Ch 05 HW

Ch 05 HW Due: 11:59pm on Thursday, September 25, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy

Boat Statics A boat owner pulls her boat into the dock shown, where there are six capstans to which to tie the boat. She has three ropes. She can tie the boat from the boat's center (A) to any of the capstans (B through G) along the dotted arrows shown. Suppose the owner has tied three ropes: one rope runs to A from B, another to A from D, and a final rope from A to F. The ropes are tied such that FAB = FAD .

The following notation is used in this problem: When a question refers to, for example, F ⃗AB , this quantity is taken to mean the force acting on the boat due to the rope running to A from B, while FAB is the magnitude of that force.

Part A What is the magnitude of the force provided by the third rope, in terms of θ?

Hint 1. Find the forces in the x direction What is the component of F ⃗AB in the x direction (call it simply Fx ), in terms of FAB and θ? Positive x is to the right in the diagram. ANSWER:

Fx = −FAB cos(θ) Hint 2. Using algebra and trigonometry Recalling that FAB

= FAD , you can find the x component of the net force due to F ⃗AB and F ⃗AD . How does this

relate to the force provided by the third rope? ANSWER:

FAB cos(θ) 2FAB cos(θ) 2FAB sin(θ) FAB sin(θ) Correct

Atwood Machine Special Cases An Atwood machine consists of two blocks (of masses m 1 and m 2 ) tied together with a massless rope that passes over a fixed, perfect (massless and frictionless) pulley. In this problem you'll investigate some special cases where physical variables describing the Atwood machine take on limiting values. Often, examining special cases will simplify a problem, so that the solution may be found from inspection or from the results of a problem you've already seen. For all parts of this problem, take upward to be the positive direction and take the gravitational constant, g, to be positive.

Part A Consider the case where m 1 and m 2 are both nonzero, and m 2

> m 1 . Let T1 be the magnitude of the tension in the

rope connected to the block of mass m 1 , and let T2 be the magnitude of the tension in the rope connected to the block of mass m 2 . Which of the following statements is true? ANSWER:

T1 is always equal to T2 . T2 is greater than T1 by an amount independent of velocity. T2 is greater than T1 but the difference decreases as the blocks increase in velocity. There is not enough information to determine the relationship between T1 and T2 . Correct

Part B

Now, consider the special case where the block of mass m 1 is not present. Find the magnitude, T , of the tension in the rope. Try to do this without equations; instead, think about the physical consequences.

Hint 1. How to approach the problem If the block of mass

m 1 is not present, and the rope connecting the two blocks is massless, will the motion of the

block of mass m 2 be any different from free fall?

Hint 2. Which physical law to use Use Newton's 2nd law on the block of mass

m2.

ANSWER:

T= 0 Correct

Part C For the same special case (the block of mass m 1 not present), what is the acceleration of the block of mass m 2 ? Express your answer in terms of g , and remember that an upward acceleration should be positive. ANSWER:

a 2 = -9.80 Correct

Part D Next, consider the special case where only the block of mass m 1 is present. Find the magnitude, T , of the tension in the rope. ANSWER:

T= 0 Correct

Part E For the same special case (the block of mass m 2 not present) what is the acceleration of the end of the rope where the block of mass m 2 would have been attached?

Express your answer in terms of g , and remember that an upward acceleration should be positive. ANSWER:

a 2 = 9.80 Correct

Part F Next, consider the special case

m 1 = m 2 = m. What is the magnitude of the tension in the rope connecting the two

blocks? Use the variable m in your answer instead of m 1 or

m2.

ANSWER:

T = mg Correct

Part G For the same special case (m 1

= m 2 = m), what is the acceleration of the block of mass m 2 ?

ANSWER:

a2 = 0 Correct

Part H Finally, suppose

m 1 →∞, while m 2 remains finite. What value does the the magnitude of the tension approach?

Hint 1. Acceleration of block of mass

m1

As m 1 becomes large, the finite tension T will have a neglible effect on the acceleration, a 1 . If you ignore T , you can pretend the rope is gone without changing your results for a 1 . As m 1 →∞, what value does a 1 approach? ANSWER:

a 1 = -9.80 Hint 2. Acceleration of block of mass

m2

As m 1 →∞, what value will the acceleration of the block of mass m 2 approach? ANSWER:

a 2 = 9.80 Hint 3. Net force on block of mass m2 What is the magnitude Fnet of the net force on the block of mass m 2 . Express your answer in terms of T ,

m 2 , g, and any other given quantities. Take the upward direction to

be positive. ANSWER:

Fnet = T − m 2 g

ANSWER:

T = 2m 2 g Correct Imagining what would happen if one or more of the variables approached infinity is often a good way to investigate the behavior of a system.

PhET Tutorial: The Ramp Learning Goal: To understand how the forces exerted on an object on an inclined plane affect the object’s motion. For this tutorial, use the PhET simulation The Ramp. This simulation allows you to place a variety of objects on an inclined ramp and look at the resulting forces and motion.

Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file. Choose to

run or open it. Select an object to place on the ramp by clicking on any object under the Choose Object section of the right panel. To change the ramp angle, you can adjust the Ramp Angle slider bar in the right panel or you can click on the ramp and drag it up or down. To turn off friction, you can click on the Frictionless option that is above the Position setting. You can have the person pushing on the object by setting an Applied Force that is nonzero (you can type in a value in the Applied Force box on the left, drag the big vertical slider bar to the left of the parallel-force graph, or you can click and drag on the object). While the simulation is running, a graph shows the parallel forces (i.e., the components of the forces along the ramp) as a function of time. Feel free to play around with the simulation. When you are done, click Reset before beginning Part A.

Part A The first thing you will investigate is static friction. The force of static friction is the parallel force exerted on a stationary object by the ramp. This force is always directed opposite the direction the object would slide if there were no friction. Select the crate as the object for the ramp. Then, slowly increase the ramp angle. The individual forces acting on the crate are shown. They’ll look something like this:

Watch what happens to the force of friction (the red vector in the picture or the red plot in the graph) before the crate starts to slide down the ramp. As the ramp angle increases, the force of static friction ANSWER: increases. remains the same. decreases.

Correct In order for the crate to remain at rest, the force of static friction must be equal in magnitude to the component of the force of gravity parallel to the ramp. As the ramp angle increases, this component of the force of gravity increases.

Part B With the crate stationary on a horizontal ramp, the force of static friction is ANSWER: directed to the left. zero. directed to the right.

Correct Since the force of gravity is vertical, it has no component parallel to the horizontal ramp. This means that there is no force along the ramp that friction has to oppose.

Part C What is the maximum ramp angle that still allows the crate to remain at rest? (Make sure the coefficient of friction is 0.7.) Express your answer to the nearest degree.

Hint 1. How to approach the problem Slowly increase the ramp angle, and look at the value of the angle once the crate begins to slide. Your graph will look something like this, and the slipping occurs when the red curve flattens out:

ANSWER:

θ = 35

∘

Correct The crate accelerates when the component of the force of gravity along the ramp is greater than the maximum force of static friction. The component of the force of gravity along the ramp is mg sin(θ) , where mg is the weight of the

θ is the ramp angle. The maximum force of static friction is µn = µmg cos(θ) , where n is the normal force and µ is the coefficient of static friction. The maximum angle can be determined by equating these two forces, which gives sin(θ) = µ cos(θ) , or θ = atan(µ) = 35∘ . object and

Part D In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of friction is less than 0.7, what happens to this angle? (Note that you can adjust the coefficient of friction by clicking on the More Features tab near the top of the window and then using the slider bar in the right panel.)

