Unformatted text preview: CHAPTER 4
Quick Quizzes
1. (a) True. Motion requires no force. Newton’s first law says an object in motion continues
to move by itself in the absence of external forces.
(b) False. It is possible for forces to act on an object with no resulting motion if the forces
are balanced. 2. (a) True. If a single force acts on an object, it must accelerate. From Newton's second law,
a = ΣF/m, and a single force must represent a nonzero net force.
(b) True. If an object accelerates, at least one force must act on it.
(c) False. If an object has no acceleration, you cannot conclude that no forces act on it. In
this case, you can only say that the net force on the object is zero. 3. False. If the object begins at rest or is moving with a velocity with only an x component,
the net force in the x direction causes the object to move in the x direction. In any other
case, however, the motion of the object involves velocity components in directions other
than x. Thus, the direction of the velocity vector is not generally along the x axis. What we
can say with confidence is that a net force in the x direction causes the object to accelerate in
the x direction. 4. Because the value of g is smaller on the Moon than on the Earth, more mass of gold would
be required to represent 1 newton of weight on the Moon. Thus, your friend on the Moon
is richer, by about a factor of 6! 5. (b). 6. (c); (d). 7. (c). The scale is in equilibrium in both situations, so it experiences a net force of zero.
Because each person pulls with a force F and there is no acceleration, each person is in
equilibrium. Therefore, the tension in the ropes must be equal to F. In case (i), the person
on the right pulls with force F on a spring mounted rigidly to a brick wall. The resulting
tension F in the rope causes the scale to read a force F. In case (ii), the person on the left
can be modeled as simply holding the rope tightly while the person on the right pulls.
Thus, the person on the left is doing the same thing that the wall does in case (i). The
resulting scale reading is the same whether there is a wall or a person holding the left side
of the scale. 8. (c). 9. (b). Friction forces are always parallel to the surfaces in contact, which, in this case, are the
wall and the cover of the book. This tells us that the friction force is either upward or
downward. Because the tendency of the book is to fall due to gravity, the friction force
must be in the upward direction.
93 CHAPTER 4 10. (b). The static friction force between the bottom surface of the crate and the surface of the
truck bed is the net horizontal force on the crate that causes it to accelerate. It is in the
same direction as the acceleration, to the east. 11. (b). It is easier to attach the rope and pull. In this case, there is a component of your
applied force that is upward. This reduces the normal force between the sled and the
snow. In turn, this reduces the friction force between the sled and the snow, making it
easier to move. If you push from behind, with a force with a downward component, the
normal force is larger, the friction force is larger, and the sled is harder to move. 94 CHAPTER 4 Problem Solutions
4.1 (a) ΣF = ma = ( 6.0 kg ) ( 2.0 (b) a = 4.2 ΣF =
m ) m s 2 = 12 N . 12 N
2
= 3.0 m s .
4.0 k g From v = v i + a t , the acceleration given to the football is a= v − v i 10 m s − 0
=
= 50 m s 2 .
t
0.20 s Then, from Newton’s 2n d law, we find ( ΣF) = m a = ( 0.50 kg ) ( 50 m s ) =
2 25 N . 4.3 ⎛ 2000 lb s ⎞ ⎛ 4.448 N ⎞ 2 × 104 N
.
w = ( 2 ton s ) ⎜
=
⎝ 1 t o n ⎟ ⎜ 1 lb ⎟
⎠⎝
⎠ 4.4 ⎛ 4.448 N ⎞ 1.7 × 102 N
.
w = ( 38 lb s ) ⎜
=
⎝ 1 lb ⎟
⎠ 4.5 ⎛ 4.448 N ⎞
The weight of the bag of sugar on Earth is w E = m gE = ( 5.00 lb s ) ⎜
= 22 .2 N . If g M
⎝ 1 lb ⎟
⎠ is the acceleration of gravity on the surface of the Moon, the ratio of the weight of an
⎛g ⎞
mg M g M
w
=
object on the Moon to its weight when on Earth is M =
, so w M = w E ⎜ M ⎟ .
w
mg
g
⎝g ⎠
E Hence, the weight of the bag of sugar on the Moon is w M E E ⎛ gJ ⎞
Jupiter, its weight would be w J = w E ⎜ ⎟ = ( 22.2 N ) ( 2.64) = 58.7 N .
⎝ gE ⎠
The mass is the same at all three locations, and is given by m= w E ( 5.00 lb ) ( 4.448 N lb )
=
= 2.27 k g .
9.80 m s 2
gE 95 E ⎛1⎞
= ( 22.2 N ) ⎜ ⎟ = 3.71 N . On
⎝6⎠ CHAPTER 4.6 a= ΣF = 7.5 × 105 N
m 1.5 × 10 k g
7 4 = 5.0 × 10−2 m s 2 , and v = v i + at gives
t= 4.7 80 k m h − 0 ⎛ 0.278 m s ⎞ ⎛ 1 m in ⎞
v − vi
=
⎜
⎟ = 7.4 m in
5.0 × 10−2 m s 2 ⎜ 1 km h ⎟ ⎝ 60 s ⎠
a
⎝
⎠ Summing the forces on the plane shown gives ( ΣFx = F − f = 10 N − f = ( 0.20 kg ) 2.0 m s 2 a = 2.0 m/s2 ) f F = 10 N From which, f = 9.6 N .
v 2 − v i2 ( 320 m s ) − 0
The acceleration of the bullet is given by a =
=
2 ( ∆x )
2 ( 0.82 m )
2 4.8 Then, ΣF = ma = ( 5.0 × 10−3 kg ) ⎢ ( ⎡ 320 m s ) 2 ⎤
⎥ = 3.1 × 102 N .
⎢ 2 ( 0.82 m ) ⎥
⎣
⎦ v 2 − v i2 (18.0 m s ) − 0
The average acceleration of the performer is a =
. Thus, the
=
2 ( ∆x )
2 ( 9.20 m )
average force acting on the performer is given by
2 4.9 ⎡ (18.0 m s ) 2 ⎤
⎥ = 1.41 × 103 N .
ΣF = ma = ( 80.0 kg ) ⎢
⎢ 2 ( 9.20 m ) ⎥
⎣
⎦ 4.10 4.11 ⎛
⎞
w
ΣFy = may becomes 4 ( 240 N ) − w = ⎜
0.504 m s 2 ) , from which w = 913 N .
2⎟(
⎝ 9.80 m s ⎠ (a) From the second law, the acceleration of the
boat is
a= ΣF = 2000 N − 1800 N
m 1000 k g f = 1800 N
2
= 0.200 m s . 96 F = 2000 N CHAPTER 4 (b) The distance moved is ( ) 1
1
2
∆x = v i t + at 2 = 0 + 0.200 m s 2 (10.0 s ) = 10.0 m .
2
2 ( ) (c) The final velocity is v = v i + at = 0 + 0.200 m s 2 (10.0 s ) = 2.00 m s .
4.12 (a) Choose the positive yaxis in the forward direction. We
resolve the forces into their components as +y N 30° The magnitude and direction of the resultant force is ΣF = ( ΣFx ) 2 + ( Fy )
Thus, 2 ⎛ ΣF ⎞
= 799 N , θ =t a n 1 ⎜ x ⎟ = 8.77° to right of yaxis.
⎝ ΣFy ⎠ Σ F = 799 N a t 8.77° to th e r igh t of th e for w a r d d ir ection (b) The acceleration is in the same direction as
a= 4.13 ΣF
m = . ΣF and is given by 799 N
2
= 0.266 m s .
3000 k g ( ) The weight is w = mg = ( 30.0 kg ) 9.80 m s 2 = 294 N directed downward and the
components of the three forces involved are:
Force
300 N
690 N
weight
Resultant xcomponent (N)
122
236
0
ΣFx = 114 N ycomponent (N)
274
648
294 N
ΣFy = 80.3 N 97 00 ycomponent
346 N
443 N
ΣFy = 790 N =4 xcomponent
200 N
–78.1 N
ΣFx = 122 N F1 Force
400 N
450 N
Resultant 0N
F2 = 45 10° +x CHAPTER ΣF = ( ΣFx ) 2 + ( Fy ) 2 4 ⎛ ΣFy ⎞
= 35.2° above horizontal.
