Download Ch. 6 #44+47,45,48,55,58,59,61,71,72,81,85 Ex)87+93 Solutions

Ch. 6 #44+47,45,48,55,58,59,61,71,72,81,85 Ex)87+93 Solutions 44. Picture the Problem: The physical apparatus is shown at right.
Strategy: Write Newton’s Second Law for each of the three blocks and add the
equations to eliminate the unknowns T1 and T2 . Solve the resulting equation for
the acceleration a. Let x be positive in the direction of each mass’s motion.
Solution: 1. Write Newton’s Second Law for each of the three blocks and add the equations: 
Fx  T1
 m1a

Fx  T1  T2
 m2 a

Fx 
block 1
block 2
 T2  m3 g  m3 a
block 3
m3 g   m1  m2  m3  a


m3
3.0 kg
a
 9.81 m/s 2   4.9 m/s2 g 


6.0
kg
m
m
m
2
3 
 1
2. Solve the resulting equation for a: Insight: Note that the blocks move as if they were a single block of mass 6.0 kg under the influence of a force equal to m3 g  29 N. 47. Picture the Problem: Refer to the figure at right.
Strategy: Write Newton’s Second Law for each of the three blocks and add the
equations to eliminate the unknowns T1 and T2 . Solve the resulting equation for
the acceleration a. Use the acceleration to find the tensions in the strings. Let x
be positive in the direction of each mass’s motion.
Solution: 1. (a) Write Newton’s Second Law for each of the three blocks and add the equations: 
Fx  T1
 m1a

Fx  T1  T2
 m2 a

Fx 
block 1
block 2
 T2  m3 g  m3 a
block 3
m3 g   m1  m2  m3  a
2. Solve the resulting equation for a: 

m3
3.0 kg
a
 9.81 m/s2   4.9 m/s2 g 


6.0
kg
m
m
m
2
3 
 1
3. Use a to find T1: T1  m1a  1.0 kg   4.9 m/s 2   4.9 N 4. (b) Now find T2 from the block 3 equation: Insight: Note that the blocks move as if they were a single block of mass 6.0 kg under the influence of a force equal to m3 g  29 N. Note also that T2 is less than 29 N because the blocks accelerate. It would be zero if m3 fell freely, 29 N if m3 T2  m3  g  a    3.0 kg   9.81  4.9 m/s 2   15 N (and the entire system) were at rest. 45. Picture the Problem: The free‐body diagrams for each mass are shown at right. Strategy: The positive axis is along the line of the string and points up the
incline for the 5.7 kg mass and in the downward direction for the hanging
mass. Let M represent the mass on the incline and m the hanging mass.
Write Newton’s Second Law for each mass and combine the equations to
find the acceleration of the hanging mass (which is the same as the
acceleration of M because they are connected by a string).
Solution: 1. (a) Write Newton’s Second Law for M: F
x
  Mg sin   T  M a
T  M  a  g sin  
2. Write Newton’s Second Law for m and substitute the expression for T found in step 1: 3. Solve for a: F
x
 T  mg  ma
  M  a  g sin     mg  ma
 Mg sin   mg   m  M  a
 3.2 kg   5.7 kg  sin 35 
 m  M sin  
2
a
  9.81 m/s  g  
3.2  5.7 kg
 mM 


a   0.076 m/s 2
4. The negative value means that the acceleration of the hanging mass is in the upward direction. 5. (b) The magnitude of the acceleration is 0.076 m/s2. Insight: Verify for yourself that the angle of the incline that balances the two masses is 34° when m = 3.2 kg.
48. Picture the Problem: Refer to the figure at right:
Strategy: Write Newton’s Second Law for each block and add the equations to
eliminate the unknown tension T. Solve the resulting equation for the
acceleration a, and use the acceleration to find the tension. Let x be positive in
the direction of each mass’s motion, m1 be the mass on the table, and m2 be the
hanging mass.
Solution: 1. (a) The tension in the string is less than the weight of the hanging
mass. If it were equal to the weight, the hanging mass would not accelerate.
2. (b) Write Newton’s Second Law for each block and add the equations: 
Fx  T

Fx  T  m2 g  m2 a
 m1a
block 1
block 2
m2 g   m1  m2  a
3. Solve the resulting equation for a:  m2 
2.80 kg
9.81 m/s 2   4.36 m/s 2 a

g 
6.30 kg
 m1  m2 
4. (c) Use the first equation to find T: T  m1a   3.50 kg   4.36 m/s 2   15.3 N Insight: Note that the blocks move as if they were a single block of mass 6.30 kg under the influence of a force equal to m2 g  27.5 N. The tension in the string would be zero if m2 fell freely, 27.5 N if m2 (and the entire system) were at rest. 55. Picture the Problem: Your car travels along a circular path at constant speed.
Strategy: Static friction between your tires and the road provides the centripetal force required to make the car travel along a circular path. Set the static friction force equal to the centripetal force and calculate its value. mv 2 1300 kg 16 m/s 

