Assignment 15 for Mastering Physics

8/17/2015 Assignment 15

https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3766134 1/15

Assignment 15Due: 11:59pm on Monday, August 17, 2015

To understand how points are awarded, read the Grading Policy for this assignment.

Finding Torque

A force of magnitude making an angle with the x axis is applied to a particle located along axis of rotation A, atCartesian coordinates in the figure. The vector lies in the xy plane, and the four axes of rotation A, B, C, andD all lie perpendicular to the xy plane.

A particle is located at a vector position with respect to an axis of rotation (thus points from the axis to the point atwhich the particle is located). The magnitude of the torque about this axis due to a force acting on the particle isgiven by

,where is the angle between and , is the magnitude of , is the magnitude of , the component of that isperpendicualr to is the moment arm, and is the component of the force that is perpendicular to .

Sign convention: You will need to determine the sign by analyzing the direction of the rotation that the torque wouldtend to produce. Recall that negative torque about an axis corresponds to clockwise rotation.In this problem, you must express the angle in the above equation in terms of , , and/or when entering youranswers. Keep in mind that and .

Part A

What is the torque about axis A due to the force ?

Express the torque about axis A at Cartesian coordinates .

Hint 1. When force is applied at the pivot point

Does a force applied at a pivot point cause an object to rotate about that pivot?

ANSWER:

F F θ

(0, 0) F

r r τ F

τ = rF sin(α) = (moment arm) ⋅ F = rF⊥

α r F r r F F r F F⊥ r

α θ ϕ ππ = 180 degrees (π/2) = 90 degrees

τA F

(0, 0)



Correct

Part B

What is the torque about axis B due to the force ? (B is the point at Cartesian coordinates , located adistance from the origin along the y axis.)

Express the torque about axis B in terms of , , , , and/or other given coordinate data.

Hint 1. Finding with respect to an axis

The vector should be drawn from the axis B to the point where is being applied. Consider both themagnitude and direction of this vector.

Hint 2. A helpful figure

The figure shows for this part of the problem. What is the value of in terms of ?

ANSWER:

ANSWER:

= τA 0

τB F (0, b)b

F θ ϕ π

r

r F

r α θ

π − θ

(π/2) − θ

(π/2) + θ

θ



Correct

Part C

What is the torque about axis C due to ? (C is the point at Cartesian coordinates , a distance alongthe x axis.)

Express the torque about axis C in terms of , , , , and/or other given coordinate data.

Hint 1. A helpful figure

The figure shows for this part of the problem. What is the value of ?

ANSWER:

Hint 2. Clockwise or counterclockwise?

Imagine a wheel of radius with its axle passing through the axis of rotation (so that the particle at point Ais on the rim of the wheel). Look at the direction of in the figure. Which way do you think will "try" toturn the wheel: clockwise or counterclockwise? Note that only the component of that is tangent to the rimof the wheel (perpendicular to ) generates torque.

ANSWER:

= τB bFsin( + θ)π2

τC F (c, 0) c

F θ ϕ π

r F⊥

F sin θ

F cos θ

r

F F

F r



Correct

Part D

What is the torque about axis D due to ? (D is the point located at a distance from the origin and makingan angle with the x axis.)

Express the torque about axis D in terms of , , , , and/or other given coordinate data.

ANSWER:

Correct

You could have found by using any of the three equations listed at the top of the page,

,and making the following associations:

and

.

Calculating Torques Using Two Standard Methods

Learning Goal:

To understand the two most common procedures for finding torques when the forces and displacements are all in oneplane: the moment arm method and the tangential force method.

The purpose of this problem is to give you further practice finding torques in twodimensional situations. In this case itis overkill to use the full cross product definition of the torque because the only nonzero component of the torque is thecomponent perpendicular to the plane containing the problem.

There are two common methods for finding torque in a twodimensional problem: the tangential force method and themoment arm method. Both of these methods will be illustrated in this problem.

Throughout the problem, torques that would cause counterclockwise rotation are considered to be positive.

Consider a uniform pole of length , attached at its base (via a pivot) to a wall. The other end of the pole is attached toa cable, so that the pole makes an angle with respect to the wall, and the cable is horizontal. The tension in the cableis . The pole is attached to the wall at point A.