= 139 N , θ =t a n 1 ⎜
⎝ ΣFx ⎟
⎠ The acceleration of the pelvis is in the direction of the resultant force and has magnitude
ΣF 139 N
a=
=
= 4.65 m s 2 , so
m
30.0 k g a = 4.65 m s 2 a t 35.2° a bov e t h e h or izon t a l t o t h e r igh t .
4.14 Since the two forces are perpendicular to each other, their resultant is: ΣF = (180 N ) 2 + ( 390 N ) 2 = 430 N , at
Thus, a = 4.15 ΣF
m = θ = t a n 1 ⎛
⎜ 390 N ⎞
= 65.2° N o f E .
⎝ 180 N ⎟
⎠ 430 N
= 1.59 m s 2 a t 65.2° N of E .
270 k g Since the burglar is held in equilibrium, the tension in the
vertical cable equals the burglar’s weight of 600 N y 37.0° T1 Now, consider the junction in the three cables: x w = 600 N ΣFy = 0 , giving T2 sin 37.0° − 600 N = 0 ,
or T2 T2 = 997 N in t h e in clin e d ca b le . Also, ΣFx = 0 which yields T2 co s 37.0° − T1 = 0 ,
or 4.16 T1 = ( 997 N ) co s 37.0° = 796 N in t h e h o r izon t al ca ble From ΣFx = 0 , T1 co s 40.0° − T2 co s 40.0° = 0 ,
or y
T2 T1 = T2 . 40.0° Then, ΣFy = 0 gives 2 ( T1 sin 40.0°) − 100 N = 0 ,
yielding T1 = T2 = 77.8 N 98 T1
40.0°
100 N x CHAPTER 4.17 4 From ΣFx = 0 , T1 co s 30.0° − T2 co s 60.0° = 0 , y
T2 or T2 = (1.73) T1 . T1 (1)
60.0° Then ΣFy = 0 becomes 30.0°
150 N x T1 sin 30.0° + (1.73 T1 ) sin 60.0° − 150 N = 0 ,
which gives T1 = 75.0 N in t h e r ig h t sid e ca b le .
Finally, Equation (1) above gives T2 = 130 N in t h e left sid e ca b le . 4.18 If the hip exerts no force on the leg, the system must be
in equilibrium with the three forces shown in the freebody diagram. y 11 0N 40° Thus ΣFx = 0 becomes w2
α
220 N w 2 cos α = (110 N ) cos 40° (1) From ΣFy = 0 , we find w 2 sin α = 220 N  (110 N ) sin 40° (2) Dividing Equation (2) by Equation (1) yields ⎛ 220 N  (110 N ) sin 40° ⎞
= 61° .
(110 N ) cos 40° ⎟
⎝
⎠ α = ta n −1 ⎜ Then, from either Equation (1) or (2), w 2 = 1.7 × 102 N . 99 x CHAPTER 4.19 4 We draw a freebody diagram of the forearm and sling as
shown. Here, F is the force exerted on the sling by the neck. ΣFx = 0 gives F sin θ = 24.0 N y
F θ (1) while ΣFy = 0 yields F co s θ = 98.0 N . 24.0 N
x (2) 98.0 N ⎛ 24.0 N ⎞
Dividing (1) by (2), we have θ = ta n −1 ⎜
= 13.8° .
⎝ 98.0 N ⎟
⎠ Then either (1) or (2) gives F = 101 N .
The force exerted on the neck by the sling is the reaction force to F and is given by
F = 101 N a t 13.8° t o t h e r ig h t o f v e r t ical a n d d o w n w a r d . 4.20 The resultant force exerted on the boat by the people is 2⎡( 600 N ) co s 30.0°⎤ = 1.04 × 103 N
⎣
⎦
in the forward direction. If the boat moves 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 = 1.04 × 103 N in t h e r ea r w a r d d ir ect io n . 4.21 m = 1.00 k g and mg = 9.80 N 25.0 m
α 0.200 m ⎞
α = tan ⎛
= 0.458°
⎜
⎝ 25.0 m ⎟
⎠ 0.200 m T −1 mg 25.0 m
α T Require that ΣFy = 0 ,
2 T s in α = m g T= 4.22 9.80 N
= 613 N
2sin α (a) An explanation proceeding from fundamental physical
principles will be best for the parents and for you.
Consider forces on the bit of string touching the weight
hanger as shown in the freebody diagram:
Horizontal Forces: ΣFx = 0 ⇒ − Tx + T co s θ = 0
Vertical Forces: ΣFy = 0 ⇒ − Fg + T sin θ = 0
100 T
θ Tx
mg CHAPTER 4 You need only the equation for the vertical forces to find that the tension in the
mg
. The force the child feels gets smaller, changing from T
string is given by T =
s in θ
to T cos θ, while the counterweight hangs on the string. On the other hand, the kite
does not notice what you are doing and the tension in the main part of the string
stays constant. You do not need a level, since you learned in physics lab to sight to a
horizontal line in a building. Share with the parents your estimate of the
experimental uncertainty, which you make by thinking critically about the
measurement, by repeating trials, practicing in advance and looking for variations
and improvements in technique, including using other observers. You will then be
glad to have the parents themselves repeat your measurements. ( (b) 4.23 ) 2
mg ( 0.132 kg ) 9.80 m s
T=
=
= 1.79 N .
sin θ
sin 46.3° The forces on the bucket are the tension in the rope and the weight of
the bucket, mg = ( 5.0 kg ) 9.80 m s 2 = 49 N . Choose the positive ( ) T direction upward and use the second law: ΣFy = may a ( T − 49 N = ( 5.0 kg ) 3.0 m s 2 )
mg T = 64 N .
4.24 (a) From the second law, we find the acceleration as
F = 10 N a= F 10 N
=
= 0.33 m s 2 .
m 30 kg To find the distance moved, we use ( ) 1
1
2
∆x = v i t + at 2 = 0 + 0.33 m s 2 ( 3.0 s ) = 1.5 m .
2
2
(b) If the shopper places her 30 N ( 3.1 k g ) child in the cart, the new acceleration will be a= F
10 N
=
= 0.30 m s 2 , and the new distance traveled in 3.0 s will be
mtotal 33 kg ∆x ′ = 0 + ( ) 1
2
0.30 m s 2 ( 3.0 s ) = 1.4 m .
2 101 CHAPTER 4.25 4 (a) The average acceleration is given by a= v f − vi
∆t = 5.00 m s − 20.0 m s
= −3.75 m s 2 .
4.00 s The average force is found from the second law as ( ΣF) = m a = ( 2000 kg ) ( −3.75 m s ) =
2 − 7.50 × 103 N . (b) The distance traveled is: ⎛ 5.00 m s + 20.0 m s ⎞
x = v ( ∆t ) = ⎜
⎟ ( 4.00 s ) = 50.0 m .
⎝
⎠
2
4.26 Let m1 = 10.0 k g , m2 = 5.00 k g , and θ = 40.0° .
Applying the second law to each object gives a
n m1 a = m1 g − T , and (1) m2 a = T − m2 g sin θ . T (2) m2 T
m1 θ m1g m 2g Adding these equations yields ⎛ m − m2 sin θ ⎞
a= ⎜ 1
g , or
⎝ m1 + m2 ⎟
⎠
⎛ 10.0 kg − ( 5.00 kg ) sin 40.0° ⎞
2
2
a= ⎜
⎟ 9.80 m s = 4.43 m s .
15.0 kg
⎝
⎠ ( ) Then, Equation (1) yields ⎡
⎤
T = m1 ( g − a) = (10.0 kg ) ⎣( 9.80 − 4.43) m s 2 ⎦ = 53.7 N .
4.27 (a) The resultant external force acting on this system having a total mass of 6.0 kg is
42 N directed horizontally toward the right. Thus, the acceleration produced is
a= ΣF =
m 42 N
2
= 7.0 m s h o r iz o nt a lly t o t h e r ig h t .