Hint 1. How to approach the problem The maximum force of static friction is given by fs, max

= µn, where n is the normal force acting on the crate

and µ is the coefficient of static friction. Think about what happens to this maximum force when the coefficient of friction decreases. ANSWER: The maximum angle remains the same. The maximum angle increases. The maximum angle decreases.

Correct Since the maximum force of static friction decreases due to the smaller coefficient of friction, a smaller component of the force of gravity along the ramp is required to make the crate accelerate.

Part E The mass of the crate can also be adjusted by clicking on the More Features tab and then using the slider bar in the right panel. How does the maximum angle for which the crate can remain at rest on the ramp depend on the mass of the crate?

Hint 1. How to approach this part Think about how the maximum force of static friction depends on the mass of the object, and compare that to how the force of gravity depends on the mass. Keep in mind that the file cabinet will begin slipping when the maximum force of static friction is equal to the component of the force of gravity along the ramp.

ANSWER: The maximum angle decreases as the mass increases. The maximum angle increases as the mass increases. The maximum angle does not depend on the mass.

Correct Although the normal force and thus the maximum force of static friction increases with increasing mass, the component of the force of gravity parallel to the ramp increases at the same rate. The maximum angle is therefore independent of the mass.

The next three parts deal with the forces on the crate while the person is pushing on it.

Part F Click Reset, and then adjust the ramp angle to 15∘ . Compare the force of static friction when there is no applied force to when there is an applied force of 100 N (pushing up the ramp). How do the two forces of static friction compare? ANSWER: The force of static friction when there is no applied force is less than the case when there is an applied force. The force of static friction when there is no applied force is greater than the case when there is an applied force. The force of static friction when there is no applied force is equal to the case when there is an applied force.

Correct In order for the crate to be stationary, the sum of the applied force and the force of static friction must have the same magnitude as the component of gravity parallel to the ramp (so that the net force is zero). Thus, the force of friction decreases by 100 N when the applied force goes from zero to 100 N.

Part G For a stationary crate (with a coefficient of friction of 0.7) on the 15∘ ramp, can the force of static friction ever be zero?

Hint 1. How to approach the problem using the simulation Adjust the applied force while watching the force of static friction in the parallel-force graph. Determine if you can make the force of static friction go to zero.

Hint 2. How to approach the problem using physics reasoning Think about what would be required for the cabinet to be at rest on a frictionless surface. In which direction is the force of gravity? In which direction would the applied force have to be exerted to balance the force of gravity

along the ramp? ANSWER: No Yes, but only for a specific applied force directed up the ramp. Yes, but only for a specific applied force directed down the ramp.

Correct When the applied force has the same strength as the component of the force of gravity parallel to the ramp, then the net force on the crate would be zero if the surface were frictionless. This means friction doesn’t have to help, and so the force of friction is zero (this is similar to Part B, where the crate is sitting on a horizontal surface with no applied forces). Notice that if the applied force is greater than this value, the force of static friction is directed down the ramp.

Part H Slowly adjust the applied force (pushing both up and down the ramp) until the crate begins to move. Determine the minimum strength of the pushing force needed to accelerate the crate up the ramp and the minimum strength of the pushing force needed to accelerate the crate down the ramp. How do these two minimum strengths compare to each other? ANSWER: The minimum push needed to get the crate to slide up the ramp is greater than that to get the crate to slide down the ramp. The minimum push needed to get the crate to slide up the ramp is less than that to get the crate to slide down the ramp. The minimum push needed to get the crate to slide up the ramp is the same as that to get the crate to slide down the ramp.

Correct When pushing up the ramp, the applied force is opposing not only the static friction force (which is directed down the ramp) but also the component of the force of gravity along the ramp. When pushing down the ramp, the applied force is being helped by the component of the force of gravity down the ramp in opposing the friction force (directed up the ramp), and so doesn’t need to be as strong.

This should be consistent with your own experiences trying to move things up and down slopes.

PhET Interactive Simulations University of Colorado http://phet.colorado.edu

Two Masses, a Pulley, and an Inclined Plane Block 1, of mass m 1 , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m 2 , as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is µ .

Part A Find the ratio of the masses m 1 /m 2 . Express your answer in terms of some or all of the variables a, µ, and θ, as well as the magnitude of the acceleration due to gravity g .

Hint 1. Draw a free-body diagram Which figure depicts the correct free-body diagrams for the blocks in this problem? Figure a)

Figure b)

Figure c)

ANSWER: a b c

Hint 2. Apply Newton's 2nd law to block 2 in the direction parallel to the incline What is Newton's 2nd law for block 2 in the direction parallel to the incline? (Assume the positive direction is going up the incline.) Express your answer in terms of m 2 , gravity g .

T , f, and θ, as well as the magnitude of the acceleration due to

ANSWER:

m 2 a = T − f − m 2 gsin(θ) Hint 3. Find an expression for the friction force What is the magnitude f of the friction force acting on block 2? ANSWER:

f = n/µ f = µ/n f = µn f ⃗ = µn⃗

Hint 4. Find the normal force

By applying Newton's 2nd law to block 2 in the direction perpendicular to the incline determine the magnitude of the normal force n. Express your answer in terms of m 2 and θ, as well as the magnitude of the acceleration due to gravity g . ANSWER:

n = m 2 gcos(θ) Hint 5. Apply Newton's 2nd law to block 1 in the vertical direction Write an expression for Newton's 2nd law in the vertical direction for block 1. Take the positive direction to point downward. Express your answer in terms of the variables due to gravity g.

m 1 and/or T , as well as the magnitude of the acceleration

ANSWER:

m1 a = m1 g − T Hint 6. Solve for the unknown tension T Using your result from the previous hint, express

T in terms of g , a, and m 1 .

ANSWER:

T = m 1 (g − a) Hint 7. Putting it all together By applying Newton's law to both block 1 and block 2, as you did in Hints 2 and 4, you found two equations where the masses m 1 , m 2 , the tension T , and the acceleration a, all appear, along with g and θ. (Note that the friction force can be expressed in terms of the normal force, which, in turn, can be written as m 2 g cos(θ), as you found in Hint 3.) Choose one of the two equation and solve for T ; substitute this result into the other equation. You will then have an equation with factors of m 1 and m 2 . You can then deduce the ratio. ANSWER:

m 1 /m 2 = a+gsin(θ)+µgcos(θ) g−a

Correct

Contact Forces Introduced Learning Goal:

To introduce contact forces (normal and friction forces) and to understand that, except for friction forces under certain circumstances, these forces must be determined from: net Force = ma. Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch, they exert repulsive normal forces on each other, as well as frictional forces that resist their slipping relative to each other. These contact forces arise from a complex interplay between the electrostatic forces between the electrons and ions in the objects and the laws of quantum mechanics. As two surfaces are pushed together these forces increase exponentially over an atomic distance scale, easily becoming strong enough to distort the bulk material in the objects if they approach too close. In everyday experience, contact forces are limited by the deformation or acceleration of the objects, rather than by the fundamental interatomic forces. Hence, we can conclude the following: The magnitude of contact forces is determined by ∑ F ⃗ = ma⃗ , that is, by the other forces on, and acceleration of, the contacting bodies. The only exception is that the frictional forces cannot exceed µn (although they can be smaller than this or even zero).

