression m g. The top of the ladder has a horizontal arrow pointing to the left out of the wall labeled N_2.
Since the floor is rough, it exerts both a normal force
N1
 and a frictional force
f1
 on the ladder. However, since the wall is frictionless, it exerts only a normal force
N2
 on the ladder. The ladder has a length of
L = 4.4 m,
 a weight of
WL = 68.5 N,
 and rests against the wall a distance
d = 3.75 m
 above the floor. If a person with a mass of
m = 90 kg
 is standing on the ladder, determine the following.
(a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.)
N1 = Incorrect: Your answer is incorrect. seenKey

951
Draw a careful diagram which includes all forces acting on the ladder. See if you can apply the first condition of equilibrium to vertical forces acting on the ladder in order to obtain
N1.
 N
N2 = Incorrect: Your answer is incorrect. seenKey

292
Determine a good point about which to sum the torques and then see if you can write a statement of the second condition of equilibrium about that point that will allow you to determine
N2,
 the normal force exerted on the ladder by the wall. You may need to use your knowledge of trigonometry to determine the angle between the ladder and the floor. N
f1 = Incorrect: Your answer is incorrect. seenKey

292
See if you can apply the first condition of equilibrium to horizontal forces acting on the ladder in order to obtain a relationship between
f1
 and
N1
 and hence determine
f1.
 N

(b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.)
N1 = Incorrect: Your answer is incorrect. seenKey

951
Draw a careful diagram which includes all forces acting on the ladder. Make sure your diagram includes the one change that distinguishes the problem in part (b) from the problem in part (a), namely the point of application of the force of gravity acting on the person. See if you can apply the first condition of equilibrium to vertical forces acting on the ladder in order to obtain
N1.
 N
N2 = Incorrect: Your answer is incorrect. seenKey

427
Using the point specified in part (a) as the point about which to sum the torques, see if you can write a statement of the second condition of equilibrium that will allow you to determine
N2,
 the normal force exerted on the ladder by the wall. N
f1 = Incorrect: Your answer is incorrect. seenKey

427
See if you can apply the first condition of equilibrium to horizontal forces acting on the ladder in order to obtain a relationship between
f1
 and
N1
 and hence determine
f1.
 N


Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

(a) The figure below shows all forces acting on the ladder when the person is halfway up the ladder.
On the xy coordinate plane there is a ladder leaning against a wall such that it makes an angle theta with the ground. The ladder reaches a distance d up on the wall. The base of the ladder has two arrows extending from it, an arrow pointing up labeled N_1 and an arrow pointing right labeled f_1. The top of the ladder has a horizontal arrow pointing to the left out of the wall labeled N_2. At a horizontal distance from the base of the ladder labeled with the expression (L/2) cos(theta) there is an arrow pointing down from the ladder labeled with the expression (W_L + m g).
As a result of the first condition of equilibrium, we know that the sum of all forces acting upward on the ladder is equal to the sum of all forces acting downward. This allows us to write
N1 = WL + mg = 68.5 N + (90.0 kg)(9.8 m/s2) = 951 N.
As a result of the first condition of equilibrium, we know that the sum of all forces acting on the ladder to the left is equal to the sum of all forces acting to the right. This allows us to write
f1 = N2.
We cannot determine f1 by the expression
f1 = μN1,
 because we do not know the coefficient of friction between the ground and ladder. As a result, let's see if we can use the second condition of equilibrium, the basic definition of torque, and the base of the ladder as the point to take the torques about in order to determine the normal force N2. From the second condition of equilibrium we can state that the sum of the clockwise torques about a point at the base of the ladder is equal to the sum of the counterclockwise torques, or
τcw = τccw.
 The weight of the ladder and the person provide clockwise torques about a point at the base of the ladder and the normal force from the wall provides the counterclockwise torque. Using the basic definition of torque in the form
τ = Fr⊥
 and inserting details we may write
(WL + mg)
L
2
cos θ = N2d,
where r⊥ for the force
(WL + mg)
 is
(L/2)cos θ
 and r⊥ for the force N2 is d. Solving this expression for N2, we obtain
N2 = 
(WL + mg)L cos θ
2d
.
Before we can obtain a numerical value for N2, we need to obtain a value for the angle θ. From the figure we see that
θ = sin−1
d
L
 = sin−1
3.75 m
4.4 m
 = 58.5°.
Inserting this and other values into the expression for N2, obtain
N2 = 
(WL + mg)L cos θ
2d
 
Inserting values we have
N2 = 
[68.5 N + (90.0 kg)(9.8 m/s2)](4.4 m)cos(58.5°)
2(3.75 m)
 = 292 N
Now that we know N2, we may write a horizontal first condition of equilibrium statement that will allow us to determine f1.
f1 = N2 = 292 N.

(b) The figure below shows all forces acting on the ladder when the person is three fourths the way up the ladder.
On the xy coordinate plane there is a ladder leaning against a wall such that it makes an angle theta with the ground. The ladder reaches a distance d up on the wall. The base of the ladder has two arrows extending from it, an arrow pointing up labeled N_1 and an arrow pointing right labeled f_1. The top of the ladder has a horizontal arrow pointing to the left out of the wall labeled N_2. At a horizontal distance from the base of the ladder labeled with the expression (L/2) cos(theta) there is an arrow pointing down from the ladder labeled with the expression W_L. At a horizontal distance from the base of the ladder labeled with the expression (3/4)L cos(theta) there is an arrow pointing down from the ladder labeled with the expression m g.
Notice that the upward and downward forces acting on the ladder are the same as for part (a). Granted the weight of the person is now acting on the ladder at a different place, but the upward and downward forces are the same, hence the vertical statement of the first condition of equilibrium is the same, and we still have
N1 = WL + mg = 68.5 N + (90.0 kg)(9.8 m/s2) = 951 N.
Looking at the figure, the only thing that has changed is the location of the weight of the person and hence r⊥ for the weight of the person. Summing torques about the base of the ladder, obtain
N2d = WL
L
2
cos θ + mg
3L
4
cos θ
N2 = 
WL + 
3mg
2
L cos θ
2d
 
N2 = 
68.5 N + 
3
2
(90.0 kg)(9.8 m/s2)
(4.4 m)cos(58.5°)
2(3.75 m)
 = 427 N.
Again, from the horizontal statement of the first condition of equilibrium we may write
f1 = N2 = 427 N.
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