6.0 k g 102 a CHAPTER 4 (b) Draw a free body diagram of the 3.0kg block and apply Newton’s second law to
the horizontal forces acting on this block: ( ) ΣFx = max ⇒ 42 N − T = ( 3.0 kg ) 7.0 m s 2 , and therefore T = 21 N
(c) The force accelerating the 2.0kg block is the force exerted on it by the 1.0kg block.
Therefore, this force is given by: ( ) F = ma = ( 2.0 kg ) 7.0 m s 2 , or F = 14 N h o r iz o nt a lly t o t h e r ig h t
4.28 The acceleration of the mass down the incline is given by 1
1
2
∆x = v i t + at 2 , or 0.80 m = 0 + a( 0.50 s ) .
2
2
This gives a = 6.4 m s 2 . ( ) Thus, the force down the incline is F = ma = ( 2 .0 kg ) 6.4 m s 2 = 13 N .
4.29 Choose the positive x axis to be up the incline.
Then, ΣFx = max ⇒ T − (mg) sin 18.5° = max ,
which gives ax = ( ) T
140 N
− g ( sin 18.5°) =
− 9.80 m s 2 sin 18.5° = 0.390 m s 2 .
m
40.0 kg The velocity after moving 80.0 m up the incline is given by ( ) v = v i2 + 2 ax ( ∆x ) = 0+2 0.390 m s 2 ( 80.0 m ) = 7.90 m s . 103 CHAPTER 4.30 4 First consider the block moving along the
horizontal. The only force in the direction
of movement is T. Thus, +x 5.00 kg ΣFx = max ⇒ T = ( 5.00 k g ) a . T T (1) +y 10.0 kg mg = 98.0 N Next consider the block which moves
vertically. The forces on it are the tension
T and its weight, 98.0 N. ΣFy = may ⇒ 98.0 N − T = (10.0 k g ) a (a) (b) (2) Note that both blocks must have the same magnitude of acceleration. Equations (1) and
(2) can be solved simultaneously to give. a = 6.53 m s 2 , and T = 32.7 N . 4.31 F = ( mtotal ) a gives mtotal = F
14.0 N
=
= 5.51 kg .
a 2.54 m s 2 But, mtotal = m forearm + mobject , so mobject = mtotal − m forearm = 5.51 kg − 4.26 kg = 1.25 k g .
4.32 If the head exerts zero force on the neck, then Newton’s third law tells us that the neck
exerts zero force on the head. Hence, the only forces acting on the head are the three
forces shown in the Figure P4.32. Consider the components of these forces.
Force
F
100
270 N xcomponent
Fx
+53.0 N
+270 N ycomponent
Fy
+84.8 N
0 Since the head is to be held in equilibrium by these forces, the resultant of these forces
must be zero. Therefore, it is necessary that ΣFx = Fx + 53.0 N + 270 N =0 and ΣFy = Fy + 84.8 N =0 . This gives the components of force F as Fx = −323 N an d Fy = −84.8 N . 104 CHAPTER Hence, F = Fx2 + Fy2 = 4 ( −323 N ) 2 + ( −84.8 N ) 2 = 334 N ⎛ Fy ⎞
⎛ −84.8 N ⎞
= 14.7° ,
and θ = ta n −1 ⎜ ⎟ = t an −1 ⎜
⎝ −323 N ⎟
⎠
⎝ Fx ⎠
F = 334 N a t 14.7° b e lo w t h e h o r iz o n t a l t o th e le ft . or,
4.33
Trailer
300 kg Car
1000 kg
Rcar
T nT wT T F
nc wc nc
θ F Choose the +x direction to be horizontal and forward with the +y vertical and upward.
The common acceleration of the car and trailer then has components of
ax = +2.15 m s 2 a n d ay = 0 .
(a) The net force on the car is horizontal and given by ( ΣFx ) car = F − T = mcar ax = (1000 kg ) ( 2.15 m s 2 ) = 2.15 × 103 N for w a r d . (b) The net force on the trailer is also horizontal and given by ( ΣFx )trailer = + T = mtrailer ax = ( 300 kg ) ( 2.15 m s 2 ) = 645 N fo r w a r d . (c) Consider the freebody diagrams of the car and trailer. The only horizontal force
acting on the trailer is T = 645 N fo r w a r d , and this is exerted on the trailer by the
car. Newton’s third law then states that the force the trailer exerts on the car is
645 N t ow a r d t h e r e a r . 105 CHAPTER 4 (d) The road exerts two forces on the car. These are F a n d n c shown in the freebody
diagram of the car.
From part (a), () Also, ΣFy car F = T + 2 .15 × 103 N = + 2 .80 × 103 N . = n c − wc = mcar ay = 0 , so n c = w c = mcar g = 9.80 × 103 N . The resultant force exerted on the car by the road is then
2
Rcar = F2 + n c = ( 2 .80 × 10 N ) + ( 9.80 × 10 N )
2 3 3 2 = 1.02 × 10 4 N ⎛n ⎞
at θ = t a n −1 ⎜ c ⎟ = t a n −1 ( 3.51) = 74.1° above the horizontal and forward. Newton’s
⎝ F⎠
third law then states that the resultant force exerted on the road by the car is
1.02 × 104 N a t 74.1° below t h e h orizon t al a n d r earw a rd .
4.34 First, consider the 3.00kg rising mass.
The forces on it are the tension, T, and
its weight, 29.4 N. With the upward
direction as positive, the second law
becomes
T − 29.4 N = ( 3.00 k g ) a . + Rising
Mass
m1 = 3.00 kg Falling
Mass + m2 = 5.00 kg (1) The forces on the falling 5.00kg mass
are its weight and T, and its
acceleration has the same magnitude as
that of the rising mass. Choosing the
positive direction down for this mass,
gives
49 N − T = ( 5.00 k g ) a . T T w1 = 29.4 N (2) Equations (1) and (2) can be solved simultaneously to give
(a) the tension as T = 36.8 N ,
(b) and the acceleration as a = 2.45 m s 2 . 106 w2 = 49.0 N CHAPTER 4 (c) Consider the 3.00kg mass. We have ( ) 1
1
2
∆y = v iy t + ay t 2 = 0 + 2 .45 m s 2 (1.00 s ) = 1.23 m .
2
2
4.35 When the block is on the verge of moving, the static friction force has a magnitude
fs = ( fs ) max = µs n .
Since equilibrium still exists and the applied force is 75 N, we have ΣFx = 75 N − fs = 0 or ( fs )max = 75 N .
In this case, the normal force is just the weight of the crate, or n = mg . Thus, the
coefficient of static friction is µs = ( fs )max ( fs )max
n = mg = 75 N
= 0.38 .
( 20 kg ) 9.80 m s 2 ( ) After motion exists, the friction force is that of kinetic friction, f k = µk n .
Since the crate moves with constant velocity when the applied force is 60 N, we find that
ΣFx = 60 N − fk = 0 or fk = 60 N . Therefore, the coefficient of kinetic friction is µk = 4.36 fk
f
60 N
= k=
= 0.31 .
n m g ( 20 k g ) 9.80 m s 2 ( ) (a) The static friction force attempting to prevent motion may reach a maximum value
of ( fs )max = µsn1 = µsm1 g = ( 0.50) (10 kg ) ( 9.80 m s 2 ) = 49 N .
This exceeds the force attempting to move the system, the weight of m2 . Hence, the
system remains at rest and the acceleration is a = 0
(b) Once motion begins, the friction force retarding the motion is fk = µk n1 = µk m1 g .
This is less than the force trying to move the system, weight of m2 . Hence, the
system gains speed at the rate ( ) 2
m2 g − µk m1 g ⎡ 4.0 kg − 0.30 (10 k g ) ⎤ 9.80 m s
Fnet
⎣
⎦
=
=
= 0.70 m s 2
a=
4.0 kg + 10 k g
mtotal
m1 + m2 107 CHAPTER 4.37 4 (a) Since the crate has constant velocity, ax = ay = 0 .
Applying Newton’s second law: ΣFx = F cos 20.0° − fk = max = 0 , or fk = ( 300 N ) cos 20.0° = 282 N
and ΣFy = n − F sin 20.0° − w = 0 , or n = ( 300 N ) sin 20.0° + 1000 N = 1.10 × 103 N .
