CH 08 HW - Chapter 8 Physics Homework for Mastering PDF

Preview

Summary

Chapter 8 Physics Homework for Mastering
...

Description

CH 08 HW Due: 3:00pm on Wednesday, November 13, 2019 You will receive no credit for items you complete after the assignment is due. Grading Policy

± Torques on a Ruler A centimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure. One coin, of mass , is placed at the zero mark; the other, of unknown mass , is placed at the 4.7 mark. The center of the ruler is at the 3.0 mark. The ruler is in equilibrium; it is perfectly balanced.

Part A Does the pivot point (i.e., the triangle in the diagram upon which the ruler balances) exert a force on the ruler? Does it exert a nonzero torque about the pivot? Choose the correct series of answers to these two questions. ANSWER:

yes/yes no/no yes/no no/yes

Correct Note that the weight of the ruler itself also does not exert a torque with respect to the pivot point since the ruler is uniform and is pivoted at its midpoint.

Part B Although the pivot exerts a force on the ruler it does not exert a torque with respect to the pivot point. Why not? ANSWER:

Because the exerted force is vertical. Because the distance from the exerted force to the pivot is zero. Because the exerted force is along the ruler.

Correct

Part C Find the mass

.

Express your answer for the unknown mass numerically, in grams, to two significant digits.

Hint 1. Sum of torques Because the ruler is in equilibrium, the sum of the torques about any point along the ruler must be zero. Specifically, the sum of the torques about the pivot point must be zero. Hint 2. Determine the moment arm of What is the moment arm of (call it ) about the ruler's pivot point? The moment arm is the distance, perpendicular to the direction of the force, from the pivot to the point of application of the force. It is what you multiply the force by to get the magnitude of the torque due to that force.

Hint 1. Finding a moment arm To find the moment arm for the force exerted by each coin on the ruler, you should measure from the pivot, not from the end of the ruler.

ANSWER: = 3

Correct

ANSWER: = 18 g

Correct

The Center of Mass of the Earth-Moon-Sun System A common, though incorrect, statement is, "The Moon orbits the Earth." That creates an image of the Moon’s orbit that looks like that shown in the figure. The Earth's gravity pulls on the Moon, causing it to orbit. However, by Newton’s third law, it is known that the Moon exerts a force back on the Earth. Therefore, the Earth should move in response to the Moon. Thus a more accurate statement is, "The Moon and the Earth both orbit the center of mass of the Earth-Moon system."

In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate the center of mass of the EarthMoon-Sun system. The mass of the Moon is 7.35×1022 , the mass of the Earth is 6.00×1024 , and the mass of the sun is 2.00×1030 . The distance between the Moon and the Earth is 3.80×105 . The distance between the Earth and the Sun is 1.50×108 .

Part A Calculate the location of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Earth is at Moon is located in the positive x direction. Express your answer in kilometers to three significant figures.

Hint 1. Calculating the center of mass The general equation for the center of mass

for a system of two particles of masses

and

is

and the

, where

and

are the locations of the particles in the given coordinate system.

While the Earth and Moon are very large bodies, treating them as particles is reasonable in this problem, because the distance between them is much greater than their radii. Hint 2. Find the coordinates of the Earth and Moon Taking the center of the Earth as the origin of your coordinate system, what is the x coordinate of the Moon

?

Express your answer in kilometers to three significant figures. ANSWER: = 3.80×105

ANSWER: = 4600

Correct

Part B Where is the center of mass of the Earth-Moon system? The radius of the Earth is 6378 answers below:

and the radius of the Moon is 1737

. Select one of the

Choose the correct description of the location of the center of mass of the Earth-Moon system. ANSWER:

The center of mass is exactly in the center between the Earth and the Moon. The center of mass is nearer to the Moon than the Earth, but outside the radius of the Moon. The center of mass is nearer to the Earth than the Moon, but outside the radius of the Earth. The center of mass is inside the Earth. The center of mass is inside the Moon.

Correct As you can see, the center of mass for the Earth-Moon system actually lies within the radius of the Earth. For this reason, saying that the Moon orbits the Earth is often a good approximation, though in fact, both the Earth and the Moon orbit that point with a synodic period of 29.5 days. The Moon makes large orbits around the center of mass of the Earth-Moon system, whereas the center of the Earth makes small orbits.

Part C Calculate the location of the center of mass of the Earth-Moon-Sun system during a full Moon. A full Moon occurs when the Earth, Moon, and Sun are lined up as shown in the figure. Use a coordinate system in which the center of the sun is at and the Earth and Moon both lie along the positive x

direction. Express your answer in kilometers to three significant figures.

Hint 1. Calculating the center of mass The general equation for the center of mass ,

, and

for a system of thee particles of masses

,

, and

is

,where

refer to the distances to each of the three particles from the origin.