 5.6 kN r
59 m
2
Solution: Set the static friction force equal to the centripetal force: f s  Fcp  macp 
Insight: The maximum static friction force is s mg   0.88 1300 kg   9.81 m/s 2   11.2 kN which corresponds to a maximum cornering speed (without skidding) of 23 m/s (50 mph). 58. Picture the Problem: The forces acting on the car are depicted at right.
Strategy: Write Newton’s Second Law in the horizontal and vertical
directions and combine them to obtain the radius of the curve. The
procedure is similar to Example 6-9 in the text.
Solution: 1. Write Newton’s Second Law in the x direction and solve for r: 2. Write Newton’s Second Law in the y direction and solve for N: 3. Substitute for N in the equation from step 1: F
x
 N sin   macp 
r
mv 2
r
mv 2
N sin 
 Fy  N cos   mg  0
N
mg
cos 
 22.7 m/s 
mv 2 cos 
v2


 79.4 m g tan   9.81 m/s 2  tan 33.5
sin  mg
2
r
Insight: Note that the car follows the curved path without any help from friction at all. This is because the inward component of the normal force is sufficient to provide the necessary centripetal force. 59. Picture the Problem: The free‐body diagram for Jill is shown at right.
Strategy: The center of Jill’s circular motion is the pivot point of the vine. There are two forces acting on Jill, the tension due to the vine upward and gravity downward. These two forces add together to produce her centripetal acceleration in the upward direction. Write Newton’s Second Law for Jill in the vertical direction and solve for T: Solution: Write Newton’s Second Law for Jill in the vertical direction and solve for T: F
y
 T  W  macp
T  W  macp  mg  m v 2 r

 2.4 m/s 
  63 kg  9.81 m/s 2 
6.9 m

T  670 N  0.67 kN
2
r 
T

v

W



Insight: The tension in the vine will be at its maximum at the bottom of her circular path because it is at that point that the vine must both support her weight and provide the upward centripetal force. Her speed is a maximum there as well, making the centripetal force the largest at that point. 61. Picture the Problem: Free‐body diagrams for the top and bottom positions on the Ferris wheel are drawn at right. Top of Ferris Wheel
Strategy: Find the speed of the seat by dividing the circumference of
its circular path by the time it takes to complete a cycle. Write
Newton’s Second Law for the passenger at the top of the Ferris wheel
and a second Newton’s Second Law for a passenger at the bottom, and
solve each for the normal force.
Solution: 1. (a) The normal force on you by the Ferris wheel seat is equal to your apparent weight. At the top of the Ferris wheel your centripetal acceleration points downward and the normal force is the smallest, but at the bottom of the Ferris wheel the normal force must both support your weight and provide the upward centripetal acceleration, so it is at its maximum there. 2. (b) Write Newton’s Second Law at the top of the Ferris wheel and solve for your apparent weight N top : 3. Find the speed of the seat: F

v

W

W
r Bottom of Ferris Wheel  N top  mg   macp   mv 2 r
y
N top  m  g  v 2 r 
2 r 2  7.2 m 

 1.6 m/s T
28 s
v
4. Calculate the apparent weight N top : 
v

N
r 
N
N top
2

1.6 m/s 

2
 m  g  v r    55 kg   9.81 m/s 

7.2 m

2


 
 520 N  0.52 kN
5. Write Newton’s Second Law at the bottom of the Ferris wheel and solve for your apparent weight N bottom : F
y
 N bottom  mg  macp  mv 2 r
2

1.6 m/s 
N bottom  m  g  v 2 r    55 kg   9.81 m/s 2 

7.2 m


 

 560 N  0.56 kN
Insight: The centripetal acceleration is 0.36 m/s2 so that the centripetal force is 3.6% of your 0.54 kN weight.
71. Picture the Problem: The block is pulled along a rough horizontal surface by a spring.
Strategy: The acceleration of the block is zero. Newton’s Second Law in the vertical direction  Fy  N  mg  0 indicates that the normal force equals the weight. Write Newton’s Second Law in the horizontal direction for the block and solve for the coefficient of friction. Let the positive x axis point along the direction of motion. F
  f k  Fspring  max  0
 k N  kx  0
kx kx  85.0 N/m  0.0620 m 
k  