The coefficient of friction is then µk = fk
282 N
=
= 0.256 .
n 1.10 × 103 N (b) In this case, ΣFy = n + F sin 20.0° − w = 0 ,
so n = w − F sin 20.0° = 897 N .
The friction force now becomes fk = µk n = ( 0.256) ( 897 N ) = 230 N . ⎛ w⎞
Therefore, ΣFx = F cos 20.0° − fk = max = ⎜ ⎟ ax and the acceleration is
⎝ g⎠ a= 4.38 (a) ax = ( F cos 20.0° − fk ) g = [ ( 300 N ) cos 20.0° − 230 N ] ( 9.80
w v f − vi
t = m s2 1000 N )= 0.509 m s 2 6.00 m s − 12.0 m s
= − 1.20 m s 2 .
5.00 s (b) From Newton’s second law, ΣFx = − fk = max , or fk = −max .
The normal force exerted on the puck by the ice is n = mg , so the coefficient of
friction is µk = (c) (
( ) 2
fk −m −1.20 m s
=
= 0.122 .
n
m 9.80 m s 2 ) ⎛ v f + v i ⎞ ⎛ 12.0 m s + 6.00 m s ⎞
∆x = v t = ⎜
t =⎜
⎟ ( 5.00 s ) = 45.0 m .
⎠
2
⎝2⎟⎝
⎠ 108 CHAPTER 4.39 4 From Newton’s third law, the forward force of the ground
on the sprinter equals the magnitude of the friction force
the sprinter exerts on the ground. If the sprinter’s shoe is
not to slip on the ground, this is a static friction force and
its maximum magnitude is n ( fs )max =µ s mg .
From Newton’s second law applied to the sprinter,
( fs )max =µ s mg = mamax where amax is the maximum forward fs acceleration the sprinter can achieve. From this, the
acceleration is seen to be amax = µ s g . Note that the mass has
canceled out. w = mg If µ s = 0.800 , ( ) amax = ( 0.800) 9.80 m s 2 = 7.84 m s 2 in d ep en d en t of t h e m a ss .
4.40 m = 20.0 k g , F = 35.0 N n
F (a) Since the velocity is constant, cos θ = θ f ΣFx = F cos θ − f = 0 , or
f 20.0 N
=
= 0.571 , θ = 55.2°
F 35.0 N mg (b) ΣFy = n + F sin θ − mg = 0 , so n = m g − F sin θ = ⎡196 − 35.0 sin ( 55.2°) ⎤ N = 167 N
⎣
⎦
4.41 The normal force acting on the crate is given by
n = F + m g co s θ . The net force tending to move the crate down F n the incline is mg sin θ − fs , where fs is the force of static
friction between the crate and the incline. If the crate is in
equilibrium, then mg sin θ − fs = 0 , so that fs = mg sin θ .
But, we also know f s ≤ µs n = µs ( F + m g co s θ ) . 109 fs
θ = 35.0° mg CHAPTER 4 Therefore, we may write mg sin θ ≤ µs ( F + m g co s θ ) , or ⎛ sin θ
⎞
⎛ sin 35.0°
⎞
− cos θ ⎟ mg = ⎜
− cos 35.0°⎟ ( 3.00 k g ) 9.80 m s 2 = 32.1 N
F≥⎜
⎝ 0.300
⎠
⎝ µs
⎠ ( 4.42 ) In the vertical direction, we have n n  300 N  (400 N)sin 35.2° = 0
from which, n = 531 N.
Therefore, m = 30.6 kg f 35.2°
0N 40
300 N f =µk n = ( 0.570 ) ( 531 N ) = 302 N . From applying the second law to the horizontal motion, we have ( 400 N ) co s 35.2° − 302 N = ( 30.6 kg ) ax , yielding ax = 0.798 m s 2 ( ) 1
1
Then, from ∆x = v ix t + ax t 2 , we have 4.00 m = 0 + 0.798 m s 2 t 2 , which gives
2
2
t = 3.17 s . 4.43 (a) The object will fall so that ma = mg − bv , o r a = ( mg − bv ) downward direction is taken as positive.
Equilibrium ( a = 0 ) is reached when ( m where the f = bv
m
mg ) 2
mg ( 50 kg ) 9.80 m s
v = v terminal =
=
= 33 m s .
15 k g s
b (b) If the initial velocity is less than 33 m/s, then a ≥ 0 and 33 m/s is the largest
velocity attained by the object. On the other hand, if the initial velocity is greater
than 33 m/s, then a ≤ 0 and 33 m/s is the smallest velocity attained by the object.
Note also that if the initial velocity is 33 m/s, then a = 0 and the object continues
falling with a constant speed of 33 m/s. 110 CHAPTER 4.44 4 (a) Find the normal force n on the 25.0 kg box: y
n ΣFy = n + ( 80.0 N ) sin 25.0° − 245 N = 0 , F= f or n = 211 N . 0N
25.0°
x 80. mg = 245 N Now find the friction force, f, as f = µk n = 0.300 ( 211 N ) = 63.4 N .
From the second law, we have ΣFx = ma , or ( 80.0 N ) cos 25.0° − 63.4 N = ( 25.0 k g ) a which yields a = 0.366 m s 2 .
(b) When the box is on the incline, y .0 n ΣFy = n + ( 80.0 N ) sin 25.0° − ( 245 N ) cos 10.0° = 0 F
f 0
=8 N
25.0°
x 245 N giving n = 207 N .
The friction force is 10.0° f = µk n = 0.300 ( 207 N ) = 62.2 N .
The net force parallel to the incline is ΣFx = ( 80.0 N ) cos 25.0° − ( 245 N ) sin 10.0° − 62 .2 N =32 .3 N .
Thus, a = 4.45 ΣFx −32 .3 N
=
= − 1.29 m s 2 , or 1.29 m s 2 d o w n t h e in clin e
25.0 kg
m The acceleration of the system is found from 1
1
2
∆y = v iy t + at 2 , or 1.00 m = 0 + a(1.20 s )
2
2 a
n = m 1g
f which gives a = 1.39 m s . 111 m2 m1
m 1g 2 T T m2g a CHAPTER 4 Using the free body diagram of m2 , the second law gives ( 5.00 kg ) ( 9.80 m s 2 ) − T = ( 5.00 kg ) (1.39 m s 2 ) ,
or T = 42 .1 N . Then applying the second law to the horizontal motion of m1 , ( ) 42 .1 N − f = (10.0 kg ) 1.39 m s 2 , or f = 28.2 N .
Since n = m1 g = 98.0 N , we have µk = 4.46 f 28.2 N
=
= 0.287 .
n 98.0 N (a) The force of friction is found as f = µk n = µk ( m g ) .
Choose the positive direction of the xaxis in the direction of motion and apply the
−f
= − µk g .
second law. We have − f = max , or ax =
m
From v 2 = v i2 + 2a( ∆x ) , with v = 0, v i = 50.0 k m h = 13.9 m s , we find 0 = (13.9 m s ) + 2 ( − µk g ) ( ∆x ) , or
2 (13.9 m s )
∆x = 2 2µk g . (1) With µk = 0.100 , this gives ∆x = 98.6 m .
(b) With µk = 0.600 , Equation (1) above gives ∆x = 16.4 m .
4.47 The normal force exerted on the block by the incline is n = mg cos 15.0° = (19.6 N ) cos 15.0° = 18.9 N ,
and the friction force is f = µk n = ( 0.250) (18.9 N ) = 4.73 N .
We choose up the incline as the positive x direction. Then, the net force parallel to the
incline while the block is moving up the incline is ΣFx = −mg sin 15.0° − f = −9.81 N
and the acceleration is given by ax = ΣFx −9.81 N
=
= −4.90 m s 2 .
m
2.00 kg
112 CHAPTER 4 The distance the block travels up the incline before stopping is found to be v 2 − v i2 0 − ( 2.50 m s )
∆x =
=
= 0.637 m .
