Download Chapter 4 Conceptual Questions: 15. (a) (b) The force F is

Chapter 4
Conceptual Questions:
15.
(a)
(b)
N
N
F
F
m
m
The force F is the combination of the pushing by the seat back and the
friction force by the seat surface.
Multiple Choice Questions:
13.
(d).
Exercises:
7.
x  (sin 30) ŷ ] = (4.76 N) ^
x + (2.75 N) ŷ ,
F 1 = (5.5 N)[(cos 30) ^
x + (sin 37) ŷ ] = (2.80 N) ^
x + (2.11 N) ŷ .
F 2 = (3.5 N)[(cos 37) ^
 F = F 1 + F 2 + F 3 = 0,
25.

F 3 = ( F 1 + F 2) = (–7.6 N) x̂ + (0.64 N) ŷ .
(a) Since the object slides upward, there is no acceleration in the horizontal
direction, so Fx = 0.
Fx = N  F cos 60 = 0,

N = F cos 60 = (60 N) cos 60 = 30 N .
(b) Fy = F sin 60  w = = F sin 60  mg = may,
so
ay =
F sin 60
(60 N) sin 60
g=
 9.80 m/s2 = 4.60 m/s2 .
m
10.0 kg
The negative sign means that the object is slowing down while it is sliding
upward.
29.
Applying Newton’s second law along the x- and y-axes gives
Fx = max = -ma cos 30° = –(2.0 kg)(1.5 m/s2) cos 30° = –2.6 N
Fy = may = ma sin 30° = (2.0 kg)(1.5 m/s2) sin 30° = 1.5 N
F = –2.6 N xˆ + 1.5 N yˆ
40.
(a) Fx = F cos 30 = (25 N) cos 30 = 21.65 N = max,
y
N
so
ax = = 0.72m/s2 .
F
30
(b) Fy = N + F sin 30  w = may = 0,
so
x
N = w  F sin 30 = (30 kg)(9.80 m/s2)  (25 N) sin 30 =
w = mg
2.8  102 N .
41.
(a) Fx = F cos = (30 N) cos 37 = 23.96 N = max,
so
y
N
23.96 N
ax =
= 0.96m / s2 .
25 kg

F
x
(b) Fy = N  F sin  w = may = 0,
N = w + F sin 37 = (25 kg)(9.80 m/s2) + (30 N) sin 37 = 2.6  102 N
43.
w = mg
The acceleration is zero. The direction of the force F must be along
and up the incline.
Fx = mg sin  F = max = 0,
so
y
N
F = mg sin  = (25.0 kg)(9.80 m/s2 ) sin 30 = 123 N up the incline .
F
x
30
m
g
44.
(a) There are (2) two forces on the skier, the weight and the
y
normal force.
N
(b) The x-component of the weight is the side opposite to the angle
 shown, so sine is used.
so
Fx = mg sin = max,
ax = g sin = (9.80 m/s2) sin 37 = 5.9 m/s2 .
(c) From v2 = vo2 + 2ax, v =

x
m
(5.0 m/s)2 + 2(5.9 m/s2)(35 m) =
21 m/s .
45.
From Exercise 44(b), ax = g sin = (9.80 m/s2) sin 30 = 4.9 m/s2 down the
slope.
If up the slope is chosen as positive, then a = 4.9 m/s2. Taking xo = 0.
v2 = vo2 + 2a(x  xo),
x=
 0 2 –  25m / s 2

2 –4.9m / s2

= 64 m .
(b)
48.
(a) There are two rings. Fy = T + T  w = 0.
So
50.

w mg (50 kg)(9.80 m/s2)
T= 2 = 2 =
= 2.7  102 N .
2
(b) In the vertical direction: Fy = T sin + T sin  mg = 0.
So
T
T=
mg
mg
(55 kg)(9.80 m/s2)
=
= 3.8  102 N .
2 sin
2 sin 45
Isolate point A (or B). The vertical component of the tension T in the topmost
cord supports half of the 10-kg weight. Therefore
T sin 45° = (5 kg)(9.80 m/s2) = 69.3 N
The horizontal forces at A balance, giving
TAB = T cos 45° = (69.3 N) cos 45° = 49 N
51.
The vertical component of the tension T in the lower cords attached at A and
B supports the 10-kg weight, so

T

2T cos 30° = mg = (10 kg)(9.80 m/s2)
T = 57 N
The tension in the vertical cord attached to the weight supports the full
weight, so its tension is mg = 98 N . From Exercise 50, the other two tensions
are 49 N and 69 N .
60.
The acceleration is zero if both are at rest.
For m1: Fx1 = T  m1 g sin = m1ax = 0,

