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More Applications of
Newton’s Laws
CHAPTER OUTLINE
5.1 Forces of Friction 5.2 Newton’s Second Law Applied to a Particle in Uniform Circular Motion 5.3 Nonuniform Circular Motion 5.4 Motion in the Presence of Velocity-Dependent Resistive Forces 5.5 The Fundamental Forces of Nature 5.6 Context ConnectionDrag Coefficients of Automobiles
ANSWERS TO QUESTIONS
Q5.1 (a) m^ r^ m
r (^) r a = R + g (b) ma = T − mg (c) ma = f − R
r r
r
r
r
r r
f
r
FIG. Q5.
Q5.2 (a) The friction of the road pushing on the tires of a car causes an automobile to move.
(b) The push of the air on the propeller moves the airplane.
(c) The push of the water on the oars causes the rowboat to move.
Q5.3 As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force of the Earth on his foot. In the second case, the action is the force exerted on the girl’s back by the snowball; the reaction is the force exerted on the snowball by the girl’s back. The third action is the force of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is the force exerted on the window by the air molecules; the reaction is the force on the air molecules exerted by the window. We could in each case interchange the terms ‘action’ and ‘reaction.’
Q5.4 The tension in the rope must be 9 200 N. Since the rope is moving at a constant speed, then the resultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers were winning with the rope steadily moving in their direction or if the contest was even, then the tension would still be 9 200 N. In all of these cases, the acceleration is zero, and so must be the resultant force on the rope. To win the tug-of-war, a team must exert a larger force on the ground than their opponents do.
Q5.5 If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between the tires and the road is less than the maximum static friction force. Anti-lock brakes work by “pumping” the brakes (much more rapidly that you can) to minimize skidding of the tires on the road.
114 More Applications of Newton’s Laws
Q5.6 With friction, it takes longer to come down than to go up. On the way up, the frictional force and the component of the weight down the plane are in the same direction, giving a large acceleration. On the way down, the forces are in opposite directions, giving a relatively smaller acceleration. If the incline is frictionless, it takes the same amount of time to go up as it does to come down.
Q5.7 As you pull away from a stoplight, friction is the force that accelerates forward a box of tissues on the level floor of the car. At the same time, friction of the ground on the tires of the car accelerates the car forward. Drop a stick into a running stream and fluid friction sets the stick into horizontal motion.
Q5.8 The speed changes. The tangential force component causes tangential acceleration.
Q5.9 A torque is exerted by the thrust force of the water times the distance between the nozzles.
Q5.10 This is the same principle as the centrifuge. All the material inside the cylinder tends to move along a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving around in a circular path.
Q5.11 The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and around in a circle.
Q5.12 Blood pressure cannot supply the force necessary both to balance the gravitational force and to provide the centripetal acceleration, to keep blood flowing up to the pilot’s brain.
Q5.13 I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” one would have to be infinitely far away from all other matter. Second, astronauts in orbit are moving in a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the space shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface. Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces and is in free fall.
Q5.14 From the proportionality of the drag force to the speed squared and from Newton’s second law, we derive the equation that describes the motion of the skydiver:
m
dv dt mg^
y D A v = − y
2
where D is the coefficient of drag of the parachutist, and A is the projected area of the parachutist’s body. At terminal speed,
a
dv y dt = y = 0 and v mg T =^ D A
F HG^
I KJ
When the parachute opens, the coefficient of drag D and the effective area A both increase, thus reducing the speed of the skydiver. Modern parachutes also add a third term, lift, to change the equation to
m
dv dt
y = mg − D^ ρ A v y − L^ ρ A vx
2 2
where v (^) y is the vertical velocity, and v (^) x is the horizontal velocity. The effect of lift is clearly seen in the “paraplane,” an ultralight airplane made from a fan, a chair, and a parachute.
116 More Applications of Newton’s Laws P5.3 If all the weight is on the rear wheels,
(a) F = ma : μ smg = ma
But ∆ x = at^ = sgt
2 2 2 2
μ so μ
s
x gt
= 2 ∆ 2 : μ s = =
mi m mi m s 2 s
a fb g e ja^ f^
(b) Time would increase, as the wheels would skid and only kinetic friction would act; or perhaps the car would flip over.