2 ax
2 − 4.90 m s 2
2 ( ) As the block starts from rest and comes back down the incline, the net accelerating force
ΣF
is ΣFx = −mg sin 15.0° + f = −0.340 N and the acceleration is ax = x = − 0.170 m s 2 .
m
Hence, the speed of the block when it returns to the bottom of the incline is given by ( ) v = v i2 + 2 ax ( ∆x ) = 0 + 2 − 0.170 m s 2 ( − 0.637 m ) = 0.465 m s .
4.48 (a) The acceleration of the car with the brakes applied is a= − f − 1200 N
=
= − 2.00 m s 2 .
600 k g
m After the car has traveled 250 m to the crossing, its speed will be
v f = v i2 + 2a( ∆x ) = ( 40 ( ) m s ) + 2 −2 .00 m s 2 ( 250 m ) = 25 m s .
2 (b) The time required for the car to reach the crossing is t= 2 ( 250 m )
∆x 2 ( ∆x )
=
=
= 7.8 s .
v
v f + v i 25 m s + 40 m s During this time, the train advances a distance of
∆x = v train t = ( 23 m s )( 7.8 s ) = 1.8 × 102 m . Since the train was originally 80 m from the crossing, we conclude that
a co llisio n w ill o ccu r if t h e t r a in is a t le a st 100 m lo n g . 113 CHAPTER 4.49 First, taking downward as positive, apply
the second law to the 12.0 kg block:
ΣFy = 118 N − T = (12 .0 k g ) a 4 +x T n
T (1)
12.0 kg For the 7.00 kg block, we have 7.00 kg +y
f n = ( 68.6 N ) co s 37.0° = 54.8 N , and 37.0° w = 118 N w = 68.6 N f = µk n = ( 0.250) ( 54.8 N ) = 13.7 N . Taking up the incline as the positive direction and applying the second law to the 7.00
kg block gives ΣFx = T − f − ( 68.6 N ) sin 37.0° = ( 7.00 k g ) a , or
T = 13.7 N + 41.3 N + ( 7.00 k g ) a (2) Solving Equations (1) and (2) simultaneously yields a = 3.30 m s 2 .
4.50 When the minimum force F is used, the block tends to
slide down the incline so the friction force, fs is directed
up the incline. +x
fs +y
m While the block is in equilibrium, we have ΣFx = F cos 60.0° + fs − (19.6 N ) sin 60.0° = 0 60.0° n F
60.0° (1) mg = 19.6 N and ΣFy = n − F sin 60.0° − (19.6 N ) co s 60.0° = 0 (2) For minimum F (impending motion), fs = ( fs ) max = µ s n = ( 0.300) n . (3) Equation (2) gives n = 0.866 F + 9.80 N . (4) (a) Equation (3) becomes: fs = 0.260 F + 2.94 N , so Equation (1) gives
0.500 F + 0.260 F + 2.94 N − 17.0 N = 0 , or F = 18.5 N . (b) Finally, Equation (4) gives the normal force 114 n = 25.8 N . CHAPTER 4 4.51
+y +y nground = w/2 = 85.0 lb
22.0° 22.0°
F2 ntip F1 f
+x 45.8 lb +x F= w = 170 lb FreeBody Diagram of Person 22.0° FreeBody Diagram of Crutch Tip From the freebody diagram of the person,
ΣFx = F sin ( 22 .0°) − F2 sin ( 22 .0°) = 0 , which gives or F = F2 = F .
1
1 Then, ΣFy = 2F cos 22 .0° + 85.0 lbs − 170 lbs = 0 yields F = 45.8 lb .
(a) Now consider the freebody diagram of a crutch tip. ΣFx = f − ( 45.8 lb ) sin 22 .0° = 0 , or f = 17.2 lb . ΣFy = nt ip − ( 45.8 lb ) cos 22 .0° = 0 , which gives ntip = 42 .5 lb .
For minimum coefficient of friction, the crutch tip will be on the verge of slipping,
f
17.2 lb
so f = ( f s ) max = µs n t ip and µs =
=
= 0.404 .
n tip 42 .5 lb
(b) As found above, the compression force in each crutch is F = F2 = F = 45.8 lb .
1
4.52 (a) First, draw a freebody diagram (Fig. 1),of the top
block. Since ay = 0, n1 = 19.6 N and, f = µk n1 = ( 0.300) (19.6 N ) = 5.88 N .
ΣFx = maT gives 10.0 N − 5.88 N = ( 2 .00 k g ) aT ,
or aT = 2.06 m s 2 . (for top block) 115 n1 = 19.6 N
Top Block
2.00 kg
f = µkn1 F mg = 19.6 N
Figure 1 CHAPTER 4 Now draw a freebody diagram (Fig. 2) of the bottom
block and observe that ΣFx = MaB gives n2 n1 f Bottom Block
8.00 kg f = 5.88 N = ( 8.00 k g ) aB , or Mg aB = 0.735 m s 2 . (for the bottom block) Figure 2 In time t, the distance each block moves (starting from
rest) is ( ) 1
dT = aT t 2 = 1.03 m s 2 t 2 , and
2 ( ) 1
dB = aB t 2 = 0.368 m s 2 t 2 .
2
For the top block to reach the right
edge of the bottom block, it is necessary
(See Fig. 3.) that
dT = dB + L , or L
Figure 3 (1.03 m s ) t = ( 0.368 m s ) t
2 dT
dB 2 2 2 + 3.00 m which gives t = 2.13 s . ( ) 1
2
(b) From above, dB = aB t 2 = 0.368 m s 2 ( 2 .13 s ) = 1.67 m .
2
4.53 Since the leg is at rest, the weights hanging from the ends of
the cables are in equilibrium. Therefore, the tension in each
cable is equal to the weight suspended from its end as shown
in the force diagram at the right. Note that in this diagram, the
xy plane is horizontal with the +y axis along the centerline
between the cables.
The net force exerted on the foot by the cables is the resultant
of the tension forces in the cables. This is computed as follows.
Force
75.0 N
45.0 N
Resultant xcomponent
+25.7 N
15.4 N
ΣFx = +10.3 N 116 ycomponent
+70.5 N
+42.3 N
ΣFy = +113 N +x 75.0 +y N
20.0°
20.0°
N
45.0 CHAPTER R= ( ΣFx ) 2 + ( ΣFy ) 2 4 = 113 N and, if we measure θ clockwise from the +y axis, ⎛ ΣFx ⎞
−1 ⎛ 10.3 N ⎞
= 5.20°
⎟ = tan ⎜
⎝ 113 N ⎟
⎠
⎝ ΣFy ⎠ θ = ta n −1 ⎜ The net force exerted on the leg by the cables is
113 N , h o r iz o n t a l a n d 5.20° fr om t h e ce n t er lin e b et w een t h e ca b les . 4.54 If the surface is inclined at angle θ to the horizontal, the component of the material’s
weight parallel to the incline is mg sin θ , directed down the slope. If we consider the
incline to be a frictionless surface, this is the only force parallel to the surface and the
acceleration down the slope is given by a= ΣFx mg sin θ
=
= g sin θ .
m
m The time for the material to slide down the incline is found from 1
1
∆x = v i t + at 2 = 0 + ( g sin θ )t 2 as
2
2 ( 9.80 m s ) sin 30.0°
2 = 12.8 s . +y 0 .
60 (a) The horizontal component of the resultant force exerted
on the light by the cables is N R x = ΣFx = ( 60.0 N ) co s 45.0° − ( 60.0 N ) co s 45.0° = 0 45.0° N 2 ( 400 m ) .0 4.55 2 ( ∆x )
=
g sin θ 60 t= 45.0°
+x The resultant y component is: Ry = ΣFy = ( 60.0 N ) sin 45.0° + ( 60.0 N ) sin 45.0° = 84.9 N .
Hence, the resultant force is 84.9 N v e r t ica lly u p w a r d . (b) The forces on the traffic light are the weight, directed downward, and the 84.9 N
vertically upward force exerted by the cables. Since the light is in equilibrium, the
resultant of these forces must be zero. Thus, w = 84.9 N .
117 CHAPTER 4.56 4 The acceleration of the ball is found from F v 2 − v i2 ( 20.0 m s ) − 0
a=
=
= 133 m s 2
2 ( ∆y )
2 (1.50 m )
2 m= 0.150 kg From the second law, ΣFy = F − w = ma , so
w = 1.47 N ( ) F = w + ma = 1.47 N + ( 0.150 kg ) 133 m s 2 = 21.5 N .
