Download Solutions #5

PHYSICS 171
AQ 2009
Homework #5
(due Friday, October 16, 2009)
1. Giancoli Chapter 5, Problem 10
A free-body diagram for the bar of soap is shown. There is no motion in the y direction and
thus no acceleration in the y direction. Write Newton’s second law for both directions, and
use those expressions to find the acceleration of the soap.
 Fx  FN  mg cos   0  FN  mg cos 
FN
F
fr
 Fx  mg sin   Ffr  ma
y
ma  mg sin   k FN  mg sin   k mg cos 
a  g  sin   k cos  
Now use Eq. 2-12b, with an initial velocity of 0, to find the final
velocity.
x  x0  v0t  12 at 2 
t
2x
a

2x
g  sin    k cos  


x

mg
2  9.0 m 
 9.80 m s   sin 8.0   0.060  cos 8.0 
2
 4.8s
2. Giancoli Chapter 5, Problem 17
a)
Since the two blocks are in contact, they can be treated as a
single object as long as no information is needed about internal forces (like the force of
one block pushing on the other block). Since there is no motion in the vertical
direction, it is apparent that
FN   m1  m2  g , and so Ffr  k FN  k  m1  m2  g. Write Newton’s second law for
the horizontal direction.
 Fx  FP  Ffr   m1  m2  a 
a
FP  Ffr
m1  m2

FP  k  m1  m2  g
m1  m2


650 N   0.18190 kg  9.80 m s 2

190 kg
 1.657 m s 2  1.7 m s 2
(b) To solve for the contact forces between the blocks, an individual block
must be analyzed. Look at the free-body diagram for the second block.
F21 is the force of the first block pushing on the second block. Again, it
is apparent that FN2  m2 g and so Ffr2  k FN2  k m2 g. Write Newton’s
second law for the horizontal direction.
F21
m2
Ffr2
FN2
m2 g
F
x
 F21  Ffr2  m2 a 




F21  k m2 g  m2a   0.18125kg  9.80 m s2  125kg  1.657 m s2  430 N
By Newton’s third law, there will also be a 430 N force to the left on block # 1 due to
block # 2.
(c)
If the crates are reversed, the acceleration of the system will remain
F12
the same – the analysis from part (a) still applies. We can also repeat the
m1
analysis from part (b) to find the force of one block on the other, if we
simply change m1 to m2 in the free-body diagram and the resulting
Ffr1
equations.
m1g
FN1
2
a  1.7 m s ;  Fx  F12  Ffr1  m1a 




F12  k m1 g  m1a   0.18 65 kg  9.80 m s 2   65 kg  1.657 m s 2  220 N
3. Giancoli Chapter 5, Problem 23
(a) For mB to not move, the tension must be equal to mB g , and so mB g  FT . For mA to not
move, the tension must be equal to the force of static friction, and so FS  FT . Note
that the normal force on mA is equal to its weight. Use these relationships to solve for
mA .
mB g  FT  Fs   s mA g  mA 
mB
s

2.0 kg
0.40
 5.0 kg  mA  5.0 kg
(b) For mB to move with constant velocity, the tension must be equal to mB g . For mA to
move with constant velocity, the tension must be equal to the force of kinetic friction.
Note that the normal force on mA is equal to its weight. Use these relationships to
solve for mA .
mB g  Fk  k mA g  mA 
mB
k

2.0 kg
0.30
 6.7 kg
4. Giancoli Chapter 5, Problem 28
We define the positive x direction to be the direction
of motion for each block. See the free-body diagrams. Write Newton’s second law in both
dimensions for both objects. Add the two x-equations to find the acceleration.
Block A:
 FyA  FNA  mA g cos  A  0  FNA  mA g cos  A
F
xA
 FT  mA g sin   FfrA  mA a
Block B:
 FyB  FNB  mB g cos  B  0  FNB  mB g cos  B
 FxB  mB g sin   FfrB  FT  mBa
y
FT
FNA
FfrA
A
A
mAg
x
FT
FNB
y
FfrB
x
B
B
m Bg
Add the final equations together from both analyses and solve for the acceleration, noting
that in both cases the friction force is found as Ffr   FN .
mA a  FT  mA g sin  A  A mA g cos  A ; mBa  mB g sin  B   BmB g cos  B  FT
mA a  mBa  FT  mA g sin  A  A mA g cos  A  mB g sin  B   BmB g cos  B  FT 
  mA  sin  A  A cos  A   mB  sin    B cos   

 mA  m B 


   2.0 kg  sin 51  0.30cos51    5.0 kg sin 21  0.30cos 21  
  9.80 m s2  

 7.0 kg 


a  g
 2.2 m s2
5. Giancoli Chapter 5, Problem 30
FT
(a) Given that mB is moving down, mA must be moving
up the incline, and so the force of kinetic friction on
mA will be directed down the incline. Since the blocks
are tied together, they will both have the same
acceleration, and so a yB  a xA  a. Write Newton’s
FN
FT
Ffr xA
m Bg
yB
m Bg


mAg
second law for each mass.
 FyB  mB g  FT  mBa  FT  mB g  mBa
F
F
xA
 FT  mA g sin   Ffr  mA a
yA
 FN  mA g cos   0  FN  mA g cos 
Take the information from the two y equations and substitute into the x equation to
solve for the acceleration.
mB g  mBa  mA g sin   k mA g cos   mAa 
a