*P5.
f
F = 45 8. lb Fg = 170 lb 22.0°
F 2 F 1
n (^) ground = Fg 2 =85 0. lb n tip
Free-Body Diagram of Person Free-Body Diagram of Crutch Tip
FIG. P5.
From the free-body diagram of the person, (^) ∑ Fx = F 1 (^) sin a 22 0. ° f − F 2 sin a22 0 .° = f 0 , which gives
F 1 = F 2 = F. Then, ∑ Fy = 2 F cos 22 0. °+ 85 0. lbs − 170 lbs = 0 yields F = 45 8. lb.
(a) Now consider the free-body diagram of a crutch tip.
∑^ F^ x =^ f −(^ 45 8.^ lb)^ sin^ 22 0.°=^0 ,
or
f = 17 2. lb.
∑^ Fy^ =^ n tip −(^ 45 8.^ lb)^ cos^ 22 0.^ ° =^0 ,
which gives
n tip = 42 5. lb.
For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so f = (^) a f f (^) s (^) max= μ (^) sn tip and μ s f n
tip
lb 42.5 lb
(b) As found above, the compression force in each crutch is
F 1 (^) = F 2 (^) = F = 45 8. lb.
Chapter 5 117
P5.5 (a) The person pushes backward on the floor. The floor pushes forward on the person with a force of friction. This is the only horizontal force on the person. If the person’s shoe is on the point of slipping the static friction force has its maximum value. F ma f n ma F ma n mg ma mg a g x x v t a t t
t
x x s x y y x s x s f i xi x
2 2
m s m s m m s
s
2 2 2
e j e j
r r
r
FIG. P5.
(b) x (^) f = (^1) sgt 2
μ 2 , t x
g
f s
= = (^ )
m m s 2 s
..
c h
P5.6 If the load is on the point of sliding forward on the bed of the slowing truck, static friction acts backward on the load with its maximum value, to give it the same acceleration as the truck Σ Fx = max : − f = m (^) load a (^) x Σ Fy = may : n − m (^) load g = 0
− μ s mg = max a x = −μ sg
v (^) xf^2 = v (^) xi^2 + 2 a (^) x d x (^) f − xi i 0 = v (^) xi^2 + 2 b − μ s g gd x (^) f − (^0) i
r
r r
FIG. P5.
(a) x v f (^) g xi s
2 2 2
m s m s
b g 2 m a.^ fe.^ j
(b) From the expression x v f (^) g xi s
2
, neither mass affects the answer.
P5.7 − f + mg sin θ =0 and + n − mg cos θ =0 with f =μ n yield μ s = tan θ c = tan 36 0. ° = 0 727.
μ k = tan θ c = tan 30 0. ° = 0 577.
P5.8 m suitcase = 20 0. kg, F = 35 0. N
F ma F F ma n F F
x x y y g
:. cos : sin
N θ
(a) F^ cos^. cos.^.
.
N
N
35.0 N
r^ r
r
r
FIG. P5.
(b) n = Fg − F sin θ = 196 − 35 0 0 821. (. ) N
n = 167 N
Chapter 5 119 P5.11 (a)
(b)
See Figure to the right.
68 0 (^2 ) 1 1
T m g m a T m g m a
(Block #2) (Block #1)
Adding,
68 0 68 0 (^) 1 29
1 2 1 2
1 2 1 1
μ μ
μ
m m g m m a a m m
g
T m a m g
b g b g
b g
m s
N
2
T
m 1^ T^ m^2 F
m 1
n 1 T
m g 1 = 118 N
f (^) 1 = μ kn 1
m 2
n 2 F
m g 2 = 176 N
f = 2 μ kn 2
r r r
r
FIG. P5.
*P5.12 Let a represent the positive magnitude of the acceleration − a $ j of m 1 , of the acceleration − a $ i of m 2 , and of the acceleration + a $ j of m 3. Call T 12 the tension in the left rope and T 23 the tension in the cord on the right.