4.57 On the level surface, the normal force exerted on the sled by the ice equals the total
weight, or n = 600 N. Thus, the friction force is f = µk n = ( 0.050) ( 600 N ) = 30 N .
Hence, the second law yields ΣFx = − f = max , or ( ) − ( 30 N ) 9.80 m s 2
−f
−f
=
=
= − 0.49 m s 2 .
ax =
m wg
600 N
The distance the sled travels on the level surface before coming to rest is v 2 − v i2 0 − ( 7.0 m s )
∆x =
=
= 50 m .
2 ax
2 − 0.49 m s 2
2 ( 4.58 ) (a) For the suspended block, ΣFy = T − 50.0 N = 0 , so the tension in the rope is
T = 50.0 N . Then, considering the horizontal forces on the 100N block, we find
ΣFx = T − fs = 0 , or fs = T = 50.0 N . (b) If the system is on the verge of slipping, fs = ( fs ) max = µs n . Therefore,
the required coefficient of friction is µs = 118 fs 50.0 N
=
= 0.500 .
n 100 N CHAPTER 4 (c) If µk = 0.250 , then the friction force acting on the 100N block is fk = µk n = ( 0.250) (100 N ) = 25.0 N .
Since the system is to move with constant velocity, the net horizontal force on the
100N block must be zero, or ΣFx = T − fk = T − 25.0 N = 0 . The required tension in
the rope is T = 25.0 N . Now, considering the forces acting on the suspended block
when it moves with constant velocity, ΣFy = T − w = 0 , giving the required weight of
this block as w = T = 25.0 N . 4.59 (a) The force that accelerates the box is the friction force between the box and the
truck bed.
(b) The maximum acceleration the truck can have before the box slides is found by
considering the maximum static friction force the truck bed can exert on the box:
( fs )max = µsn = µs ( mg ) .
Thus, from the second law, amax = 4.60 ( fs )max
m = µs ( mg )
m ( ) = µs g = ( 0.300) 9.80 m s 2 = 2.94 m s 2 . Consider the vertical forces acting on the block: ΣFy = ( 85.0 N ) sin 55.0° − 39.2 N n = may = 0 , 4.00 kg
55.0° so the normal force is n = 30.4 N . 85.0 N Now, consider the horizontal forces: ( ΣFx = ( 85.0 N ) cos 55.0° − fk = max = ( 4.00 kg ) 6.00 m s 2
or n fk ) fk = ( 85.0 N ) cos 55.0° − 24.0 N = 24.8 N . The coefficient of kinetic friction is then µk = 119 fk 24.8 N
=
= 0.814 .
n 30.4 N mg = 39.2 N CHAPTER 4.61 4 When an object of mass m is on this frictionless incline, the only force acting parallel to
the incline is the parallel component of weight, mg sin θ directed down the incline. The
acceleration is then a= F mg sin θ
=
= g sin θ = 9.80 m s 2 sin 35.0° = 5.62 m s 2
m
m ( ) directed down the incline.
(a) The time for the sled projected up the incline to come to rest is given by t= v f − vi
a = 0 − 5.00 m s
= 0.890 s .
− 5.62 m s 2 The distance the sled travels up the incline in this time is
⎛ v f + vi ⎞
⎛ 0 + 5.00 m s ⎞
∆s = v t = ⎜
⎟ ( 0.890 s ) = 2.22 m .
⎟t =⎜
⎝
⎠
2
⎝2⎠ (b) The time required for the first sled to return to the bottom of the incline is the same
as the time needed to go up, i.e., t = 0.890 s . In this time, the second sled must travel
down the entire 10.0 m length of the incline. The needed initial velocity is found
1
from ∆s = v i t + at 2 as
2 vi = ( ) −5.62 m s 2 ( 0.890 s )
∆s at −10.0 m
−=
−
= −8.74 m s ,
t
2
0.890 s
2 or 8.74 m s d o w n t h e in clin e .
4.62 Let m1 = 5.00 k g , m 2 = 4.00 k g , a n d m 3 = 3.00 k g . Let T1 be the tension in the string
between m1 a n d m2 , and T2 the tension in the string between m2 an d m3 . 120 CHAPTER 4 (a) We may apply Newton’s second law to each of the masses.
for m1: m1 a = T1 − m1 g (1) for m2 : m2 a = T2 + m 2 g − T1 (2) for m3 : m 3 a = m 3 g − T2 (3) Adding these equations yields ( m1 + m2 + m3 ) a = ( −m1 + m2 + m3 ) g , so ⎛ 2.00 k g ⎞
⎛ −m + m 2 + m 3 ⎞
2
2
a= ⎜ 1
⎟ g = ⎜ 12.0 kg ⎟ 9.80 m s = 1.63 m s .
⎝ m1 + m2 + m3 ⎠
⎝
⎠ ( ) ( ) ( ) (b) From Equation (1), T1 = m1 ( a + g ) = ( 5.00 kg ) 11.4 m s 2 = 57.2 N , and
from Equation (3), T2 = m3 ( g − a) = ( 3.00 kg ) 8.17 m s 2 = 24.5 N . 4.63 (a) 1
1
∆x = v i t + ax t 2 = 0 + ax t 2 gives:
2
2 ax = 2 ( ∆x ) 2 ( 2 .00 m )
2
=
= 1.78 m s .
2
t2
1.50 s )
( (b) Considering forces parallel to the incline, the
second law yields +y µ
fk = n ( ) ΣFx = ( 29.4 N ) sin 30.0° − fk = ( 3.00 kg ) 1.78 m s 2 ,
or fk = 9.37 N . +x
30.0° Perpendicular to the plane, we have equilibrium,
so 30.0° ΣFy = n − ( 29.4 N ) co s 30.0° = 0 , or n = 25.5 N
Then, µk = fk 9.37 N
=
= 0.368 .
n 25.5 N (c) From part (b) above, fk = 9.37 N .
(d) Finally, v 2 = v i2 + 2ax ( ∆x ) gives ( ) v = v i2 + 2ax ( ∆x ) = 0 + 2 1.78 m s 2 ( 2.00 m ) = 2.67 m s . 121 w = mg = 29.4 N kn CHAPTER 4.64 4 (a) Force diagrams for penguin and sled are shown. The primed forces are reaction
forces for the corresponding unprimed forces. WALL n1
T′ n1 ′ T n2 f1′ F = 45 N f1
f2 w = 49 N w2 = 98 N F′
(force
on Pulling
Agent) n2′
w2′ w′ f2′ Earth Earth (b) The weight of the penguin is 49 N, and hence the normal force exerted on him by
the sled, n1 , is also 49 N. Thus, the friction force acting on the penguin is:
f1 = µk n1 = 0.20 ( 49 N ) = 9.8 N .
Since the penguin is in equilibrium, the tension in the cord attached to the wall and
the friction force f1 must be equal: T = 9.8 N
(c) The normal force exerted on the sled by the Earth is the weight of the penguin
(49 N) plus the weight of the sled (98 N). Thus, the net normal force, n2 equals
147 N, and the friction force between sled and ground is:
f2 = µk n 2 = 0.20 (147 N ) = 29.4 N .
Applying the second law to the horizontal motion of the sled gives: 45 N − f1′ − f2 = ( 10 k g ) a ,
4.65 2
or a = 0.58 m s . Figure 1 is a freebody diagram for the system
consisting of both blocks. The friction forces are
f1 = µ k n1 = µk ( m1 g ) and f 2 = µk ( m 2 g ) . For this a
n1 system, the tension in the connecting rope is an
internal force and is not included in second law
calculations. The second law gives
ΣFx = 50 N − f1 − f2 = ( m1 + m2 ) a , which reduces to a= 50 N
− µk g .
m1 + m2 (1) 122 n2 m1
f1 m2 m1g f2 Figure 1 m2g 50 N CHAPTER 4 a Figure 2 gives a freebody diagram of m1 alone. For
this system, the tension is an external force and
must be included in the second law. We find:
ΣFx = T − f1 = m1 a , or
T = m1 ( a + µk g ) . n1
m1
f1 (2) T m1 g
Figure 2 (a) If the surface is frictionless, µk = 0 . Then, Equation (1) gives a= 50 N
50 N
−0=
= 1.7 m s 2
m1 + m2
30 k g ( ) and Equation (2) yields T = (10 kg ) 1.7 m s 2 + 0 = 17 N .