Normal and friction forces Two types of contact forces operate in typical mechanics problems, the normal and frictional forces, usually designated by n and f (or Ffric , or something similar) respectively. These are the components of the overall contact force: n perpendicular to and f parallel to the plane of contact.

Kinetic friction when surfaces slide When one surface is sliding past the other, experiments show three things about the friction force (denoted fk ): 1. The frictional force opposes the relative motion at the point of contact, 2. fk is proportional to the normal force, and 3. the ratio of the magnitude of the frictional force to that of the normal force is fairly constant over a wide range of speeds.

The constant of proportionality is called the coefficient of kinetic friction, often designated µk . As long as the sliding continues, the frictional force is then

fk = µk n (valid when the surfaces slide by each other). Static friction when surfaces don't slide When there is no relative motion of the surfaces, the frictional force can assume any value from zero up to a maximum µs n, where µs is the coefficient of static friction. Invariably, µs is larger than µk , in agreement with the observation that when a force is large enough that something breaks loose and starts to slide, it often accelerates. The frictional force for surfaces with no relative motion is therefore

fs ≤ µs n (valid when the contacting surfaces have no relative motion). The actual magnitude and direction of the static friction force are such that it (together with other forces on the object) causes the object to remain motionless with respect to the contacting surface as long as the static friction force required does not exceed µs n. The equation fs = µs n is valid only when the surfaces are on the verge of sliding.

Part A When two objects slide by one another, which of the following statements about the force of friction between them, is true?

ANSWER: The frictional force is always equal to µk n. The frictional force is always less than

µk n.

The frictional force is determined by other forces on the objects so it can be either equal to or less than µk n .

Correct

Part B When two objects are in contact with no relative motion, which of the following statements about the frictional force between them, is true? ANSWER: The frictional force is always equal to µs n. The frictional force is always less than

µs n.

The frictional force is determined by other forces on the objects so it can be either equal to or less than µs n .

Correct For static friction, the actual magnitude and direction of the friction force are such that it, together with any other forces present, will cause the object to have the observed acceleration. The magnitude of the force cannot exceed µs n. If the magnitude of static friction needed to keep acceleration equal to zero exceeds µs n , then the object will

slide subject to the resistance of kinetic friction. Do not automatically assume that fs = µs n unless you are considering a situation in which the magnitude of the static friction force is as large as possible (i.e., when determining at what point an object will just begin to slip). Whether the actual magnitude of the friction force is 0, less than µs n, or equal to µs n depends on the magnitude of the other forces (if any) as well as the acceleration of the object through

∑ F ⃗ = ma⃗ .

Part C When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some point "break loose" and start to accelerate down the board. The box begins to slide once the component of gravity acting parallel to the board Fg just begins to exceeds the maximum force of static friction. Which of the following is the most general explanation for why the box accelerates down the board? ANSWER:

The force of kinetic friction is smaller than that of maximum static friction, but

Fg remains the same.

Once the box is moving, Fg is smaller than the force of maximum static friction but larger than the force of kinetic friction. Once the box is moving, Fg is larger than the force of maximum static friction.

When the box is stationary, Fg equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction.

Correct At the point when the box finally does "break loose," you know that the component of the box's weight that is parallel to the board just exceeds µs n (i.e., this component of gravitational force on the box has just reached a magnitude such that the force of static friction, which has a maximum value of µs n , can no longer oppose it.) For the box to then accelerate, there must be a net force on the box along the board. Thus, the component of the box's weight parallel to the board must be greater than the force of kinetic friction. Therefore the force of kinetic friction µk n must be less than the force of static friction µs n which implies µk < µs , as expected.

Part D Consider a problem in which a car of mass M is on a road tilted at an angle θ . The normal force Select the best answer. ANSWER:

n = Mg n = Mg cos(θ)

n=

Mg cos(θ)

is found using ∑ F ⃗ =

M a⃗

Correct The key point is that contact forces must be determined from Newton's equation. In the problem described above, there is not enough information given to determine the normal force (e.g., the acceleration is unknown). Each of the answer options is valid under some conditions (θ = 0, the car is sliding down an icy incline, or the car is going around a banked turn), but in fact none is likely to be correct if there are other forces on the car or if the car is accelerating. Do not memorize values for the normal force valid in different problems--you must determine n⃗ from

∑ F ⃗ = ma⃗ .

Centripetal Acceleration Explained Learning Goal: To understand that centripetal acceleration is the acceleration that causes motion in a circle. Acceleration is the time derivative of velocity. Because velocity is a vector, it can change in two ways: the length (magnitude)

can change and/or the direction can change. The latter type of change has a special name, the centripetal acceleration. In this problem we consider a mass moving in a circle of radius R with angular velocity ω,

r ⃗(t) = R[cos(ωt)^i + sin(ωt)^j ] = R cos(ωt)^i + R sin(ωt)^j . The main point of the problem is to compute the acceleration using geometric arguments.

Part A What is the velocity of the mass at a time t? You can work this out geometrically with the help of the hints, or by differentiating the expression for r ⃗(t) given in the introduction. Express this velocity in terms of R , ω, t , and the unit vectors ^ i and ^j .

Hint 1. Direction of the velocity What is the angle between r ⃗(t

= 0) and r ⃗(t) ? As shown in the figure, this angle is directly related to the direction of the velocity vector. Keep in mind that when t = 0 , r ⃗(t = 0) = R^ i. Express your result in terms of quantities given in the problem introduction.

ANSWER:

θ = ωt Hint 2. Speed What is v(t), the speed (magnitude of velocity) of the mass at time t? Express v(t) in terms of ω and R . ANSWER:

v(t) = ωR

ANSWER:

v⃗ (t) = Rω(−sin(ωt)^i + cos(ωt)^j ) Correct

Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch

when it crosses the positive x axis, an instant that corresponds to t = 0 . [Notice that when t = 0 , r ⃗(t = 0) = R^ i .] For the remainder of this problem, assume that the time t is measured from the moment you start timing the motion. Then the time − t refers to the moment a time t before you start your stopwatch.

Part B What is the velocity of the mass at a time − t ? Express this velocity in terms of R , ω, t , and the unit vectors ^ i and ^j . ANSWER:

v⃗ (−t) = Rωsin(ωt)^i + Rωcos(ωt)^j Correct

Part C What is the average acceleration of the mass during the time interval from − Express this acceleration in terms of R ,

t to t?

ω, t, and the unit vectors ^i and ^j .

Hint 1. Average acceleration The definition of average acceleration over the interval from t1 to t2 is

a⃗ avg [t1 , t2 ] =

v ⃗(t2)−v ⃗(t1) . t2−t1

ANSWER: ^ a⃗ avg [−t, t] = −2Rωsin(ωt)i t+t

Correct

Part D What is the magnitude of this acceleration in the limit of small t ? In this limit, the average acceleration becomes the instantaneous acceleration. Express your answer in terms of R and ω.