ANSWER: = 456

Correct The equatorial radius of the Sun is 695,000 . As you can see, the center of mass for the Sun-Earth-Moon system is well within the Sun. However, if you were to find the center of mass of the Jupiter-Sun system, you would find that it is slightly above the surface of the Sun at 780,000 from the center of the Sun. A distant alien civilization would not be able to see Jupiter directly, because it is far too faint, but they would be able to see the Sun move back and forth as it orbited the center of mass with Jupiter. Because the sun is "wobbling," alien scientists would be able to infer that there was a planet around the Sun. This is one of the methods that human scientists are using to identify planets around other stars.

Testing Experiment Table 8.4 A tentative rule was given about tipping in Testing Experiment Table 8.4, where if a vertical line passing through the object's center of mass is within the object's area of support the object does not tip. According to the rule, the full box of crackers in the following experiment, which has a width of 13 and a height of 20 , was seen to tip at an angle of: .

Part A According to the rule, at what angle will a box with the left half filled with crackers begin to tip? Assume the box is light-weight and that the contents remain in the configuration in .

Hint 1. Identifying the center of mass Since the box is only half-filled, the center of mass will be shifted to the left half of the box. Hint 2. Determining the tipping angle Draw a vertical line starting from the pivot point to the center of mass of this half-filled box. From there, it is possible to make this line the hypotenuse of a right triangle, where is the angle with respect to this vertical line and the left side of the box.

ANSWER:

= 18

Correct

Part B Now the same box is tilted the other way with the crackers staying on the left side of the box, as shown in . Which of the following statements are based on the rule?

ANSWER: It will be equivalent to tip to this side, because for a given angle the center of mass is at the same height. It will be harder to tip to this side, because for a given angle the center of mass is shifted sideways. It will be easier to tip to this side, because the center of mass moves more when tilted. It will be easier to tip to this side, because for a given angle the center of mass is higher up. None of these explanations are correct.

Correct While the center of mass may be higher, according to the rule what matters is whether the center is on one side of the pivot or the other.

Part C Based on the diagram in , which of the following expressions for

will determine when the box will tip?

Hint 1. How to approach the problem As in Part A, allow the vertical line drawn in the diagram to represent the hypotenuse of a right triangle. From there, use the overall dimensions of the box to determine the corresponding tipping angle .

ANSWER:

Correct A right triangle can be made that has its hypotenuse from the pivot to the center of mass. Then half way up the side is one edge and from the center of mass to that side is the other edge. These have lengths of 0.75 times the width and 0.5 times the height.

Torque Magnitude Ranking Task The wrench in the figure has six forces of equal magnitude acting on it.

Part A Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measured about an axis centered on the bolt. Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Definition of torque Torque is a measure of the "twist" that an applied force exerts on an object. Mathematically, torque is defined as , where is the magnitude of the displacement vector from the rotation axis to the point of application of the force of magnitude between this displacement and the applied force, as shown in the figure.

, and

is the angle

The direction of a torque can be either counterclockwise (as above) or clockwise. This is determined by the direction the object will rotate under the action of the force. Hint 2. Maximum torque Based on the mathematical definition of torque, torque is maximized when the force is large in magnitude, located a large distance from the axis of interest, and oriented perpendicular to the displacement , which is often referred to as the lever arm of the force.

ANSWER:

Reset

Help

B D

F

A

C

E

The correct ranking cannot be determined.

Correct

Torques on a Seesaw: A Tutorial Learning Goal: To make the connection between intuitive understanding of a seesaw and the standard formalism for torque. This problem deals with the concept of torque, the "twist" that an off-center force applies to a body that tends to make it rotate.

Use your intuition to try to answer the following question. If your intuition fails, work the rest of the problem and return here when you feel that you are more comfortable with torques.

Part A Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose weight is , is sitting at distance to the left of the pivot, at what distance should Marcel place Gilles, whose weight is , to the right of the pivot to balance the seesaw? Express your answer in terms of

,

, and

.

Hint 1. How to approach the problem Consider whether

increases or decreases as each of the variables that it depends on,

,

, and

, is made larger or smaller.

ANSWER: =

Correct

Now consider this problem as a more formal introduction to torque. The torque of each child about the pivot point is the product of the child's weight and the distance from the pivot to the child's center of mass (center of gravity.) The sign of the torque is taken to be positive by convention if it would cause a counterclockwise rotation of the seesaw. The distance is measured perpendicular to the line of force and is called the moment arm.

Part B Find the torque

about the pivot due to the weight

Express your answer in terms of

and

of Gilles on the seesaw.

.

ANSWER: =

Correct

Marcel wants the seesaw to balance, which means that there can be no angular acceleration about the pivot. For the angular acceleration to be zero, the sum of the torques about the pivot must equal zero: .

Part C Determine

, the sum of the torques on the seesaw. Consider only the torques exerted by the children.

Express your answer in terms of

,

,

, and

.