(b) If µk = 0.10 , Equation (1) gives the acceleration as a= ( ) 50 N
− ( 0.10) 9.80 m s 2 = 0.69 m s 2 ,
30 kg while Equation (2) gives the tension as ( ) T = (10 k g ) ⎡0.69 m s 2 + ( 0.10) 9.80 m s 2 ⎤ = 17 N .
⎣
⎦ 4.66 Before he enters the water, the diver is in freefall with an acceleration of 9.80 m s 2
downward. Taking downward as the positive direction, his velocity when he reaches the
water is given by ( ) v = v i2 + 2a( ∆y ) = 0 + 2 9.80 m s 2 (10.0 m ) = 14.0 m s .
His average acceleration during the 2.00 s after he enters the water is a= v f − vi
t = 0 − (14.0 m s )
2.00 s = − 7.00 m s 2 . Continuing to take downward as the positive direction, the average upward force by the
water is found as ΣFy = F + m g = m a , o r
F = m ( a − g ) = ( 70.0 k g ) ⎡( −7.00 m s 2 ) − 9.80 m s 2 ⎤ = − 1.18 × 10 3 N ,
⎣
⎦ or F = 1.18 × 103 N u p w a r d .
123 CHAPTER 4.67 4 We shall choose the positive direction to be to the right and call the forces exerted by
each of the people F1 and F2. Thus, when pulling in the same direction, Newton’s second
law becomes ( ) F + F2 = ( 200 kg ) 1.52 m s 2 , or
1 F + F2 = 304 N .
1 (1) F − F2 = −104 N .
1 (2) When pulling in opposite directions, ( ) F − F2 = ( 200 kg ) −0.518 m s 2 , or
1
Solving simultaneously, we find:
4.68 F = 100 N , and F2 = 204 N .
1 In the vertical direction, we have ΣFy = T cos 4.0° − mg = 0 , or +y T= mg
.
co s 4.0° 4.0°
T In the horizontal direction, the second law becomes:
mg ΣFx = T sin 4.0° = m a , so a= +x T sin 4.0°
= g tan 4.0° = 0.69 m s 2 .
m 124 CHAPTER 4.69 4 The magnitude of the acceleration is a = 2.00 m s 2 for all three blocks and applying
Newton’s second law to the 10.0kg block gives (10.0 kg ) ( 9.80 m s 2 ) − T1 = (10.0 kg ) ( 2.00 m s 2 ) , or T1 = 78.0 N .
Applying the second law to the 5.00kg block gives: ( ( ) ) T1 − T2 − µk ⎡( 5.00 k g ) 9.80 m s 2 ⎤ = ( 5.00 k g ) 2 .00 m s 2 .
⎣
⎦ T2 = 68.0 N  ( 49.0 N ) µk With T1 = 78.0 N , this simplifies to: (1) For the 3.00kg block, the second law gives T2 − µk n − ⎡mg sin 25.0°⎤ = ma .
⎣
⎦
With m = 3.00 kg, a = 2.00 m s 2 , g = 9.80 m s 2 , a n d n = mg cos 25.0° , this reduces to: T2 − ( 26.6 N ) µk = 18.4 N (2) Solving Equations (1) and (2) simultaneously, and using the value of T1 from above, we
find that
(a) T1 = 78.0 N , T2 = 35.9 N , and (b) µk = 0.656 4.70 The scale simply reads the magnitude of the normal force exerted on the student by the
seat. The seat is parallel to the track, and hence inclined at 30.0° to the horizontal. Thus,
the magnitude of this normal force and the scale reading is
n = mg cos θ = ( 200 lb ) cos 30.0° = 173 lb . 4.71 Choose the positive x axis to be down the incline and the y
axis perpendicular to this as shown in the freebody
diagram of the toy. The acceleration of the toy then has
components of ay = 0, a n d ax = ∆v x + 30.0 m s
=
= + 5.00 m s 2 .
6.00 s
∆t +y
T +x Applying the second law to the toy gives:
(a) ⎛a ⎞
⎛ 5.00 m s 2 ⎞
= 30.7° ,
ΣFx = mg sin θ = max , θ = sin −1 ⎜ x ⎟ = sin −1 ⎜
2
⎝ g⎠
⎝ 9.80 m s ⎟
⎠
and
125 m a
θ mg θ CHAPTER 4 (b) ΣFy = T − mg cos θ = may = 0 , or ( ) T = mg cos θ = ( 0.100 kg ) 9.80 m s 2 cos 30.7° = 0.843 N .
4.72 Taking the downward direction as positive, applying the second law to the falling
person yields ΣFy = mg − f = m ay , or ay = g − ⎛ 100 N ⎞
f
= 9.80 m s 2 − ⎜
= 8.6 m s 2 .
m
⎝ 80 k g ⎟
⎠ 2
Then, v y = v i2y + 2 ay ( ∆y ) gives the velocity just before hitting the net as ( ) v y = v i2y + 2 ay ( ∆y ) = 0 + 2 8.6 m s 2 ( 30 m ) = 23 m s . 4.73 The acceleration the car has as it is coming to a stop is
a= v 2 − v i2
f
2 ( ∆x ) = 0 − ( 35 m s ) 2 2 (1000 m ) = − 0.61 m s 2 . Thus, the magnitude of the total retarding force acting on the car is ⎛ w⎞
⎛ 8800 N ⎞
F =m a =⎜ ⎟ a =⎜
0.61 m s 2 = 5.5 × 10 2 N .
2
⎝ g⎠
⎝ 9.80 m s ⎟
⎠ ( 4.74 ) (a) In the vertical direction, we have 8000 N ΣFy = ( 8000 N ) sin 65.0° − w = m ay = 0 , 65.0° so w = ( 8000 N ) sin 65.0° = 7.25 × 103 N . w = mg (b) Along the horizontal, the second law yields ⎛ w⎞
ΣFx = ( 8000 N ) cos 65.0° = max = ⎜ ⎟ ax , or
⎝ g⎠ ax = g ⎡( 8000 N ) co s 65.0°⎤
⎣
⎦
w ( 9.80 m s ) ( 8000 N ) cos 65.0° =
=
2 7.25 × 10 N
3 126 4.57 m s 2 CHAPTER 4.75 4 First, we will compute the needed accelerations: ay = 0 . (1) Before it starts to move: (2) During the first 0.80 s: ay = (3) While moving at constant velocity: ay = 0 . (4) During the last 1.5 s: ay = v y − v iy
t v y − v iy
t = = 1.2 m s − 0
= 1.5 m s 2 .
0.80 s 0 − 1.2 m s
= −0.80 m s 2 .
1.5 s Applying Newton’s second law to the vertical motion of the man gives: ( ) ΣFy = n − mg = may , or n = m g + ay . ( ) n = ( 72 kg ) 9.80 m s 2 + 0 = 7.1 × 102 N . (a) When ay = 0 , (b) When ay = 1.5 m s 2 , n = 8.1 × 102 N . n = 7.1 × 102 N . (c) When ay = 0 , (d) When ay = −0.80 m s 2 , n = 6.5 × 102 N .
4.76 Consider the two freebody
diagrams, one of the penguin alone
and one of the combined system
consisting of penguin plus sled. n1 The normal force exerted on the
penguin by the sled is n2 F f1 f2 w1 = 70.0 N n 1 = w 1 = m1 g wtotal = 130 N and the normal force exerted on the combined system by the ground is
n 2 = w total = mt ot al g = 130 N . The penguin is accelerated forward by the static friction force exerted on it by the sled.
When the penguin is on the verge of slipping, this acceleration is amax = ( f1 )max
m1 = µs ( m1 g )
m1 ( ) = µs g = ( 0.700) 9.80 m s 2 = 6.86 m s 2 .