For m 1 , ∑ Fy = may + T 12 − m g 1 = − m a 1
For m 2 , ∑ Fx = max − T 12 + μ kn + T 23 = − m a 2
and ∑ Fy = may n − m g 2 = 0
for m 3 , ∑ Fy = may T 23 − m g 3 = + m a 3
we have three simultaneous equations
− + =
T a T T a T a
12 12 23 23
N kg N kg N kg
b g a f b g b g
n
T 12 T 23
m g 2
f = μ kn
m g 1
T 12
m g 3
T 23
FIG. P5.
(a) Add them up:
- 39 2. N − 3 43. N − 19 6. N =a7 00. kgf a
a = 2 31. m s 2 , down for m (^) 1 , left for m (^) 2 ,and up for m 3.
(b) Now − T 12 + 39 2. N = a4 00. kg (^) fc 2 31. m s (^2) h
T 12 = 30 0. N
and T 23 − 19 6. N = a2 00. kg fc 2 31. m s (^2) h
T 23 = 24 2. N.
120 More Applications of Newton’s Laws
P5.13 (Case 1, impending upward motion) Setting F P n f n f P P P
x s s s s
∑ =^ °−^ =
: cos. : cos.
...
, max μ ,max μ a f Setting F P P P
∑ y =^ °−^ −^ =
: sin.... max.
a (^) f N
(Case 2, impending downward motion) As in Case 1, f (^) s , max = 0 161. P
Setting F P P P
∑ y =^ °+^ −^ =
: sin.... min.
a f N
r
r r
r
r r
FIG. P5.
P5.14 We must consider separately the disk when it is in contact with the roof and when it has gone over the top into free fall. In the first case, we take x and y as parallel and perpendicular to the surface of the roof:
F ma n mg n mg
∑ y =^ y +^ −^ =
: cos cos
θ θ
then friction is f (^) k = μ (^) kn =μ kmg cos θ
r
r r
FIG. P5.
F ma f mg ma a g g
x x k x x k
∑ =^ −^ −^ =
: sin cos sin. cos sin..
θ μ θ θ (^) a 0 4 (^37 37) f 9 8 m s 2 9 03m s^2
The Frisbee goes ballistic with speed given by
v v a x x v
xf xi x f i xf
d i b^ m sg^ e m s^ ja^ m^ f m^ s m s
. 2.^2
For the free fall, we take x and y horizontal and vertical:
v v a y y y
y
yf yi y f i f
f
2 2 2 2
d i b g (^) e jd i b g
. sin. sin
.
m s m s m
m
m s m s
m
2
2
122 More Applications of Newton’s Laws
P5.19 (^) T cos 5 00. ° = mg = (^) b80 0. kg (^) ge9 80. m s (^2) j
(a) T = 787 N :
r T = a68 6. N f i $^ +a 784 Nf$ j
(b) T sin 5 00. ° = ma (^) c : a (^) c = 0 857. m s 2 toward the center of the circle. The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion.
r
r
FIG. P5.
*P5.20 Fg = mg = b 4 kg ge9 8. m s (^2) j=39 2. N
sin. .
m 2 m
r = a 2 mf cos 48 6. ° = 1 32. m
F ma mv r
T T
T T
x x
a b
a b
∑ =^ =
2
2 48 6 48 6
cos. cos. .
cos.
kg m s m N (^) N
b gb g
F ma T T T T
y y a b a b
sin. sin.. . sin.
N
N N
39.2 N
Ta
Tb forces
a (^) c (^) v
motion
FIG. P5.
(a) To solve simultaneously, we add the equations in Ta and Tb :
T (^) a + T (^) b + T (^) a − Tb = 165 N +52 3. N
Ta = 217 N= 108 2
N
(b) T (^) b = 165 N − Ta = 165 N − 108 N = 56 2. N
Chapter 5 123
Section 5.3 Nonuniform Circular Motion
P5.21 Let the tension at the lowest point be T.