Hint 1. Torque from the weight of the seesaw The seesaw is symmetric about the pivot, and so the gravitational force on the seesaw produces no net torque. More generally, when determining torques, the gravitational force on an object in a uniform gravitational field can be taken to act at the object's center of mass. Here, the center of mass is directly above the pivot, so the weight of the seesaw has zero moment arm and produces no torque about the pivot.

ANSWER: =

Correct If you did not solve for the distance

required to balance the seesaw in Part A, do so now.

The equation applies to any body that is not rotationally accelerating. Combining this equation with (which applies to any body that is not accelerating linearly) gives a pair of equations that are sufficient to form the basis of statics. The art of applying these equations to large or complicated structures constitutes a significant part of mechanical and civil engineering.

Gilles has an identical twin, Jean, also of weight distance .

. The two twins now sit on the same side of the seesaw, with Gilles at distance

from the pivot and Jean at

Part D Where should Marcel position Jacques to balance the seesaw? Express your answer in terms of

,

,

, and

.

Hint 1. Find the sum of the torques when the seesaw is balanced For the seesaw to balance, the sum of the torques must be zero. What is the sum of the torques due to the three children? Express your answer in terms of

,

,

,

, and

.

ANSWER: =

ANSWER: =

Correct

Bad news! When Marcel finds the distance from the previous part, it turns out to be greater than , the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height above the pivot.

Part E With what force in the rightward direction, the force should be toward the left.

, should Marcel push? If your expression would give a negative result (using actual values) that just means

Express your answer in terms of

,

,

,

,

, and .

Hint 1. Sign conventions It is easy to make sign errors in torque problems, and experience shows that it is better to use standard conventions (here, that goes to the right, the direction of positive x) than to change the direction of positive torque or displacement to suit your convenience. (You are likely to forget your unconventional choice at a later point in the problem.) In this case, your intuition correctly expects that Marcel must push to the left to make things balance; the equations will confirm this by giving a negative result for . (A positive result would mean that the force would be directed to the right in the figure.) Hint 2. Find the torque due to Marcel's push The sum of all four torques (due to each of the three children, plus Marcel's) must be zero. What is the torque due to Marcel's push? Keep in mind that a positive torque will cause counterclockwise rotation of the seesaw. Express your answer in terms of the unknown force

and the height

at which it is applied.

ANSWER: =

ANSWER:

=

Correct Notice that this answer will necessarily be negative, because you were told that to balance the seesaw with the twins on the right, Jacques had to be "beyond the end" of the seesaw. Therefore, when indicates Jacque's position, will be less than zero. Hence Marcel must push to the left, as you would expect.

A Bar Suspended by Two Vertical Strings The figure shows a model of a crane that may be mounted on a truck.A rigid uniform horizontal bar of mass = 90.0 and length = 6.00 is supported by two vertical massless strings. String A is attached at a distance = 1.10 from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass = 3500 is supported by the crane at a distance = 5.80 from the left end of the bar. Throughout this problem, positive torque is counterclockwise and use 9.80

for the magnitude of the acceleration due to gravity.

Part A Find

, the tension in string A.

Express your answer in newtons using three significant figures.

Hint 1. Choosing an axis Choose a rotation axis p about which to apply the requirement . Since the system is in static equilibrium, the choice of rotation axis is arbitrary. However, there is a convenient choice of p that allows you to find by eliminating the torque created by an unknown force. Hint 2. Find the torque around the convenient axis It is convenient to choose the rotation axis through the point at which string B is attached to the bar. With respect to this axis, the torque created by the tension in string B is zero. Which of the following is a correct expression for the total torque about this point? ANSWER:

=

Hint 3. Applying the equilibrium condition For a static system,

. Solve for

.

ANSWER: = 1.83×105

Correct

Part B Find

, the magnitude of the tension in string B.

Express your answer in newtons using three significant figures.

Hint 1. Two different methods to find There are two ways to find . One way is to balance the torques as you did in the calculation of using a different rotation axis. In this case, a convenient axis passes through the point at which string A is attached to the bar. The second, easier, method is to use the second equation for static equilibrium,

.

Hint 2. Direction of forces Since both strings are vertical, all forces acting on the bar are vertical.

ANSWER: = 1.48×105

Correct

Part C If the bar and block are too heavy, one of the strings may break. Assuming that the two strings are identical, which one will break first? ANSWER: string A string B

Correct The strings must be carefully chosen after the maximum possible tension in A has been calculated, allowing for a safety margin.

Balancing Torques Ranking Task A sign is to be hung from the end of a thin pole, and the pole supported by a single cable. Your design firm brainstorms the six scenarios shown below. In scenarios A, B, and D, the cable is attached halfway between the midpoint and end of the pole. In C, the cable is attached to the mid-point of the pole. In E and F, the cable is attached to the end of the pole.