127 CHAPTER 4 Since the penguin does not slip on the sled, the combined system must have the same
acceleration as the penguin. Hence, applying the second law to the combined system
gives ΣFx = F − f2 = mtotal amax , or ⎛w ⎞
F = f2 + mt otal amax = µk ( w t ot al ) + ⎜ total ⎟ amax .
⎝g⎠
⎛ 130 N ⎞
6.86 m s 2 = 104 N .
This yields F = ( 0.100) (130 N ) + ⎜
2⎟
⎝ 9.80 m s ⎠ ( 4.77 ) Since the board is in equilibrium, ΣFx = 0 and we see that
the normal forces must be the same on both sides of the
board. Also, if the minimum normal forces (compression
forces) are being applied, the board is on the verge of
slipping and the friction force on each side is
f = ( f s ) max = µs n . n f f n The board is also in equilibrium in the vertical direction, so
w = 95.5 N w
ΣFy = 2 f − w = 0, o r f = .
2
The minimum compression force needed is then n= 4.78 f µs = 95.5 N
w
=
= 72 .0 N .
2 µs 2 ( 0.663) The net force exerted on the leg by the cable is a horizontal force
directed to the left and equal to the tension in the cable. This
tension may be determined by applying the conditions of
equilibrium to the point where the weight is attached to the cable. T1 30° 30° T2 ΣFx = 0 ⇒ T2 sin 30° − T1 sin 30° = 0 , so T2 = T1 = T .
Then, ΣFy = 0 ⇒ T cos 30° + T cos 30° − w = 0 , or the force applied
to the leg is T= w
55 N
=
= 32 N .
2co s 30° 2co s 30° 128 w = 55 N CHAPTER 4.79 (a) Consider the first freebody
diagram in which Chris and the
chair treated as a combined system.
The weight of this system is
w total = 480 N , and its mass is
w
mtotal = total = 49.0 kg .
g 4 n T = 250 N T = 250 N T = 250 N
a Taking upward as positive, the
acceleration of this system is found
from the second law as
wChris = 320 N wtotal = 320 N + 160 N ΣFy = 2T − wtotal = mtotal ay .
Thus,
500 N − 480 N
ay =
= + 0.408 m s 2 ,
49.0 kg
or 0.408 m s 2 u pw a r d . (b) The downward force that Chris exerts on the chair has the same magnitude as the
upward normal force exerted on Chris by the chair. This is found from the freebody diagram of Chris alone as ΣFy = T + n − wChris = mChris ay , n = mChris ay + wChris − T . ⎛ 320 N ⎞
0.408 m s 2 + 320 N − 250 N = 83.3 N .
Hence, n = ⎜
2
⎝ 9.80 m s ⎟
⎠ ( 4.80 ) Let R represent the horizontal force of air resistance. Since the helicopter and bucket
move at constant velocity, ax = ay = 0 . The second law then gives:
+y ΣFy = T cos 40.0° − mg = 0 , or T = mg
.
cos 40.0° Also, ΣFx = T sin 40.0° − R = 0 , or R = T sin 40.0° .
Thus,
R = mg ta n 40.0° = ( 620 kg ) 9.80 m s 2 t a n 40.0° = 5.10 × 103 N . ( ) 129 40.0° T
R +x
w = mg CHAPTER 4 Answers to Even Numbered Conceptual Questions
2. If the car is traveling at constant velocity, it has zero acceleration. Hence, the resultant
force acting on it is zero. 4. The force causing the ball to rebound upward is the normal force exerted on the ball by
the floor. 6. w = mg and g decreases with altitude. Thus, to get a good buy, purchase it in Denver. If
gold was sold by mass, it would not matter where you bought it. 8. If it has a large mass, it will take a large force to alter its motion even when floating in
space. Thus, to avoid injuring herself, she should push it gently toward the storage
compartment. 10. The net force acting on the object decreases as the resistive force increases. Eventually, the
resistive force becomes equal to the weight of the object, and the net force goes to zero. In
this condition, the object stops accelerating, and the velocity stays constant. The rock has
reached its terminal velocity. 12. The barbell always exerts a downward force on the lifter equal in magnitude to the
upward force that he exerts on the barbell. Since the lifter is in equilibrium, the magnitude
of the upward force exerted on him by the scale (i.e., the scale reading) equals the sum of
his weight and the downward force exerted by the barbell. As the barbell goes through the
bottom of the cycle and is being lifted upward, the scale reading exceeds the combined
weights of the lifter and the barbell. At the top of the motion and as the barbell is allowed
to move back downward, the scale reading is less than the combined weights. If the
barbell is moving upward, the lifter can declare he has thrown it just by letting go of it for
a moment. Thus, the case is included in the previous answer. 14. While the engines operate, their total upward thrust exceeds the weight of the rocket, and
the rocket experiences a net upward force. This net force causes the upward velocity of the
rocket to increase in magnitude (speed). The upward thrust of the engines is constant, but
the remaining mass of the rocket (and hence, the downward gravitational force or weight)
decreases as the rocket consumes its fuel. Thus, there is an increasing net upward force
acting on a diminishing mass. This yields an acceleration that increases in time. 16. The truck’s skidding distance can be shown to be x = 18. Because the mass of the truck is decreasing, the acceleration will increase. v i2 where µ k is the coefficient of
2 µk g
kinetic friction and v i is the initial velocity of the truck. This equation demonstrates that
the mass of the truck does not affect the skidding distance, but halving the velocity will
decrease the skidding distance by onefourth. 130 CHAPTER 20. 4 Walking is a familiar example. To walk, you push backward on the ground, and the
friction force pushes you forward, in the direction of your walking motion. In this case the
impending motion of the object, your foot, is backward relative to the ground, and the
friction force exerted on your foot is forward. Another example is that of a package on the
back of a pickup truck. As the truck accelerates forward, the inertia of the package tends
to cause it to be left behind. Thus, the impending motion relative to the floor of the truck is
toward the rear, and the friction force exerted on the package is forward, in the same
direction as its motion. 131 CHAPTER 4 Answers to Even Numbered Problems
2. 25 N 4. 1.7 × 102 N 6. 7.4 min 8. 3.1 × 102 N 10. 913 N 12. (a) 799 N at 8.77° to the right of forward direction
(b) 0.266 m s 2 in the direction of the resultant force 14. 1.59 m s 2 at 65.2° N of E 16. 77.8 N in each wire 18. 1.7 × 102 N , 61° 20. 1.04 × 103 N r ea r w a r d mg
sin θ 22. (a) T= 24. (a) 1.5 m 26. 4.43 m s 2 up the incline, 53.7 N 28. 13 N down the incline 30. 6.53 m s 2 , 32.7 N 32. 334 N at 14.7° below the horizontal to the left 34. (a) 36.8 N (b) 2 .45 m s 2 36. (a) 0 (b) 0.70 m s 2 38. (a) −1.20 m s 2 (b) 0.122 40. (a) 55.2° (b) 167 N 42. 3.17 s (b) 1.79 N (b) 1.4 m 132 (c) 1.23 m (c) 45.0 m CHAPTER 4 44. (a) 0.366 m s 2 (b) 1.29 m s 2 down the incline 46. (a) 98.6 m (b) 16.4 m 48. (a) 25 m s (b) a collision occurs if the train is at least 100 m long 50. (a) 18.5 N (b) 25.8 N 52. (a) 2.13 s (b) 1.67 m 54. 12.8 s 56. 21.5 N 58. (a) (b) 0.500 60. 0.814 62. (a)
(b) 1.63 m s 2
57.2 N tension in string connecting 5kg and 4kg, 24.5 N tension in string connecting
4kg and 3kg 64. (b) 9.8 N 66. 1.18 × 103 N upward 68. 0.69 m s 2 70. 173 lb 72. 23 m s 74. (a) 76. 104 N 78. 32 N 80. 5.10 × 103 N 50.0 N 7.25 × 103 N (c) 0.58 m s 2 (b) 4.57 m s 2 133 (c) 25.0 N 134 ...
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This note was uploaded on 03/03/2010 for the course PHY P221 taught by Professor Dr.ha during the Spring '05 term at Indiana University South Bend.
 Spring '05
 Dr.Ha
 Physics, Force

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