 0.140
N mg
 3.85 kg   9.81 m/s2 
Solution: Write Newton’s Second Law in the x direction and solve for k : Insight: Since the kinetic friction force remains constant, the block will accelerate if it is pulled with a force greater than kx   85.0 N/m  0.0620 m   5.27 N. x
72. Picture the Problem: The free‐body diagram of the child is shown at right.
Strategy: Choose coordinate axes with x pointing parallel to the slide and down along the 
direction of motion, and y pointing parallel to N . Write Newton’s Second Law in the y and x directions and combine the equations to solve for a. Solution: 1. Write Newton’s Second Law in the y direction: F
2. Write Newton’s Second Law in the x direction: F
y  N  mg cos   0
N  mg cos 
 mg sin   f k  ma
ma  mg sin    k N
x
ma  mg sin    k  mg cos  
a  g  sin    k cos  
3. Substitute the expression for N from step 1 and solve for a:   9.81 m/s 2  sin 26.5   0.315  cos 26.5
a  1.61 m/s 2
Insight: As expected, increasing θ will increase a, but increasing k will decrease a. 81. Picture the Problem: The force vectors acting on blocks A and B as well as the rope knot are shown in the diagram at right. Strategy: Write Newton’s Second Law for blocks A and B, as well as
Newton’s Second Law for the rope knot. In all cases the acceleration is

zero. Combine the equations to solve for fs , which acts on block A and
points toward the left. Let the x direction point to the right for block A and
down for block B.
Solution: 1. (a) Write Newton’s Second Law for block A: 2. Write Newton’s Second Law for block B: 3. Write Newton’s Second Law for the rope knot: 
Fx   fs  TA  0 
Fx  TB  mB g  0 Block A
Block B
F
 TA  T3 cos 45  0
F
 TB  T3 sin 45  0
x
knot
y
knot
4. Divide the y equation for the knot by the x equation: 5. Substitute TA  TB into the equation from step 2: T3 sin 45
T
 tan 45  1  B T3 cos 45
TA
TA  mB g f s  TA  mB g   2.33 kg   9.81 m/s 2   22.9 N  23 N 6. Substitute the result into the equation from step 1: 7. (b) As long as mass A is heavy enough that f s, max  s mA g   0.320  8.82 kg   9.81 m/s 2   27.7 N  22.9 N, the friction force is not affected by changes in mass A. It will stay the same if the mass of block A is doubled. Insight: The minimum mass of block A that will satisfy the criteria of step (b) is 7.29 kg. The answer to (a) is reported with only two significant figures because the angle 45° is only given to two significant figures. 85. Picture the Problem: The free‐body diagram of the airplane is depicted at right.
Strategy: Let the x axis point horizontally from the airplane toward the center of its circular motion, and let the y axis point straight upward. Write Newton’s Second Law in both the horizontal and vertical directions and use the resulting equations to find θ and the tension T. Solution: 1. (a) Write Newton’s Second Law in the x and y directions: 2. Solve the y equation for T and substitute the result into the x equation, and solve for θ: F
F
x
 T sin   macp  m v 2 r
y
 T cos   mg  0
T  mg cos 
 mg 
2
T sin   
 sin   m v r
 cos  
tan   v 2 rg


1.21 m/s 
  19
2
  0.44 m   9.81 m/s  
2
  tan 1 
 0.075 kg   9.81 m/s2 
mg
T

 0.78 N
cos 
cos19
3. (b) Calculate the tension from the equation in step 2: Insight: This airplane is pretty small. The toy weighs only 0.74 N = 2.6 ounces and flies in a circle of diameter 2.9 ft.
87. Picture the Problem: The forces that are exerted on each box are depicted at right.
Strategy: Write Newton’s Second Law in the horizontal direction for the top box in order to show
that it is accelerated by the static friction force alone. Find the maximum acceleration that the static
friction force can produce, and then determine the force F that will produce that acceleration for
both boxes.
Solution: 1. (a) Write Newton’s Second Law for the top box and solve for the maximum acceleration: 2. Write Newton’s Second Law for both boxes together, substitute the maximum acceleration from step 1, and solve for F: F
x
 fs, max  s mtop g  mtop a
a  s g
F
x
 F   mtop  mbottom  a
F   mtop  mbottom  s g
  2.0  5.0 kg  0.47   9.81 m/s 2 
F  32 N
3. (b) If mtop is increased, F will increase as well. The maximum acceleration remains the same but a larger force would be needed to accelerate the larger mass at the same rate. Insight: Note that the maximum acceleration is independent of the mass of the top block. A more massive top block will increase the friction force, but it will also have more inertia and therefore require a larger force in order to accelerate it. 93. Picture the Problem: The free‐body diagram of the cereal box is depicted at right.
Strategy: The forwards acceleration produces a normal force on the cereal box that allows the static friction force to balance the weight of the box. Write Newton’s Second Law in the vertical direction to find the required magnitude of the normal force and then write Newton’s Second Law in the horizontal direction to find the acceleration. Solution: 1. Write Newton’s Second Law in the vertical direction and solve for N: 2. Write Newton’s Second Law in the horizontal direction and solve for a: F
y
 s N  mg  0
N  mg s
F
x

fs  s N yˆ

N

mg
 N  ma
a
N mg
g 9.81 m/s 2



m s m s
0.38
 26 m/s 2
Insight: If you sustain this acceleration for 5 seconds in order to impress fellow grocery shoppers with your gravity‐defying stunt, your cart will be traveling 130 m/s or 290 mi/h! You’re bound to get the attention of the store manager!