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EXAM III - PHYS1321 University Physics I (W.K. Chu) 04/20/2012 Friday 1:00 2:00 PM

Student Name: ...

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[Answer all Questions Show necessary calculations for partial credits] [Total points 100]

1. [25 points] In Figure 1, a climber leans out against a vertical ice wall that has negligible

friction. Distance a is 0.9 m and distance L is 2.0 m. His center of mass is distance d = 1.0 m

from the feetground contact point. If he is on the verge of sliding, what is the coefficient of

static friction between feet and ground?

Figure 1

As shown in the free-body diagram, the forces on the climber consist of the normal forces 1NF on his

hands from the ground and 2NF on his feet from the wall, static frictional force ,sf and downward

gravitational force mg . Since the climber is in static equilibrium, the net force acting on him is zero.

Applying Newtons second law to the vertical

and horizontal directions, we have

net, 2

net, 1

0

0 .

x N s

y N

F F f

F F mg

In addition, the net torque about O (contact

point between his feet and the wall) must

also vanish:

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net 20 cos sinNO

mgd F L .

The torque equation gives

2 cos / sin cot /NF mgd L mgd L .

On the other hand, from the force equation we have 2N sF f and 1 .NF mg These expressions can

be combined to yield

2 1 cots N Nd

f F FL

.

On the other hand, the frictional force can also be written as 1s s Nf F , where s is the coefficient of

static friction between his feet and the ground. From the above equation and the values given in the

problem statement, we find s to be

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2. [25 points] In Figure 2, a cube of edge length L = 0.600 m and mass 450 kg is suspended by

a rope in an open tank of liquid of density 1030 kg/m3. Find (a) the magnitude of the total

downward force on the top of the cube from the liquid and the atmosphere, assuming

atmospheric pressure is 1.00 atm, (b) the magnitude of the total upward force on the bottom

of the cube, and (c) the tension in the rope. (d) Calculate the magnitude of the buoyant force

on the cube using Archimedes' principle. What relation exists among all these quantities?

Figure 2

(a) The pressure (including the contribution from the atmosphere) at a depth of htop = L/2

(corresponding to the top of the block) is

5 3 2 5

top atm top 1.01 10 Pa (1030 kg/m )(9.8 m/s )(0.300 m) 1.04 10 Pap p gh

where the unit Pa (pascal) is equivalent to N/m2. The force on the top surface (of area A = L2 = 0.36 m2) is

Ftop = ptop A = 3.75 104 N.

(b) The pressure at a depth of hbot = 3L/2 (that of the bottom of the block) is

5 3 2

bot atm bot

5

1.01 10 Pa (1030 kg/m )(9.8 m/s ) (0.900 m)

1.10 10 Pa

p p gh

where we recall that the unit Pa (pascal) is equivalent to N/m2. The force on the bottom surface is

Fbot = pbot A = 3.96 104 N.

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(c) Taking the difference Fbot Ftop cancels the contribution from the atmosphere (including any

numerical uncertainties associated with that value) and leads to

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bot top bot top( ) 2.18 10 NF F g h h A gL

which is to be expected on the basis of Archimedes principle. Two other forces act on the block: an

upward tension T and a downward pull of gravity mg. To remain stationary, the tension must be

2 3 3

bot top( ) (450 kg)(9.80 m/s ) 2.18 10 N 2.23 10 N.T mg F F

(d) This has already been noted in the previous part: 32.18 10 NbF , and T + Fb = mg.

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3. [25 points] In Figure 3, a 1.0 g bullet is fired into a 0.50 kg block attached to the end of a 0.60 m non-uniform rod of mass 0.50 kg. The blockrodbullet system then rotates in the

plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about that

axis at A is 0.060 kgm2. Treat the block as a particle. (a) What then is the rotational inertia

of the blockrodbullet system about point A? (b) If the angular speed of the system

about A just after impact is 4.5 rad/s, what is the bullet's speed just before impact?

Figure 3

(a) With r = 0.60 m, we obtain I = 0.060 + (0.501)r2 = 0.24 kg m2.

(b) Invoking angular momentum conservation, with SI units understood,

0 0 00.001 0.60 0.24 4.5fL mv r I v l

which leads to v0 = 1.8 103 m/s.

61. We make the unconventional choice of clockwise sense as positive, so that the angular velocities in

this problem are positive. With r = 0.60 m and I0 = 0.12 kg m2, the rotational inertia of the putty-rod

system (after the collision) is

I = I0 + (0.20)r2 = 0.19 kg m2.

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Invoking angular momentum conservation 0 fL L or 0 0I I , we have

2

00 2

0.12 kg m2.4rad/s 1.5rad/s.

0.19 kg m

I

I

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4. [25 points] A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a

30 kg box as shown in Figure 4. The outer radius R of the device is 0.50 m, and the

radius r of the hub is 0.20 m. When a constant horizontal force of magnitude 140 N is applied to a rope wrapped around the outside of the device, the box, which is suspended from

a rope wrapped around the hub, has an upward acceleration of magnitude 0.80 m/s2. What is

the rotational inertia of the device about its axis of rotation?

Figure 4

Let T be the tension on the rope. From Newtons second law, we have

( )T mg ma T m g a .

Since the box has an upward acceleration a = 0.80 m/s2, the tension is given by

2 2(30 kg)(9.8 m/s 0.8 m/s ) 318 N.T

The rotation of the device is described by app /F R Tr I Ia r . The moment of inertia can then be

obtained as

app 2

2

( ) (0.20 m)[(140 N)(0.50 m) (318 N)(0.20 m)]1.6 kg m

0.80 m/s

r F R TrI

a