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```					Homework Assignment 08: ..........40 points total

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4.16     Find the tension in the two wires that support the 100 N light fixture in Figure
P4.16

Solution:
From Fx  0 , T1 cos40.   T2 cos40.   0
0             0

or         T1  T2

Then, Fy  0 gives 2 T1 si 40.   100 N  0
n 0

yielding T1  T2  77. N
8

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4.20: Two people are pulling a boat through the water as in Figure P4.20. Each exerts a
force of 600 N directed at a 30 degree angle relative to the forward motion of the boat. If
the boat moves with constant velocity, find the resistive force F exerted by the water on
the boat.

Solution:

The resultant force exerted on the boat by the people is
2 600 N  cos30.   1.  103 N in the forward direction. If the boat moves
               0      04
with constant velocity, the total force acting on it must be zero. Hence, the
resistive force exerted on the boat by the water must be

f  1.  103 N i t r w ar di ecton
04        n he ear d r i

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4.32 a) an elevator of mass m moving upward has two forces acting on it, the upward
force of tension in the cable and the downward force due to gravity. When the elevator is
accelerating upward, which is greater, T or W?
b) When the elevator is moving at a constant velocity upward, which is greater, T or W?
c) When the elevator is moving upward, but the acceleration is downward, which is
greater, T or W?
d) Let the elevator have a mass of 1500 kg and an upward acceleration of 2.5 m/s2. Find
e) The elevator of part d) is moving with constant upward velocity of 10 m/s. Find T. Is
f) Having initially moved upward with a constant velocity, the elevator begins to
part c)?

Solution:

(a) When the acceleration is upward, the total upward force T must exceed

             
the total downward force w  m g  1500 kg 9. m s2  1.  104 N
80        47

(b) When the velocity is constant, the acceleration is zero. The total upward
force T and the total downward force w must be equali m agniude .
n        t

(c) If the acceleration is directed downward, the total downward force w
must exceed the total upward force T.

(d)
                      
Fy  m ay  T  m g  m ay  1500 kg 9. m s2  2. m s2  1.  104 N
80        50        85

Yes , T  w .

(e)                                                        
Fy  m ay  T  m g  m ay  1500 kg 9. m s2  0  1.  104 N
80            47

Yes , T  w .

(f)
                      
Fy  m ay  T  m g  m ay  1500 kg 9. m s  1. m s  1.  104 N
80   2
50   2
25

Yes , T  w .
4.62 Three objects are connected by light strings as shown in Fig. P4.62. The string
connecting the 4 kg object and the 5 kg object passes over a light frictionless pulley.
Determine a) the acceleration of each object and the b) tension in the two strings.

Solution:

Let m 1  5. kg,m 2  4. kg,and m 3  3. kg . Let T1 be the tension in the
00          00                  00
string between m 1 and m 2 , and T2 the tension in the string between m 2 and m 3 .

(a) We may apply Newton’s second law to each of the masses.

for m1:           m 1a  T1  m 1g                                                (1)

for m 2 :         m 2a  T2  m 2 g  T1                                          (2)

for m 3 :         m 3a  m 3 g  T2                                               (3)

Adding these equations yields  m 1  m 2  m 3  a    m 1  m 2  m 3  g , so

 m 1  m 2  m 3       2. kg 
00
a 
 m1  m2  m3           0 
80 
 g   12. kg  9. m s  1. m s
2
63  2

           
(b) From Equation (1), T1  m 1  a g   5. kg 11. m s2  57. N , and
00      4          2

17 
from Equation (3), T2  m 3  g  a   3. kg 8. m s2  24. N
00                5   
Homework Assignment: 09..................... 40 points total

Conceptual Question 17:

4.17 Suppose you are driving a car at a high speed. Why should you avoid slamming
on your brakes when you want to stop in the shortest possible distance?

The reason for antilock brakes is that based on comparison of the static and kinetic
coefficients of friction. For most materials, the static coefficient is slightly
larger than the kinetic coefficient of friction, which means that a larger
friction force will be developed is two bodies are not sliding over each
other. If you hit the brakes, you start sliding and the friction force is
smaller than if your tires roll, but do not slide.

Since the friction force is larger for rolling tires than sliding tires, you should be able
to stop in a shorter distance. Realistically, as long as your tires don’t slide
you also can steer the car, which you can’t do if you start sliding, so
antilock brakes provide you with a way to stop the car faster and maintain
control of the car as you stop.

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Problems:
4.35 A dock worker loading crates on a ship finds that a 20 kg crate initially at rest on a
horizontal surface requires a 75 N horizontal force to set it in motion. After the crate is
in motion, a horizontal force of 60 N is required to keep it moving at constant speed.
Find the coefficients of static and kinetic friction between the crate and the floor.

