PHY 206 Spring 2003, HW #5 Solutions
18. **A 3-kg box rests on a horizontal shelf and is attached to a 2-kg box which hangs over a
frictionless small pulley. What minimum coefficient of static friction is needed to keep the boxes from
moving? If they start moving, and the coefficient of kinetic friction is 0.3, how long does it take the 2-kg
box to hit the floor from an initial height of 2 m?
For the hanging mass, the forces acting are m2 g  T  0 . (If the box is not moving , the
acceleration is zero.) For the box on the table, we can write Newton’s Second Law as
F1, x  T  s Fn  0 and
F1, y   m1g  Fn  0 .


From the second of these conditions we can find the normal force, and from our equation for the hanging
mass we can find the tension. Using these in the x-direction component of Newton’s Second Law for the
mass on the shelf, we find
 2kg   9.81 sm   s  3kg   9.81 sm  , so s = 0.67.
2
2
If the blocks start to slide, we can write the above equations as
F1, x  T  k Fn  m1a and
F1, y   m1g  Fn  0 and


F
2
 m2 g  T  m2 a . Combining
m2   k m1
g  2.16 sm2 . Once we have the acceleration, we can use our
m1  m2
1
kinematics equation to find the time to travel 2m, y  at 2  t  1.36s.
2
these equations gives us a 
23. The coefficient of static friction between the bed of a truck and a box is s = 0.30. The truck travels
at 80 km/h. If the box is not to slide, what is the least distance the truck needs to stop?
The only force acting on the box is friction, acting backwards against the tendency to slide. For
the x-direction we can write
Fx   f s  ma . The friction force is given be f s  s Fn  s mg . The

maximum acceleration is then a   f s / m  s g  2.94 sm2 . Now we can go back to our kinematics
equation, v 2  v02  2ax and solve for x. Converting the initial speed to m/s first, we then find x =
84 m.
29. Two blocks connected by a string slide down a
20 incline. The lower block has m1 = 0.25 kg
and k = 0.2, while the upper block has m2 = 0.8
kg and k = 0.3. Find (a) the acceleration of the
blocks and (b) the tension in the string .
We can set up Newton’s Second Law
equations for each block, with the assumption that the
accelerations are the same for the blocks. Along the
plane, we have
m1 g sin   1k m1 g cos  T  m1a
m2 g sin    2 k m1 g cos  T  m2 a
Adding the two equations and solving for the
acceleration a, we find a = 0.81 m/s2. Since I defined
“down the plane” as positive, this acceleration is down the plane; the answer in the text implies that he
defined the coordinate system the other way around. Using this value for a in the first equation, we solve
for T = 0.18 N, acting against the motion.
37. A 2-kg block sits on a 4-kg block, which is in turn on a frictionless surface. The coefficients of
friction between the blocks are s = 0.3 and k = 0.2. a) What is the maximum force F that can be
applied to the 4-kg block if the 2-kg block is not to slide? b) If F is half this value, find the
acceleration of each block and the forces of friction. c) If F is twice the value found in (a), find the
acceleration of each block. (Problem is very similar to Example 5-6.)
Using the free-body diagrams, the only force acting horizontally on the 2-kg mass is the friction
force. Acting on the 4-kg mass is the external force F and the friction force f. Using Newton’s Second
Law for the horizontal motion
F f  m a
 F F  f   m a
1
2
1 1
2 2
a) For the case in which the two masses move together, that is, the friction force is strong enough to prevent
relative slippage, we have f = f and a1 = a2 . Thus, after adding the two equations, we find F = (m1 + m2 )a.
The maximum acceleration for the top box is that for which the frictional force is maximum (see first
equation above), fs,max = s F1n = s m1g, so the maximum acceleration is amax = sg. Thus the maximum
applied force will be Fmax = (m1 + m2 )a = (m1 + m2 ) sg = 17.7 N. b) For a force F = 1/2 Fmax , the
acceleration is a = ½Fmax/(m1 +m2) = 1.48 m/s2. The frictional force is f = m1a = 2.96 N(acts forward on
the top block and backward on the bottom block. c) For F = 2Fmax the blocks no longer slide together,
and the acceleration is no longer common to both. The frictional force is still the common action-reaction
pair, but corresponds now to kinetic friction, f = k m1g = 3.9 N. For the top mass, using the first of the
two equations above, a1 = 3.9N/2 kg = 1.95 m/s2. From the second equation, a2 = (35.4 N – 3.9 N)/4 kg
=7.88 m/s2.
48. **A 0.2-kg stone on a 0.8-m string rotates in a horizontal plane. The string’s angle with the
horizontal is 20o. What is the speed of the stone?
The circle radius is Lcos m. The forces acting in the x- and y-directions are
 Fy T sin   mg  0 and
 Fx  T cos 
mv 2
. From the first of these we find T = 5.7 N.
r
From the second equation we have v  rg cot  = 4.5 m/s.
49. A 0.75-kg stone attached to a string rotates in a circle of radius 0.35 m. The string angle with the
vertical is 30o. Find the speed of the stone and the tension.
The forces acting in the x- and y-directions are
 Fy T cos  mg  0 and
 Fx  T sin  
mv 2
. From the first of these we find T = 8.5 N.
r
From the second equation we have v  rg tan  = 1.4 m/s.
60. **A 100-g disk sits on a turntable making one rotation per second. The disk is located at 10 cm
from the axis of rotation. a) What is the frictional force? b) If the disk slides off when at a radius of 16
cm or greater, what is the coefficient of friction?
Using the definition of the period, T 
the forces acting radially as
F
r
2 r
we can find the speed of the disk. We write the sum of
v
 f s (there is only one force acting). The sum of radial forces is
mv 2 m  2 r 
mv 2
 