Hint 1. Expansion of

sin(x)

For small times t (or more precisely when

sin(ωt)?

ωt ≪ 1), what is the first term in the Taylor series expansion for

Express your answer in terms of ω and t.

Hint 1. Taylor series expansion The Taylor series expansion of sin(x) is

sin x = x −

x3 3!

+ x5! − 5

x7 7!

+ ….

ANSWER:

sin(ωt) = ωt Answer Requested

ANSWER:

a centripηl = ω2 R Answer Requested

Part E Consider the following statements: a. The centripetal acceleration might better be expressed as

−ω2 r ⃗(t) because it is a vector.

b. The magnitude of the centripetal acceleration is v2radial /R .

c. The magnitude of the centripetal acceleration is v2tangential /R .

d. A particle that is going along a path with local radius of curvature R at speed s experiences a centripetal acceleration

−s 2 /R.

e. If you are in a car turning left, the force you feel pushing you to the right is the force that causes the centripetal acceleration. In these statements vradial refers to the component of the velocity of an object in the direction toward or away from the origin of the coordinate system or the rotation axis. Conversely, vtangential refers to the component of the velocity perpendicular to vradial .

Identify the statement or statements that are false. ANSWER: a only b only c only d only e only b and e c and e d and e

Correct That's right; the true statements are therefore: a. The centripetal acceleration might better be expressed as −ω2 r ⃗ (t) because it is a vector. c. The magnitude of the centripetal acceleration is v2tangential /R .

d. A particle that is going along a path with local radius of curvature R at speed s experiences an acceleration

−s 2 /R.

There is so much confusion about centripetal force that you should probably ban this term from your vocabulary and thought processes. If you are in a car turning left, your centripetal acceleration is to the left (i.e., inward) and some real force must be applied to you to give you this acceleration--typically this would be provided by friction with the seat. The force you "feel" pushing you to the right is not a real force but rather a "fictitious force" that is present if you are in an accelerating coordinate system (in this case the car). It is best to stick to inertial (i.e., nonaccelerating) coordinate systems when doing kinematics and dynamics (i.e., F ⃗

= ma⃗ calculations).

Video Tutor: Ball Leaves Circular Track First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.

Part A Consider the video demonstration that you just watched. Which of the following changes would make it more likely for the ball to hit both the white can and the green can?

Hint 1. How to approach the problem To answer this question, you first have to decide whether changing the ball’s mass or its speed can change the path it follows after it leaves the track. Newton’s second law says that a net force acting on the ball will change the ball’s motion—that is, its speed and/or direction. Newton’s first law says that, in the absence of a net force, the ball’s motion won’t change. After the ball leaves the track, does a net force act on it? Draw a free-body diagram for the ball if you’re not sure. To hit the green can, the ball must continue following a curved path. What would be needed to make that happen?

ANSWER: Roll the ball faster. Use a ball that is heavier than the original ball. Roll the ball slower. Use a ball that is lighter than the original ball, but still heavier than an empty can. None of the above

Correct By Newton’s first law, after it has left the circular track, the ball will travel in a straight line until it is subject to a nonzero net force. Thus, the ball can only hit the white can, because that is the only can in the ball’s straight-line path.

± Banked Frictionless Curve, and Flat Curve with Friction A car of mass M = 1100kg traveling at 40.0km/hour enters a banked turn covered with ice. The road is banked at an angle θ , and there is no friction between the road and the car's tires. . Use g = 9.80m/s 2 throughout this problem.

Part A What is the radius

r of the turn if θ = 20.0∘ (assuming the car continues in uniform circular motion around the turn)?

Express your answer in meters.

Hint 1. How to approach the problem You need to apply Newton's 2nd law to the car. Because you do not want the car to slip as it goes around the curve, the car needs to have a net acceleration of magnitude v2 /r pointing radially inward (toward the center of the curve).

Hint 2. Identify the free-body diagram and coordinate system Which of the following diagrams represents the forces acting on the car and the most appropriate choice of coordinate axes?

ANSWER: Figure A Figure B Figure C

Hint 3. Calculate the normal force Find n, the magnitude of the normal force between the car and the road. Take the positive x axis to point horizontally toward the center of the curve and the positive y axis to point vertically upward. Express your answer in newtons.

Hint 1. Consider the net force The only forces acting on the car are the normal force and gravity. There must be a net acceleration in the horizontal direction, but because the car does not slip, the net acceleration in the vertical direction must be zero. Use this fact to find n.

Hint 2. Apply Newton's 2nd law to the car in the y direction Which equation accurately describes the equation for the net force acting on the car in the y direction? ANSWER:

∑ Fy ∑ Fy ∑ Fy ∑ Fy

= n cos θ + Mg = n sin θ + Mg = n cos θ − Mg = n sin θ − Mg

ANSWER:

n = 1.15×104 N

Hint 4. Determine the acceleration in the horizontal plane Take the y axis to be vertical and let the x axis point horizontally toward the center of the curve. By applying ∑ Fx = Ma in the horizontal direction, determine a, the magnitude of the acceleration, using your result for the normal force. Express your answer in meters per second squared.

Hint 1. Apply Newton's 2nd law to the car in the x direction Which equation accurately describes the equation for the net force acting on the car in the x direction? ANSWER:

∑ Fx = n cos θ ∑ Fx = n sin θ

∑ F x = n cos θ + Mv r ∑ F x = n cos θ − Mv r

2 2

ANSWER:

a = 3.57 m/s 2

ANSWER:

r = 34.6 m Correct

Part B

= 0) and that the ice has melted, so that there is a coefficient of static friction µ between the road and the car's tires. What is µmin , the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 40.0km/hour and that the radius of the curve is given by the value you found for r in Part A. Now, suppose that the curve is level (θ

Express your answer numerically.

Hint 1. How to approach the problem You need to apply Newton's 2nd law to the car. Because you do not want the car to slip as it goes around the curve, the car needs to have a net acceleration of magnitude v2 /r pointing radially inward (toward the center of the curve).

Hint 2. Identify the correct free-body diagram Which of the following diagrams represents the forces acting on the car as it goes around the curve? Ffr represents the friction force.

ANSWER: Figure A Figure B Figure C Figure D

Hint 3. Calculate the net force What is the net force Fnet that acts on the car? Express your answer in newtons.

Hint 1. How to determine the net force Newton's 2nd law tells you that

∑ F ⃗ = ma⃗ .

Because you do not want the car to slip as it goes around the curve, the car needs to have a net acceleration of magnitude v2 /r pointing radially inward (toward the center of the curve). ANSWER:

Fnet = 3920 N Hint 4. Calculate the friction force If the coefficient of friction were equal to µmin , what would be Ffr , the magnitude of the force provided by friction? Let

m be the mass of the car and g be the acceleration due to gravity.

Hint 1. Equation for the force of friction The force of friction is given by

Ffr = µn. Hint 2. Find the normal force What is the normal force n acting on the car? Enter your answer in newtons.