(b)
mB g  mA g sin   mA g k g cos 
 mA  mB 
1
2
 12 g 1  sin   k g cos  
 9.80 m s  1  sin 34  0.15cos 34  1.6 m s
2
2
To have an acceleration of zero, the expression for the acceleration must be zero.
a  12 g 1  sin   k cos    0  1  sin   k cos   0 
k 
1  sin 
cos 

1  sin 34
cos 34
 0.53
6. Giancoli Chapter 5, Problem 33
To find the limiting value, we assume that the blocks are NOT slipping, but that the force of
static friction on the smaller block is at its maximum value, so that Ffr   FN . For the twoblock system, there is no friction on the system, and so F   M  m  a describes the
horizontal motion of the system. Thus the upper block has a vertical acceleration of 0 and a
yA
FN
Ffr
horizontal acceleration of
F
 M  m

. Write Newton’s second law for the
mg
upper block, using the force diagram, and solve for the applied force F. Note
that the static friction force will be DOWN the plane, since the block is on the verge of
sliding UP the plane.
F
 FN cos   Ffr sin   mg  FN  cos    sin    mg  0  FN 
F
 FN sin   Ffr cos   FN  sin    cos    ma  m
y
x
F  FN  sin    cos  

M m
mg

 cos    sin  
 sin    cos  
 M  m g
 cos    sin  
m
F
M m
 sin    cos  
mg
 cos    sin  

M m
m
7. Giancoli Chapter 5, Problem 44
FT
(a) At the bottom of the motion, a free-body diagram of the bucket would be as
shown. Since the bucket is moving in a circle, there must be a net force on it
towards the center of the circle, and a centripetal acceleration. Write
Newton’s second law for the bucket, with up as the positive direction.
 FR  FT  mg  ma  m v 2 r 
v
r  FT  mg 
m

1.10 m  25.0 N   2.00 kg   9.80 m

s 2 
2.00 kg
 1.723  1.7 m s
(b) A free-body diagram of the bucket at the top of the motion is shown. Since the
bucket is moving in a circle, there must be a net force on it towards the center
of the circle, and a centripetal acceleration. Write Newton’s second law for the
bucket, with down as the positive direction.
F
R
 FT  mg  ma  m v 2 r
 v
mg
FT
r  FT  mg 
m
If the tension is to be zero, then
v
r  0  mg 
 rg 
1.10 m   9.80 m

s 2  3.28 m s
m
The bucket must move faster than 3.28 m/s in order for the rope not to go slack.
8. Giancoli Chapter 5, Problem 40
At the top of a circle, a free-body diagram for the passengers would be as shown, assuming
the passengers are upside down. Then the car’s normal force would be pushing DOWN on
the passengers, as shown in the diagram. We assume no safety devices are present. Choose
the positive direction to be down, and write Newton’s second law for the passengers.
mg
y
x
F  F
N
 mg  ma  m v 2 r

 FN  m v 2 r  g

We see from this expression that for a high speed, the normal force is
positive, meaning the passengers are in contact with the car. But as the
speed decreases, the normal force also decreases. If the normal force
becomes 0, the passengers are no longer in contact with the car – they are in
free fall. The limiting condition is as follows.
2
vmin
r  g  0  vmin  rg 
FN
mg
 9.80 m s  7.6 m   8.6 m s
2
9. Giancoli Chapter 5, Problem 48
To experience a gravity-type force, objects must be on the inside of the outer
wall of the tube, so that there can be a centripetal force to move the objects in
a circle. See the free-body diagram for an object on the inside of the outer
wall, and a portion of the tube. The normal force of contact between the
object and the wall must be maintaining the circular motion. Write
Newton’s second law for the radial direction.
 FR  FN  ma  mv2 r
FN
If this is to have the same effect as Earth gravity, then we must also have that
FN  mg . Equate the two expressions for normal force and solve for the speed.
FN  m v 2 r  mg  v 

gr 
 9.80 m s  550 m   73.42 m s
2
  86, 400 s 
3
  1836 rev d  1.8  10 rev d

 2  550 m    1 d 
 73.42 m s  
1 rev
10. Giancoli Chapter 5, Problem 54
If the masses are in line and both have the same frequency of
rotation, then they will always stay in line. Consider a free-body
FNB
diagram for both masses, from a side view, at the instant that they
FTB
are to the left of the post. Note that the same tension that pulls
mB
inward on mass 2 pulls outward on mass 1, by Newton’s third law.
Also notice that since there is no vertical acceleration, the normal
m Bg
force on each mass is equal to its weight. Write Newton’s second
law for the horizontal direction for both masses, noting that they are
in uniform circular motion.
 FRA  FTA  FTB  mAaA  mA vA2 rA  FRB  FTB  mBaB  mB vB2 rB
The speeds can be expressed in terms of the frequency as follows:
 rev   2 r   2 rf .
v f


 sec   1 rev 
FTB  mB vB2 rB  mB  2 rB f  rB  4 2mBrB f 2
2
FTA  FTB  mA vA2 rA  4 mBrB f 2  mA  2 rA f  rA  4 2 f 2  mA rA  mBrB 
2
FNA
FTB
FTA
mA
mAg