= τC −cFsin(θ)

τD F dϕ

F θ ϕ π

= τD dFsin(ϕ − θ)

τD

τ = rF sin(α) = (moment arm) ⋅ F = rF⊥

r = d sin(α) = sin(π − ϕ + θ) = sin(ϕ − θ)moment arm = = d sin(ϕ − θ)r⊥

= F sin(ϕ − θ)F⊥

Lθ

T



Tangential force method

The tangential force method involves finding the component of the applied force that is perpendicular to thedisplacement from the pivot point to where the force is applied. This perpendicular component of the force is called thetangential force.

Part A

What is , the magnitude of the tangential force that acts on the pole due to the tension in the rope?

Express your answer in terms of and .

ANSWER:

Correct

When using the tangential force method, you calculate the torque using the equation,

Ft

T θ

= Ft cos(θ)T

τ = dFt



where is the distance from the pivot to the point where the force is applied. The sign of the torque can be determinedby checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotationimplies positive torque).

Part B

What is the magnitude of the torque on the pole, about point A, due to the tension in the rope?

Express your answer in terms of , , and .

ANSWER:

Correct

Moment arm method

The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel tothe force, running through the point at which the force is applied, and extending off to infinity in either direction. Youmay shift the force vector anywhere you like along this line without changing the torque, provided you do not changethe direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point wherethe displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called themoment arm.

For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at apoint directly above the point A in the figure. The moment arm of the force is the distance between the pivot and the tailof the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force,and the sign of the torque is again determined by the sense of the rotation of the pole it would cause.

Part C

Find , the length of the moment arm of the force.

Express your answer in terms of and .

ANSWER:

d

τ

T L θ

= τ TLcos(θ)

Rm

L θ



Correct

To calculate the torque using the moment arm method, use the equation,

where is the moment arm perpendicular to the applied force.

Part D

Find the magnitude of the torque on the pole, about point A, due to the tension in the rope.


ANSWER:

Correct

For this problem, the two methods of finding torque involved nearly the same of amount of algebra, and eithermethod could be used. Of course, both methods lead to the same final result.

Now consider a woman standing on the ball of her foot as shown . A normal force of magnitude acts upward on theball of her foot. The Achilles' tendon is attached to the backof the foot. The tendon pulls on the small bone in the rear ofthe foot with a force . This small bone has a length , andthe angle between this bone and the Achilles' tendon is .The horizontal displacement between the ball of the foot andthe point P is .

Part E

Suppose you were asked to find the torque about point P due to the normal force in terms of given quantities.Which method of finding the torque would be the easiest to use?

ANSWER:

= Rm cos(θ)L

τ = FRmRm

τ

T L θ

= τ cos(θ)TL

N

F xϕ

D

N



Correct

Part F

Find , the torque about point P due to the normal force.

Express your answer in terms of and any of the other quantities given in the figure.

ANSWER:

Correct

Part G

Suppose you were asked to find the torque about point P due to the force of magnitude in the Achilles' tendon.Which of the following statements is correct?

ANSWER:

Correct

Part H

Find , the torque about point P due to the force applied by the Achilles' tendon.


ANSWER:

Correct

tangential force methodmoment arm method

τN

N

= τN −ND

F

The tangential force method must be used.The moment arm method must be used.Either method may be used.Neither method can be used.

τF

F ϕ x

= τF sin(ϕ)Fx



PSS 11.1 Equilibrium of a Rigid Body

Learning Goal:

To practice ProblemSolving Strategy 11.1 Equilibrium of a Rigid Body.

A horizontal uniform bar of mass 2.8 and length 3.0 is hung horizontally on two vertical strings. String 1 isattached to the end of the bar, and string 2 is attached a distance 0.65 from the other end. A monkey of mass 1.4

walks from one end of the bar to the other. Find the tension in string 1 at the moment that the monkey ishalfway between the ends of the bar.

ProblemSolving Strategy 11.1 Equilibrium of a Rigid Body

IDENTIFY the relevant concepts: The first and second conditions for equilibrium are useful whenever there is a rigid body that is not rotating and notaccelerating in space.