Hint 1. Acceleration in the y direction Because the car is neither sinking into the road nor levitating, you can conclude that a y ANSWER:

n = 1.08×104 N

ANSWER:

= 0.

F fr =

µmin Mg

Ffr = µmin Mg

ANSWER:

µmin = 0.364 Correct

Problem 5.56 An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope . The rope will break if the tension in it exceeds 2.50×104N , and our hero's mass is 81.3 kg .

Part A If the angle between the rope and the horizontal is θ = 11.2∘ , find the tension in the rope. ANSWER: 2050

N

Correct

Part B What is the smallest value the angle θ can have if the rope is not to break? ANSWER:

0.913

∘

Correct

Exercise 5.15 A load of bricks with mass m 1 = 14.6kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m 2 = 27.6kg is suspended from the other end of the rope, as shown in the figure. The system is released from rest. Use g = 9.80m/s 2 for the magnitude of the acceleration due to gravity.

Part A What is the magnitude of the upward acceleration of the load of bricks? ANSWER: 3.02

m/s 2

Correct

Part B What is the tension in the rope while the load is moving? ANSWER: 187

N

Correct

Enhanced EOC: Exercise 5.39 A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

floor are µs and µk , respectively. A woman pushes downward on the crate at an angle θ below the horizontal with a force F .⃗ You may want to review (

pages 146 - 154) .

For help with math skills, you may want to review: Vector Components Limits Involving Infinity For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Minimizing kinetic friction.

Part A What is the magnitude of the force vector

F ⃗ required to keep the crate moving at constant velocity?

Express your answer in terms of m , g, θ , and µk .

Hint 1. How to approach the problem Consider that the crate is moving at constant velocity. What does this imply about the forces on the crate in any direction? Start by drawing a free-body diagram showing all the forces acting on the crate, including a coordinate system of your choice. Resolve these forces into components in accordance with your coordinate system, being careful to ensure that the directions (i.e., signs) are correct. ANSWER: µ mg

F = cosθ−k µ sinθ k Correct

Part B If µs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of µs .

Express your answer in terms of θ .

Hint 1. How to approach the problem Since the woman cannot move the crate, no matter how large a force she applies, the variable F must not appear in the solution.

How can you eliminate the force from the set of equations you identified in Part A? How can you find the value of µs at which the horizontal component of the force exerted on the crate by the woman is always smaller in magnitude than the maximum frictional force exerted on the crate by the floor? ANSWER:

µs = cotθ Correct

Exercise 5.46 The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 m from the central shaft.

Part A Find the time of one revolution of the swing if the cable supporting a seat makes an angle of 30.0 ∘ with the vertical. ANSWER:

T = 6.19 s Correct

Part B Does the angle depend on the weight of the passenger for a given rate of revolution? ANSWER:

Yes. No.

Correct

Problem 5.73 Block A in the figure weighs 1.30 N and block B weighs 3.57N . The coefficient of kinetic friction between all surfaces is 0.304.

Part A Find the magnitude of the horizontal force moves with it (figure (a)).

F ⃗ necessary to drag block B to the left at constant speed if A rests on B and

ANSWER:

F = 1.48 N Correct

Part B Find the magnitude of the horizontal force (figure (b)). ANSWER:

F = 1.88 N

F ⃗ necessary to drag block B to the left at constant speed if A is held at rest

Correct

Problem 5.120 A small remote-control car with a mass of 1.56 kg moves at a constant speed of hollow metal cylinder that has a radius of 5.00m .

v = 12.0m/s in a vertical circle inside a

Part A What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point A (at the bottom of the vertical circle)? ANSWER: 60.2

N

Correct

Part B What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point B (at the top of the vertical circle)? ANSWER: 29.6

N

Correct Score Summary: Your score on this assignment is 96.6%. You received 13.52 out of a possible total of 14 points.

Ch 05 HW Due: 11:59pm on Thursday, September 25, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy

Boat Statics A boat owner pulls her boat into the dock shown, where there are six capstans to which to tie the boat. She has three ropes. She can tie the boat from the boat's center (A) to any of the capstans (B through G) along the dotted arrows shown. Suppose the owner has tied three ropes: one rope runs to A from B, another to A from D, and a final rope from A to F. The ropes are tied such that \(F_{\rm AB} = F_{\rm AD}\).

The following notation is used in this problem: When a question refers to, for example, \(\texttip{\vec{F}_{\rm AB}} {F_AB_vec}\), this quantity is taken to mean the force acting on the boat due to the rope running to A from B, while \ (\texttip{F_{\rm AB}}{F_AB}\) is the magnitude of that force.

Part A What is the magnitude of the force provided by the third rope, in terms of \(\texttip{\theta }{theta}\)?

Hint 1. Find the forces in the x direction What is the component of \(\texttip{\vec{F}_{\rm AB}}{F_AB_vec}\) in the x direction (call it simply \(\texttip{F_{\mit x}}{F_x}\)), in terms of \(\texttip{F_{\rm AB}}{F_AB}\) and \(\texttip{\theta }{theta}\)? Positive x is to the right in the diagram. ANSWER: \(\texttip{F_{\mit x}}{F_x}\) = \(-F_{\rm{AB}} {\cos}\left({\theta}\right)\)

Hint 2. Using algebra and trigonometry Recalling that \(F_{\rm AB} = F_{\rm AD}\), you can find the x component of the net force due to \(\texttip{\vec{F}_{\rm AB}}{F_AB_vec}\) and \(\texttip{\vec{F}_{\rm AD}}{F_AD_vec}\). How does this relate to the force provided by the third rope? ANSWER:

\(F_{\rm AB}\cos(\theta)\) \(2 F_{\rm AB}\cos(\theta)\) \(2 F_{\rm AB}\sin(\theta)\) \(F_{\rm AB}\sin(\theta)\)

Correct

Atwood Machine Special Cases An Atwood machine consists of two blocks (of masses \ (\texttip{m_{\rm 1}}{m_1}\) and \(\texttip{m_{\rm 2}}{m_2}\)) tied together with a massless rope that passes over a fixed, perfect (massless and frictionless) pulley. In this problem you'll investigate some special cases where physical variables describing the Atwood machine take on limiting values. Often, examining special cases will simplify a problem, so that the solution may be found from inspection or from the results of a problem you've already seen. For all parts of this problem, take upward to be the positive direction and take the gravitational constant, \(\texttip{g}{g}\), to be positive.

Part A Consider the case where \(\texttip{m_{\rm 1}}{m_1}\) and \(\texttip{m_{\rm 2}}{m_2}\) are both nonzero, and \(m_2 > m_1\). Let \(\texttip{T_{\rm 1}}{T_1}\) be the magnitude of the tension in the rope connected to the block of mass \(\texttip{m_{\rm 1}}{m_1}\), and let \(\texttip{T_{\rm 2}}{T_2}\) be the magnitude of the tension in the rope connected to the block of mass \(\texttip{m_{\rm 2}}{m_2}\). Which of the following statements is true? ANSWER: \(T_1\) is always equal to \(T_2\). \(T_2\) is greater than \(T_1\) by an amount independent of velocity. \(T_2\) is greater than \(T_1\) but the difference decreases as the blocks increase in velocity. There is not enough information to determine the relationship between \(T_1\) and \(T_2\).