T = m1 g sin.
For m2: Fy2 = m2 g  T = m2 ay = 0,

So
m2 g = T = m1 gsin.
m2 = m1 sin = (2.0 kg) sin 37 = 1.2 kg .
If both are moving at constant velocity, the answer is the same 1.2 kg
because the acceleration is still zero and the forces must still balance out.
61.
(a) For m1:
Fx1 = T  m1 g sin = m1ax. (1)
For m2:
Fy2 = m2 g  T = m2 ay.
(2)
ax = ay = a in magnitude.
Equation (1) + Equation (2) gives (m2  m1 sin)g = (m1 + m2)a.
m2  m1 sin
2.50 kg  (3.00 kg) sin 37
 (9.80 m/s2) = 1.24 m/s2
a=
g=
m1 + m 2
2.50 kg + 3.00 kg
=.1.2 m/s2 (m1 up and m2 down)
73.
Yes , the coefficient of kinetic friction can be found.
Fy = N  mg cos = may = 0,
For constant velocity,
Or
N = mg cos.

ax = 0.
So
mg sin = k N = k (mg cos).
Fx = mg sin  fk = 0.
Therefore
k =
sin
= tan .
cos
75.
(a) Use the result from Exercise 66(a).
s = tan, independent of m, size,
etc. So it is still 30 .
(b) s = tan 30 = 0.58 .
77.
(a) If there were no friction, the minimum pushing force is equal to F = mg
sin  = (35.0 kg)(9.80 m/s2) sin 20 = 117 N.
Since the person has to use a 150 N force, there is friction.
So the answer is no , the incline is not frictionless.
(b) fk = 150 N  117 N = 33 N .
79.
(a) First assume that m1 has the tendency to move up the incline.
For m1: Fy1 = N1  m1 g cos = 0,

N1 = m1 g cos.
Fx1 = T  m1 g sin  fsmax = T  m1 g sin  s N1
= T  m1 g sin  s m1 g cos = 0.
So
T = m1 g sin + s m1 g cos = m1 g(sin + s cos).
For m2: Fx2 = m2 g  T = 0,

m2 g = T = m1 g(sin + s cos).
Therefore
y
N
m2 = m1 (sin + s cos) = (2.0 kg)(sin 37 + 0.30 cos 37) =
F
1.7 kg.
fk
Next assume that m1 has the tendency to move down the
incline.
The static friction force will now be pointing up the incline,
so the term s m1 g cos becomes negative. Repeating the
calculation: m2 = m1 (sin  s cos) = (2.0 kg)(sin 37 
x

m
0.30 cos 37) = 0.72 kg.
Thus m2 can be anywhere between 0.72 kg and 1.7 kg .
(b) When both are moving at constant velocity, the acceleration is still zero.
However, s is replaced by k.
Assume that m1 has the tendency to move up the incline.
m2 = m1 (sin + k cos) = (2.0 kg)(sin 37 + 0.20 cos 37) = 1.5 kg.
Assume that m1 has the tendency to move down the incline.
m2 = m1 (sin  k cos) = (2.0 kg)(sin 37  0.20 cos 37) = 0.88 kg.
Therefore m2 can be anywhere between 0.88 kg and 1.5 kg .
84.
(a) As the boat rises, the cables will go flatter (the angle  decreases). The
sum of the vertical components of the tensions in the cable must support the
weight of the boat. The weight of the boat remains the same. As  decreases,
the tension must (1) increase .
(b) Fy = 2T sin   mg = ma = 0.
At 45, T =
mg
(500 kg)(9.80 m/s2)
=
= 3.46  103 N .
2 sin 
2 sin 45
At 30, T =
(500 kg)(9.80 m/s2)
= 4.90  103 N .
2 sin 30
As we see, the decrease in angle increases the tension.
88.
(a)
x
y
fk
N
mg

Since the block is moving downward on the incline but slowing down
(because it comes to rest), its acceleration must be up the incline; call this the
+x-direction, and call the +y-direction perpendicular to this pointing upward
from the surface of the incline. (It would also be possible to call the +x-axis
pointing down the plane.)
(b) Forces along the y-axis balance.
F
y
 N  mg cos   0
Acceleration is along the x-axis.
F
x
 f k  mg sin   ma
(c) Using kinematics to find the acceleration gives
v2 = v02 + 2a(x – x0)
0 = (1.60 m/s)2 + 2a(1.10 m)

a = 1.164 m/s2 up the incline
fk = µkN = µkmg cos 
Putting this into the result from Newton’s second law gives
µkmg cos –- mg sin  = ma
Solving for µk gives
1.164 m/s2
 sin 30
2
a / g  sin 
9.80
m/s
k 

 0.714
cos 
cos 30