Title Homework6_Chap5.pdf 1.6 MB 47
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Homework6_Chap5

Due: 11:45pm on Thursday, November 20, 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 bollards 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 bollards (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 .

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

Part A

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

Hint 1. Find the forces in the x direction

What is the component of in the x direction (call it simply ), in terms of and ?

Positive x is to the right in the diagram.

F

AB
FAB

θ

F

AB Fx FAB θ

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Hint 2. Using algebra and trigonometry

Recalling that , you can find the x component of the net force due to and . How does this relate to the force provided by the third

rope?

Correct

Atwood Machine Special Cases

An Atwood machine consists of two blocks (of masses and ) 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, , to be positive.

= Fx − cos(θ)FAB

AB F

cos(θ)FAB
2 cos(θ)FAB
2 sin(θ)FAB

sin(θ)FAB

m1 m2

g

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The box begins to slide once the component of gravity acting parallel to the board 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?

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 (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 , 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 must be less than the force of static friction which implies

, as expected.

Part D

Consider a problem in which a car of mass is on a road tilted at an angle . The normal force

Fg

The force of kinetic friction is smaller than that of maximum static friction, but remains the same.

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

Once the box is moving, is larger than the force of maximum static friction.

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

reduces the friction.

Fg

Fg

Fg

Fg

nμs
nμs

nμk nμs
<μk μs

M θ

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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 ( , 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 from .

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 with angular velocity ,

.

The main point of the problem is to compute the acceleration using geometric arguments.

is found using

n = Mg
n = Mg cos(θ)

n = Mg
cos(θ)

∑ = MF

a

θ = 0

n

∑ = mF

a

R ω

(t) = R[cos(ωt) + sin(ωt) ]r→ î ĵ
= Rcos(ωt) + Rsin(ωt)î ĵ

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Correct

Part B

Find the magnitude of the horizontal force necessary to drag block to the left at constant speed if is held at rest (figure (b)).

Correct

Problem 5.120

A small remote-control car with a mass of 1.70 moves at a constant speed of = 12.0 in a vertical circle inside a hollow metal cylinder that has a radius of 5.00

.

F

B A

= 1.87 F N

kg v m/s m

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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)?

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)?