Correct

Part B

Now, consider the special case where the block of mass \(m_1\) is not present. Find the magnitude, \(\texttip{T}{T}\), of the tension in the rope. Try to do this without equations; instead, think about the physical consequences.

Hint 1. How to approach the problem If the block of mass \(m_1\) is not present, and the rope connecting the two blocks is massless, will the motion of the block of mass \(\texttip{m_{\rm 2}}{m_2}\) be any different from free fall?

Hint 2. Which physical law to use Use Newton's 2nd law on the block of mass \(\texttip{m_{\rm 2}}{m_2}\).

ANSWER: \(\texttip{T}{T}\) = 0

Correct

Part C For the same special case (the block of mass \(m_1\) not present), what is the acceleration of the block of mass \(\texttip{m_{\rm 2}}{m_2}\)? Express your answer in terms of \(\texttip{g}{g}\), and remember that an upward acceleration should be positive. ANSWER: \(\texttip{a_{\rm 2}}{a_2}\) = -9.80

Correct

Part D Next, consider the special case where only the block of mass \(m_1\) is present. Find the magnitude, \(\texttip{T}{T}\), of the tension in the rope. ANSWER: \(\texttip{T}{T}\) = 0

Correct

Part E For the same special case (the block of mass \(m_2\) not present) what is the acceleration of the end of the rope where

the block of mass \(m_2\) would have been attached? Express your answer in terms of \(\texttip{g}{g}\), and remember that an upward acceleration should be positive. ANSWER: \(\texttip{a_{\rm 2}}{a_2}\) = 9.80

Correct

Part F Next, consider the special case \(m_1 = m_2 = m\). What is the magnitude of the tension in the rope connecting the two blocks? Use the variable \(\texttip{m}{m}\) in your answer instead of \(\texttip{m_{\rm 1}}{m_1}\) or \(\texttip{m_{\rm 2}}{m_2}\). ANSWER: \(\texttip{T}{T}\) = \(mg\)

Correct

Part G For the same special case (\(m_1 = m_2 = m\)), what is the acceleration of the block of mass \(\texttip{m_{\rm 2}}{m_2}\) ? ANSWER: \(\texttip{a_{\rm 2}}{a_2}\) = 0

Correct

Part H Finally, suppose \(m_1 \rightarrow \infty\), while \(\texttip{m_{\rm 2}}{m_2}\) remains finite. What value does the the magnitude of the tension approach?

Hint 1. Acceleration of block of mass \(\texttip{m_{\rm 1}}{m_1}\) As \(\texttip{m_{\rm 1}}{m_1}\) becomes large, the finite tension \(\texttip{T}{T}\) will have a neglible effect on the acceleration, \(\texttip{a_{\rm 1}}{a_1}\). If you ignore \(\texttip{T}{T}\), you can pretend the rope is gone without changing your results for \(\texttip{a_{\rm 1}}{a_1}\). As \(m_1 \rightarrow \infty\), what value does \(\texttip{a_{\rm 1}}{a_1}\) approach?

ANSWER: \(\texttip{a_{\rm 1}}{a_1}\) = -9.80

Hint 2. Acceleration of block of mass \(\texttip{m_{\rm 2}}{m_2}\) As \(m_1 \rightarrow \infty\), what value will the acceleration of the block of mass \(\texttip{m_{\rm 2}}{m_2}\) approach? ANSWER: \(\texttip{a_{\rm 2}}{a_2}\) = 9.80

Hint 3. Net force on block of mass \(\texttip{m_{\rm 2}}{m_2}\) What is the magnitude \(\texttip{F_{\rm net}}{F_net}\) of the net force on the block of mass \(\texttip{m_{\rm 2}}{m_2}\). Express your answer in terms of \(\texttip{T}{T}\), \(\texttip{m_{\rm 2}}{m_2}\), \(\texttip{g}{g}\), and any other given quantities. Take the upward direction to be positive. ANSWER: \(\texttip{F_{\rm net}}{F_net}\) = \(T-m_{2} g\)

ANSWER: \(\texttip{T}{T}\) = \(2m_{2} g\)

Correct Imagining what would happen if one or more of the variables approached infinity is often a good way to investigate the behavior of a system.

PhET Tutorial: The Ramp Learning Goal: To understand how the forces exerted on an object on an inclined plane affect the object’s motion. For this tutorial, use the PhET simulation The Ramp. This simulation allows you to place a variety of objects on an inclined ramp and look at the resulting forces and motion.

Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file. Choose to run or open it. Select an object to place on the ramp by clicking on any object under the Choose Object section of the right panel. To change the ramp angle, you can adjust the Ramp Angle slider bar in the right panel or you can click on the ramp and drag it up or down. To turn off friction, you can click on the Frictionless option that is above the Position setting. You can have the person pushing on the object by setting an Applied Force that is nonzero (you can type in a value in the Applied Force box on the left, drag the big vertical slider bar to the left of the parallel-force graph, or you can click and drag on the object). While the simulation is running, a graph shows the parallel forces (i.e., the components of the forces along the ramp) as a function of time. Feel free to play around with the simulation. When you are done, click Reset before beginning Part A.

Part A The first thing you will investigate is static friction. The force of static friction is the parallel force exerted on a stationary object by the ramp. This force is always directed opposite the direction the object would slide if there were no friction. Select the crate as the object for the ramp. Then, slowly increase the ramp angle. The individual forces acting on the crate are shown. They’ll look something like this:

Watch what happens to the force of friction (the red vector in the picture or the red plot in the graph) before the crate starts to slide down the ramp.

As the ramp angle increases, the force of static friction ANSWER: increases. remains the same. decreases.

Correct In order for the crate to remain at rest, the force of static friction must be equal in magnitude to the component of the force of gravity parallel to the ramp. As the ramp angle increases, this component of the force of gravity increases.

Part B With the crate stationary on a horizontal ramp, the force of static friction is ANSWER: directed to the left. zero. directed to the right.

Correct Since the force of gravity is vertical, it has no component parallel to the horizontal ramp. This means that there is no force along the ramp that friction has to oppose.

Part C What is the maximum ramp angle that still allows the crate to remain at rest? (Make sure the coefficient of friction is 0.7.) Express your answer to the nearest degree.

Hint 1. How to approach the problem Slowly increase the ramp angle, and look at the value of the angle once the crate begins to slide. Your graph will look something like this, and the slipping occurs when the red curve flattens out:

ANSWER: \(\texttip{\theta }{theta}\) = 35

\(\degree\)

Correct The crate accelerates when the component of the force of gravity along the ramp is greater than the maximum force of static friction. The component of the force of gravity along the ramp is \(mg~\rm sin (\theta)\), where \(mg\) is the weight of the object and \(\texttip{\theta }{theta}\) is the ramp angle. The maximum force of static friction is \(\mu n =\mu mg~\rm cos(\theta)\), where \(\texttip{n}{n}\) is the normal force and \(\texttip{\mu }{mu}\) is the coefficient of static friction. The maximum angle can be determined by equating these two forces, which gives \(\rm sin (\theta) =\) \(\mu \cos(\theta)\), or \(\theta = \rm atan(\mu)=35\degree.\)

Part D In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of friction is less than 0.7, what happens to this angle? (Note that you can adjust the coefficient of friction by clicking on the More Features tab near the top of the window and then using the slider bar in the right panel.)