SET UP the problem using the following steps:

1. Draw a sketch of the physical situation, including dimensions, and select the body in equilibrium to beanalyzed.

2. Draw a freebody diagram showing the forces acting on the selected body and no others. Do not includeforces exerted by this body on other bodies. Be careful to show correctly the point at which each forceacts.

3. Choose coordinate axes, and specify a positive direction of rotation for torques. Represent forces in termsof their components with respect to the axes you have chosen.

4. In choosing a point about which to compute torques, note that if a force has a line of action that goesthrough a particular point, the torque of the force with respect to that point is zero. You can often eliminateunknown forces or components from the torque equation by a clever choice of point for your calculation.The body doesn't actually have to be pivoted about an axis through the chosen point.

EXECUTE the solution as follows:

1. Write equations expressing the equilibrium conditions. Keep in mind that , , and are always separate equations; never add x and y components in a single equation. Also,

recall that when a force is represented in terms of its components, you can compute the torque of thatforce by finding the torque of each component separately, each with its appropriate lever arm and sign,and adding the results.

2. You always need as many equations as you have unknowns. Depending on the number of unknowns, youmay need to compute torques with respect to two or more axes to obtain enough equations. Often, thereare several equally good sets of force and torque equations for a particular problem.

EVALUATE your answer: A useful way to check your results is to rewrite the second condition for equilibrium, , using a differentchoice of origin. If you've done everything correctly, you'll get the same answers using this new choice of origin as youdid with your original choice.

IDENTIFY the relevant concepts

The rigid body under consideration is the bar. Because there is no indication to suggest otherwise, it is reasonable toassume that the bar remains at rest as the monkey walks from one end of the bar to the other. This means that theconditions for equilibrium hold at any time, including the instant at which the monkey is halfway between its ends.Thus, the above strategy can be applied.

SET UP the problem using the following steps

kg mm

kg T1

∑ = 0Fx ∑ = 0Fy

∑ = 0τz

∑ = 0τz



Part A

Which of the following diagrams correctly represents the forces acting on the bar at the moment described in theproblem introduction? (Note that the forces are not necessarily drawn to scale.)

ANSWER:

Correct

Four vertical forces act upon the bar: the upward tension in both strings, the weight of the bar, and thedownward force exerted by the monkey. Note that whereas the weight of the bar is represented as a forceacting at the center of gravity of the bar (which coincides with the center of mass the bar), the force exertedby the monkey acts at the point of contact between the monkey and the bar, which is a point on the topsurface of the bar halfway between its ends. This may be clearer if you draw your own freebody diagram. Besure to label all the forces in your diagram.

Part B

One of the equilibrium conditions that should be applied in this problem requires that you write a torque equation forthe bar. Which of the following choices of reference point for calculating torques would lead to a torque equation inwhich the only unknown quantity is ?

Hint 1. The point of reference for calculating torques

In equilibrium problems, the choice of reference point for calculating torques is completely arbitrary. Ingeneral, it helps to pick the point so as to simplify the calculations as much as possible. For example, if aforce has a line of action that goes through a particular point, the torque of the force with respect to thatpoint is zero and its contribution would not appear in the torque equation .

diagram Adiagram Bdiagram Cdiagram D

T1

∑ = 0τz



ANSWER:

CorrectIf you choose the point of attachment of string 2 as the reference point for calculating torques, the torqueequation will not depend on , the (unknown) tension in string 2, because the line of action of goesthrough the point of attachment of string 2, and the lever arm of the force in this case would be zero.Eliminating terms from the torque equation, particularly those involving unknown quantities, generally makes iteasier to solve. Other choices of reference point would also work, but they might lead to more complicatedmathematics. Keep in mind that once you make your choice, you must use the same point to calculate all thetorques on a body.

Now, choose a set of coordinate axes, and specify a positive direction of rotation for torques.

EXECUTE the solution as follows

Part C

Find , the magnitude of the force of tension in string 1, at the moment that the monkey is halfway between theends of the bar.

Express your answer in newtons using three significant figures.

You did not open hints for this part.

ANSWER:

EVALUATE your answer

Part D

This question will be shown after you complete previous question(s).