. Thus we have f s 
  0.39N . For the
r
r T 
r
second part, we consider the y-direction forces,  Fy  Fn  mg  0 , since there is no motion in the
2
equal to the centripetal force, Fc 
y-direction. The maximum static frictional force is defined as f s   s Fn   s mg . Returning to part
one, if the disk is at a radius of 16 cm, the frictional force needed to hold it in place becomes 0.632 N.
Since we know the mass, we find  s 
0.632N
 0.64 .
mg
61. A tether ball of mass 0.25 kg is attached to a pole by a 1.2-m cord at an angle of 20. The ball
moves in a horizontal circle. a) Find the tension in the cord. b) Find the speed of the ball.
From the free-body diagram, we have
T cos  mg  0
T sin  
mv 2
r
. From the first
equation we find the tension to be T =
2.61 N. The radius is given by r = L sin
= 0.41 m. We can use the second
equation to find the speed, v = 1.21 m/s.
63. **A bead is free to slide on a semicircular section of wire of radius 10 cm.
The wire rotates about an axis at a rate of 2
rev/sec. The mass of the bead is 100 g. At
what angle  with respect to the vertical
axis will the bead remain stationary?
After some thought, we realize that this
problem is very much like the previous one. The force keeping the bead from flying off is exerted by
the wire instead of by a cord or a gravitational pull. We can then use
Fy F cos  mg  0 for

the forces in the y-direction, since there is no motion vertically if the bead is stable. For the radial
direction, the only force acting is F sin , so we can write F sin  
mv 2 m 4 2 r 2

. The period
r
r P2
is defined as P  2 r / v . The radius of the circle in which the bead is spinning is given by
r  L sin , with L = 10 cm. If we put all of these pieces together we find cos 
gP 2
 0.62 or
4 2 L
  51.6 .
67. The loop-the-loop travels with constant speed. At the top, the normal force exerted by the seat is
equal to the weight mg. What is the normal force exerted by the seat at the bottom of the loop?
mv 2
From the free-body diagram we have at the top of the loop mg  Fn  
where the “-“ sign
r
for the centripetal force comes because it is pointed radially inward, or in the negative y-direction.
We are given that Fn,top = mg at the top. At the bottom, we can write mg  Fn ,bottom  
mv 2
.
r
Adding the two equations, we find Fn,bottom = 3 mg, answer (d).
97. **An 800-N box rests on a 30-incline. It is possible to prevent the box from sliding by exerting a
200-N force parallel to the plane. a) What is the coefficient of static friction? b) What is the
maximum force that can be applied before the box starts to slide up the incline?
a) The box initially wants to slide down the incline, and the friction force acting up the plane
prevents this, along with the external applied force. Therefore we can
write F  f s  mg sin   0 . The frictional force is given by fs = Fn =  mg cos. We can
calculate  from this, and find = 0.29. b) If the student wants to push the box up the plane, the
frictional force will oppose her, and we can write, for the forces acting parallel to the plane
F  f s  mg sin   0 , which leads to F = 600 N, using the value for  found from part a).
99.
**You want to put a crate of books up on a truck with the help of a ramp set up with an angle of 30o.
The mass of the crate is 100 kg and k = 0.5. You push horizontally to get the crate up the incline.
After it has started moving, how large must your force be to keep it moving at constant speed?
Now we have an extra, external force, and we want to keep all forces balanced. We can write
F
F
x
  mg sin   f k  F cos   ma  0
y
  mg cos   F sin   Fn  ma  0
The frictional force is given by f k  k Fn . In contrast to other problems, here the normal force is
found from the second equation to be Fn  mg cos  F sin  . Combining these equations we
find F  cos   sin    mg  sin   cos  and we solve to find F = 1485 N.
101. A 5.5-kg block slides from rest 72 m down an incline of 30o. The coefficient of sliding friction is
0.35. What is the speed of the object at the bottom?
The forces acting along the plane can be written
Fx  mg sin   f k  ma , with positive being

down the plane. The friction force is f k   k mg cos . Putting these together, we find
a   sin   k cos  g  1.93 sm2 . From v 2  v02  2ax we find v = 17 m/s (d).