Hint 1. How to approach the problem The maximum force of static friction is given by \(f_{\rm s, ~ max} = \mu n\), where \(\texttip{n}{n}\) is the normal force acting on the crate and \(\texttip{\mu }{mu}\) is the coefficient of static friction. Think about what happens to this maximum force when the coefficient of friction decreases. ANSWER: The maximum angle remains the same. The maximum angle increases. The maximum angle decreases.

Correct Since the maximum force of static friction decreases due to the smaller coefficient of friction, a smaller component of the force of gravity along the ramp is required to make the crate accelerate.

Part E The mass of the crate can also be adjusted by clicking on the More Features tab and then using the slider bar in the right panel. How does the maximum angle for which the crate can remain at rest on the ramp depend on the mass of the crate?

Hint 1. How to approach this part Think about how the maximum force of static friction depends on the mass of the object, and compare that to how the force of gravity depends on the mass. Keep in mind that the file cabinet will begin slipping when the maximum force of static friction is equal to the component of the force of gravity along the ramp. ANSWER: The maximum angle decreases as the mass increases. The maximum angle increases as the mass increases. The maximum angle does not depend on the mass.

Correct Although the normal force and thus the maximum force of static friction increases with increasing mass, the component of the force of gravity parallel to the ramp increases at the same rate. The maximum angle is therefore independent of the mass.

The next three parts deal with the forces on the crate while the person is pushing on it.

Part F Click Reset, and then adjust the ramp angle to \(15\degree\). Compare the force of static friction when there is no applied force to when there is an applied force of 100 \(\rm N\) (pushing up the ramp). How do the two forces of static friction compare? ANSWER: The force of static friction when there is no applied force is less than the case when there is an applied force. The force of static friction when there is no applied force is greater than the case when there is an applied force. The force of static friction when there is no applied force is equal to the case when there is an applied force.

Correct In order for the crate to be stationary, the sum of the applied force and the force of static friction must have the same magnitude as the component of gravity parallel to the ramp (so that the net force is zero). Thus, the force of friction decreases by 100 \(\rm N\) when the applied force goes from zero to 100 \(\rm N\).

Part G For a stationary crate (with a coefficient of friction of 0.7) on the \(15\degree\) ramp, can the force of static friction ever be zero?

Hint 1. How to approach the problem using the simulation Adjust the applied force while watching the force of static friction in the parallel-force graph. Determine if you can make the force of static friction go to zero.

Hint 2. How to approach the problem using physics reasoning Think about what would be required for the cabinet to be at rest on a frictionless surface. In which direction is the force of gravity? In which direction would the applied force have to be exerted to balance the force of gravity along the ramp? ANSWER: No Yes, but only for a specific applied force directed up the ramp. Yes, but only for a specific applied force directed down the ramp.

Correct When the applied force has the same strength as the component of the force of gravity parallel to the ramp, then the net force on the crate would be zero if the surface were frictionless. This means friction doesn’t have to help, and so the force of friction is zero (this is similar to Part B, where the crate is sitting on a horizontal surface with no applied forces). Notice that if the applied force is greater than this value, the force of static friction is directed down the ramp.

Part H Slowly adjust the applied force (pushing both up and down the ramp) until the crate begins to move. Determine the minimum strength of the pushing force needed to accelerate the crate up the ramp and the minimum strength of the pushing force needed to accelerate the crate down the ramp. How do these two minimum strengths compare to each other? ANSWER:

The minimum push needed to get the crate to slide up the ramp is greater than that to get the crate to slide down the ramp. The minimum push needed to get the crate to slide up the ramp is less than that to get the crate to slide down the ramp. The minimum push needed to get the crate to slide up the ramp is the same as that to get the crate to slide down the ramp.

Correct When pushing up the ramp, the applied force is opposing not only the static friction force (which is directed down the ramp) but also the component of the force of gravity along the ramp. When pushing down the ramp, the applied force is being helped by the component of the force of gravity down the ramp in opposing the friction force (directed up the ramp), and so doesn’t need to be as strong.

This should be consistent with your own experiences trying to move things up and down slopes.

PhET Interactive Simulations University of Colorado http://phet.colorado.edu

Two Masses, a Pulley, and an Inclined Plane

Block 1, of mass \(\texttip{m_{\rm 1}}{m_1}\), is connected over an ideal (massless and frictionless) pulley to block 2, of mass \ (\texttip{m_{\rm 2}}{m_2}\), as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude \(\texttip{a} {a}\) and that the coefficient of kinetic friction between block 2 and the plane is \(\texttip{\mu }{mu}\).

Part A Find the ratio of the masses \(m_1/m_2\). Express your answer in terms of some or all of the variables \(\texttip{a}{a}\), \(\texttip{\mu }{mu}\), and \(\texttip{\theta }{theta}\), as well as the magnitude of the acceleration due to gravity \(\texttip{g}{g}\).

Hint 1. Draw a free-body diagram Which figure depicts the correct free-body diagrams for the blocks in this problem? Figure a)

Figure b)

Figure c)

ANSWER: a b c

Hint 2. Apply Newton's 2nd law to block 2 in the direction parallel to the incline What is Newton's 2nd law for block 2 in the direction parallel to the incline? (Assume the positive direction is going up the incline.) Express your answer in terms of \(\texttip{m_{\rm 2}}{m_2}\), \(\texttip{T}{T}\), \(\texttip{f}{f}\), and \(\texttip{\theta }{theta}\), as well as the magnitude of the acceleration due to gravity \(\texttip{g}{g}\). ANSWER:

\(m_2 a\) = \(T-f-m_{2} g {\sin}\left({\theta}\right)\)

Hint 3. Find an expression for the friction force What is the magnitude \(\texttip{f}{f}\) of the friction force acting on block 2? ANSWER: \(f=n/\mu\) \(f=\mu/n\) \(f=\mu n\) \(\vec{f}=\mu\vec{n}\)

Hint 4. Find the normal force By applying Newton's 2nd law to block 2 in the direction perpendicular to the incline determine the magnitude of the normal force \(\texttip{n}{n}\). Express your answer in terms of \(\texttip{m_{\rm 2}}{m_2}\) and \(\texttip{\theta }{theta}\), as well as the magnitude of the acceleration due to gravity \(\texttip{g}{g}\). ANSWER: \(\texttip{n}{n}\) = \(m_{2} g {\cos}\left({\theta}\right)\)

Hint 5. Apply Newton's 2nd law to block 1 in the vertical direction Write an expression for Newton's 2nd law in the vertical direction for block 1. Take the positive direction to point downward. Express your answer in terms of the variables \(\texttip{m_{\rm 1}}{m_1}\) and/or \(\texttip{T}{T}\), as well as the magnitude of the acceleration due to gravity \(\texttip{g}{g}\). ANSWER: \(m_1a\) = \(m_{1} g-T\)

Hint 6. Solve for the unknown tension \(\texttip{T}{T}\) Using your result from the previous hint, express \(\texttip{T}{T}\) in terms of \(\texttip{g}{g}\), \(\texttip{a}{a}\), and \(\texttip{m_{\rm 1}}{m_1}\). ANSWER: \(\texttip{T}{T}\) = \(m_{1} \left(g-a\right)\)

Hint 7. Putting it all together By applying Newton's law to both block 1 and block 2, as you did in Hints 2 and 4, you found two equations where the masses \(\texttip{m_{\rm 1}}{m_1}\), \(\texttip{m_{\rm 2}}{m_2}\), the tension \(\texttip{T}{T}\), and the acceleration \(\texttip{a}{a}\), all appear, along with \(\texttip{g}{g}\) and \(\texttip{\theta }{theta}\). (Note that the friction force can be expressed in terms of the normal force, which, in turn, can be written as \(m_2g\cos(\theta)\), as you found in Hint 3.) Choose one of the two equation and solve for \(\texttip{T}{T}\); substitute this result into the other equation. You will then have an equation with factors of \(\texttip{m_{\rm 1}}{m_1}\) and \(\texttip{m_{\rm 2}}{m_2}\). You can then deduce the ratio.