Forces on a Bridge

the center of mass of the barthe point of attachment of string 1the point of attachment of string 2the end of the bar closest to string 2

T2 T2

T1

= T1 N



A bridge, constructed of 11 beams of equal length and negligible mass, supports an object of mass as shown.Real bridges of this sort have steel rockers at the ends(check one out sometime), these assure that the supportforces on the bridge are vertical even when it expands orcontracts thermally.

Throughout this problem, use for the magnitude of theacceleration due to gravity.

Part A

Find , the vertical force that pier P exerts on the left end of the bridge.

Express the vertical force at P in terms of and .

Hint 1. How to approach the problem

View the entire bridge as the system, and find the torque about a pivot chosen to eliminate the torque fromone of the unknowns.

Hint 2. A suggested origin for determining torques

Find the sum of the torques about the right pier Q of the bridge. Remember that counterclockwisetorque is positive.

Answer in terms of , , , and .

ANSWER:

Hint 3. Find the sum of the torques

The sum of the torques is _________.

ANSWER:

L M

g

FP

M g

Σ τQ

M g L FP

= ∑ τQ 2MgL − 3 LFP

positivenegativezero



ANSWER:

Correct

Part B

Find , the vertical force that pier Q exerts on the right end of the bridge.

Express the vertical force at Q in terms of and .

Hint 1. Two different methods

One way to solve this problem is to add up the torques about the left pier P using the method analogous toPart A.Another is to use Newton's 2nd law to find the total vertical force on the bridge, .

Answer in terms of , , , and . Take the upward direction to be positive.

ANSWER:

Hint 2. Total force

Because this is a statics problem (i.e., nothing is moving), the total force is zero.

ANSWER:

Correct

Part C

Assuming that the bridge segments are free to pivot at each intersection point, what is the tension in thehorizontal segment directly above the point where the object is attached? If you find that the horizontal segmentdirectly above the point where the object is attached is being stretched, indicate this with a positive value for . Ifthe segment is being compressed, indicate this with a negative value for .

Express the tension in terms of and .

= FP Mg23

FQ

M g

Ftotal

FP FQ M g

= Ftotal + − MgFP FQ

Ftotal

= FQ Mg13

T

TT

M g



Hint 1. One way to start

Since the bridge and all segments of it are static, the sum of the torques acting on any portion of the bridgeyou choose is zero for any pivot point you may choose. See if you can find a rigid portion of the bridge anda wisely chosen pivot to which you can apply this powerful fact.

Hint 2. Find the torque about a special point

Consider the triangular portion shown in bold and let x be the pivot. (This choice eliminates the torques dueto the tensions in the beams that attach at pointx.) Find the torques on this lefthand triangle(which can be considered a solid piece becauseof the connections). Remember thatcounterclockwise torque is positive. Assume thatthe horizontal segment above is beingstretched, so that the force that the tension in thissegment exerts on the bold triangle is directed tothe right.

Express the torque in terms of , , and .

Hint 1. Forces required to find torque

The only forces not passing through this pivot point are and .

Hint 2. Torque from tension

Find the magnitude of the torque about this pivot point due to the tension in the top member. Themoment arm must be found from trigonometry. (The triangle is equilateral, with angles.)

Answer in terms of and .

ANSWER:

M

T L FP

T FP

τT

60∘

L T



ANSWER:

Hint 3. How to find from the torque

Since must equal zero (for a statics problem), you can solve the equation you found in the previoushint for and then substitute for using the expression you found in Part A.

ANSWER:

Correct

The negative value of the tension shows that the segment is actually under a compressive load.

This should be clear if you think of the bridge as beingcomposed of the two shaded sections shown: The compressive load in the top beam prevents the load fromfalling into the river.

>By repeating this process, you should be able to solve for the tension or compression of every segment inthe bridge. You could then go to the Machinist's Handbook or an equivalent reference and find out what crosssectional shape of each segment would be sufficient to bear the maximum anticipated load times the requiredsafety factor.

Score Summary:Your score on this assignment is 75.0%.You received 30 out of a possible total of 40 points.

= τTTL⋅ 3√

2

= ∑ τx − L−TL⋅ 3√

2 FP

T

∑ τxT FP

= T− Mg2

3

.866

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Assignment 15 for Mastering Physics