Solution:
When the block is on the verge of moving, the static friction force has a
magnitude f   f  m ax   sn .
s    s

Since equilibrium still exists and the applied force is 75 N, we have

Fx  75 N  f  0 or  f  m ax  75 N
s          s

In this case, the normal force is just the weight of the crate, or n  m g . Thus, the
coefficient of static friction is

s 
 f
s m ax

 f
s m ax

75 N
 0.
38
n             mg                  
 20 kg 9. m s2
80   
After motion exists, the friction force is that of kinetic friction, f  kn
k

Since the crate moves with constant velocity when the applied force is 60 N, we
find that Fx  60 N  f  0 or f  60 N . Therefore, the coefficient of kinetic
k        k

friction is

f    f        60 N
k     k
 k                 0.
31
n m g  20 kg 9. m s2
80                    

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4.44 A student decides to move a box of books into her dormitory room by pulling a
rope attached to the box. She pulls with a force of 80 N at an angle of 25
degrees above the horizontal. The box has a mass of 25 kg and the coef. of
kinetic friction between the floor and the box is 0.3. a) find the acceleration
of the box. B) the student now starts moving the box up a 10 degree incline,
keeping her 80 N force directed at 25 degrees (a) above the line of the
incline. If the coefficient of friction is unchanged, what is the new
acceleration of the box?

Solution:

Find the normal force n on the 25.0 kg box:

Fy  n   80. N  si 25.   245 N  0
0      n 0

or               n  211 N

Now find the friction force, f, as

f  kn  0.  211 N   63. N
300             4

From the second law, we have Fx  m a, or

 80. N  cos25.   63. N =  25. kg a
0          0       4         0

which yields a 0.
366 m s2

(b) When the box is on the incline,

Fy  n   80. N  si 25.    245 N  cos10.   0
0      n 0                      0

giving n  207 N

The friction force is

f  kn  0.  207 N   62. N
300             2

The net force parallel to the incline is

Fx   80. N  cos25.    245 N  s n10.   62. N =- 3 N
0          0                i   0       2     32.

Fx 32. N
3
Thus, a                 1. m s2 , or 1. m s2 dow n t i i
29           29            he nclne
m     0
25. kg
4.53 In figure P4.53, the coefficient of kinetic friction between the two blocks is 0.3. The
surface of the table and the pulleys are frictionless. A) Draw the FBD for each block.
B) Determine the acceleration of each block c) find the tension in
the strings.

Solution:
(a)

(b) For the 10-kg object:
Fy  m ay  T2  98 N  10 kg   a               or    T2  98 N  10 kg a     (1)

For the 2.0-kg object: Fy  m ay  n1  20 N  0       or    n1  20 N

so     f  kn1   0.  20 N   6. N
k            30             0

Also, Fx  m ax  6. N  T1   2. kg    a
0             0                     or    T1  6. N   2. kg  a
0        0          (2)

Finally, for the 3.0-kg object:

Fx  m ax  T2  T1  6. N   3. kg    a
0        0                      or
T2  T1  6. N   3. kg  a
0        0                              (3)

Substituting Equations (1) and (2) into Equation (3) yields:

98 N  10 kg a  6. N   2. kg a  6. N   3. kg a or
0        0          0        0
86 N
a          5. m s2
7
15 kg

(c) Substituting the computed value for the magnitude of the acceleration into
7  
Equations (1) and (2) gives: T1  6. N   2. kg 5. m s2  17 N
0        0                     
          
and T2  98 N  10 kg 5. m s2  41 N
7
Homework Assignment: 10 .................30 points total

3.54 A boy and girl are tossing an apple back and forth between them. Figure P3.54
shows one path the apple follows when watched by an observer looking on from the side.
The apple is moving from left to right. Five points mark the path. Ignore air resistance.
A) Make a copy of this figure. At each of the marked points draw an arrow that shows
the magnitude and direction of the apple’s velocity when it passes through that point.
b) make another copy of this figure. This time at each marked point place an arrow
indicating the magnitude and direction of any acceleration the apple exhibits at that point.

solution:

(a) Velocity vector at several points                     (b) Acceleration vector at several
points

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4.41 The coefficient of static friction between the 3 kg crate and the 35 degree incline it
rests on is 0.300 What minimum force F directed perpendicular to the inclined plane is
needed to prevent the crate from sliding down the incline?

Solution:

The normal force acting on the crate is given by
n  F  m gcos . The net force tending to move the crate down
the incline is m g si   f , where f is the force of static
n      s         s

friction between the crate and the incline. If the crate is in
equilibrium, then m gsi   f  0 , so that f  Fg si 
n      s              s       n

But, we also know f  sn  s  F  m g cos 
s

Therefore, we may write m g si    s  F  m g cos  , or
n

 si                   n35. 
 si                
F
n
 s
 cos  m g  
        0.
0
300
 cos35.   3. kg 9. m s2  32. N
0

00     80        1          

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4.68 A 3 kg object hangs at one end of a rope that is attached to a support on a railroad
car. When the car accelerates to the right, the rope makes an angle of 4 degrees with the
vertical. Find the acceleration of the car.

Solution:

In the vertical direction, we have

mg
Fy  T cos4.   m g  0 , or
0                       T
cos4. 
0

In the horizontal direction, the second law becomes:

Fx  T s n 4.   m a, so
i 0

T s n 4. 
i 0
a               gt 4.   0. m s2
an 0     69
m

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