ANSWER: \(m_1/m_2\) = \(\large{\frac{a+g {\sin}\left({\theta}\right)+{\mu}g {\cos}\left({\theta}\right)}{g-a}}\)

Correct

Contact Forces Introduced Learning Goal: To introduce contact forces (normal and friction forces) and to understand that, except for friction forces under certain circumstances, these forces must be determined from: net Force = ma. Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch, they exert repulsive normal forces on each other, as well as frictional forces that resist their slipping relative to each other. These contact forces arise from a complex interplay between the electrostatic forces between the electrons and ions in the objects and the laws of quantum mechanics. As two surfaces are pushed together these forces increase exponentially over an atomic distance scale, easily becoming strong enough to distort the bulk material in the objects if they approach too close. In everyday experience, contact forces are limited by the deformation or acceleration of the objects, rather than by the fundamental interatomic forces. Hence, we can conclude the following: The magnitude of contact forces is determined by \(\sum\vec{F}=m\vec{a}\), that is, by the other forces on, and acceleration of, the contacting bodies. The only exception is that the frictional forces cannot exceed \(\mu n\) (although they can be smaller than this or even zero).

Normal and friction forces Two types of contact forces operate in typical mechanics problems, the normal and frictional forces, usually designated by \ (\texttip{n}{n}\) and \(\texttip{f}{f}\) (or \(\texttip{F_{\rm fric}}{F_fric}\), or something similar) respectively. These are the components of the overall contact force: \(\texttip{n}{n}\) perpendicular to and \(\texttip{f}{f}\) parallel to the plane of contact.

Kinetic friction when surfaces slide When one surface is sliding past the other, experiments show three things about the friction force (denoted \(\texttip{f_{\rm k}} {f_k}\)): 1. The frictional force opposes the relative motion at the point of contact, 2. \(\texttip{f_{\rm k}}{f_k}\) is proportional to the normal force, and 3. the ratio of the magnitude of the frictional force to that of the normal force is fairly constant over a wide range of speeds.

The constant of proportionality is called the coefficient of kinetic friction, often designated \(\texttip{\mu _{\rm k}}{mu_k}\). As long as the sliding continues, the frictional force is then \(f_{\rm k} = \mu_{\rm k} n\) (valid when the surfaces slide by each other).

Static friction when surfaces don't slide When there is no relative motion of the surfaces, the frictional force can assume any value from zero up to a maximum \ (\mu_{\rm s}n\), where \(\texttip{\mu _{\rm s}}{mu_s}\) is the coefficient of static friction. Invariably, \(\texttip{\mu _{\rm s}} {mu_s}\) is larger than \(\texttip{\mu _{\rm k}}{mu_k}\), in agreement with the observation that when a force is large enough that something breaks loose and starts to slide, it often accelerates. The frictional force for surfaces with no relative motion is therefore \(f_{\rm s} \leq \mu_{\rm s}n\) (valid when the contacting surfaces have no relative motion). The actual magnitude and direction of the static friction force are such that it (together with other forces on the object) causes the object to remain motionless with respect to the contacting surface as long as the static friction force required does not exceed \(\mu_{\rm s} n\). The equation \(f_{\rm s} = \mu_{\rm s} n\) is valid only when the surfaces are on the verge of sliding.

Part A When two objects slide by one another, which of the following statements about the force of friction between them, is true? ANSWER: The frictional force\(\) is always equal to \(\mu_{\rm k}n\). The frictional force\(\) is always less than \(\mu_{\rm k}n\). The frictional force\(\) is determined by other forces on the objects so it can be either equal to or less than \(\mu_{\rm k}n\).

Correct

Part B When two objects are in contact with no relative motion, which of the following statements about \(\) the frictional force between them, is true? ANSWER: The frictional force\(\) is always equal to \(\mu_{\rm s}n\). The frictional force is always less than \(\mu_{\rm s}n\). The frictional force is determined by other forces on the objects so it can be either equal to or less than \(\mu_{\rm s}n\).

Correct For static friction, the actual magnitude and direction of the friction force are such that it, together with any other forces present, will cause the object to have the observed acceleration. The magnitude of the force cannot exceed \(\mu_{\rm s} n\). If the magnitude of static friction needed to keep acceleration equal to zero exceeds \(\mu_{\rm s} n\), then the object will slide subject to the resistance of kinetic friction. Do not automatically assume that \(f_{\rm s} = \mu_{\rm s}n\) unless you are considering a situation in which the magnitude of the static friction force is as large as possible (i.e., when determining at what point an object will just begin to slip). Whether the actual magnitude of the friction force is 0, less than \(\mu_{\rm s}n\), or equal to \(\mu_{\rm s} n\) depends on the magnitude of the other forces (if any) as well as the acceleration of the object through \(\sum \vec{F} = m \vec{a}\).

Part C When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some point "break loose" and start to accelerate down the board. The box begins to slide once the component of gravity acting parallel to the board \(\texttip{F_{\rm g}}{F_g}\) just begins to exceeds the maximum force of static friction. Which of the following is the most general explanation for why the box accelerates down the board? ANSWER: The force of kinetic friction is smaller than that of maximum static friction, but \(\texttip{F_{\rm g}}{F_g}\) remains the same. Once the box is moving, \(\texttip{F_{\rm g}}{F_g}\) is smaller than the force of maximum static friction but larger than the force of kinetic friction. Once the box is moving, \(\texttip{F_{\rm g}}{F_g}\) is larger than the force of maximum static friction. When the box is stationary, \(\texttip{F_{\rm g}}{F_g}\) equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction.

Correct At the point when the box finally does "break loose," you know that the component of the box's weight that is parallel to the board just exceeds \(\mu_{\rm s}n\) (i.e., this component of gravitational force on the box has just reached a magnitude such that the force of static friction, which has a maximum value of \(\mu_{\rm s} n\) , can no longer oppose it.) For the box to then accelerate, there must be a net force on the box along the board. Thus, the component of the box's weight parallel to the board must be greater than the force of kinetic friction. Therefore the force of kinetic friction \(\mu_{\rm k} n\) must be less than the force of static friction \(\mu_{\rm s} n\) which implies \(\mu_{\rm k}