F ma T mg ma mvr
T m g v r
T
∑ =^ −^ =^ c =
F HG^
I KJ
= +
L
N
M M
O
Q
P P
2
2
2 85 0 9 80
kg m s (^) 10 0 1 38 1 000
m s m kN^ N b g 2 b^ g
He doesn’t make it across the river because the vine breaks.
r
r
r r
FIG. P5.
P5.22 (a) ∑ Fy = ma y = mvR
2
mg n mv R
2 n mg mv R
2
(b) When n = 0 , mg mv R
2
Then, v = gR.
P5.23 F mv r
∑ y =^ =^ mg^ + n
2
But n = 0 at this minimum speed condition, so
mv r
mg v gr
2 = ⇒ = = (^) e9 80. m s (^2) ja1 00. m (^) f = 3 13. m s.
r
r
FIG. P5.
P5.24 (a) a (^) c = vr
2 r va c
2 13 0^2
m s m s 2 m
b g e j
(b) Let n be the force exerted by the rail. Newton’s law gives
Mg + n = Mvr
2
r r
FIG. P5.
n = M vr − g M g g Mg
F HG^
I KJ^
2 b 2 g , downward
continued on next page
Chapter 5 125
P5.27 (a) At terminal velocity, R = v bT = mg
∴ = =
×
×
− b (^) − mg v (^) T
3 2
kg m s m s
N s m
e je 2 j
(b) In the equation describing the time variation of the velocity, we have
v = v T e 1 − e − bt m j v = 0 632. v T when e −^ bt m = 0 368.
or at time t m b
= −FHG IKJ ln a0 368. f = 2 04. × 10 −^3 s
(c) At terminal velocity, R = v bT = mg = 2 94. × 10 −^2 N
P5.28 v mg b
bt m
= FHG IKJ L − FHG − IKJ
NM^
O
QP
1 exp where exp a f x = e x is the exponential function.
At t → ∞ , v v mg T b
At t = 5 54. s 0 500 1 5 54 9 00
. exp. .
v T = v T − F− b
HG^
I
KJ
L
N
M
M
O
Q
P
P
s kg
a f
exp. .
ln.. ;
.. .
F −
HG^
I
KJ^
b
b
b
s kg s kg kg s
kg s
b g
b g
b gb g
(a) v mg T (^) b = v (^) T = =
kg m s kg s
m s
b ge 2 j
(b) 0 750 1 1 13 9 00
. exp. .
v T = v T L − FHG − t IKJ
NM^
O
s QP
exp. .
FHG −^ 1 13IKJ =.
t 0 250 s
t = −
. ln. 11 1 .
a f s. s
(c) dxdt = FHG mgb IKJ L − FHG − btm IKJ
NM^
O
QP
1 exp ; dx mgb mbt dt x
x t
0
0
z =^ z
F
HG^
I
KJ^ −^
F−
HG^
I
KJ
L
NM^
O
QP
exp
x x mgt b
m g b
bt m
mgt b
m g b
bt m
t − = +
F
HG^
I
KJ^
F−
HG^
I
KJ^ =^ +^
F
HG^
I
KJ^
F−
HG^
I
KJ^ −
L
NM^
O
0 QP
2 2 0
2 exp 2 exp 1
At t = 5 54. s , x = +
F
H
G
G
I
K
J
J
2
. (^2)
..^.
kg 9.80 m s s exp. 1.13 kg s
kg m s kg s
2
2
e j
b g e j
b g
b g
x = 434 m + 626 ma − 0 500. f= 121 m
126 More Applications of Newton’s Laws
P5.29 (a) v t a f = v ei − ct^ v a 20 0. sf= 5 00. = v ei −20 0^. c , v i = 10 0. m s.
So 5 00. = 10 0. e − 20 0. c^ and − 20 0 = FHG^1 IKJ
. c ln c = − = × −^ −
ln .
(^12) 2 1 20 0
c h 3 47 10 s
(b) At t = 40 0. s v = b 10 0. m s g e −40 0. c = b10 0. m s ga0 250. f= 2 50. m s
(c) v = v ei −^ ct a dv dt
= = − cv e (^) i − ct = − cv
P5.30 ∑ F = ma
z z −
−
kmv m dvdt
kdt dv v k dt v dv
k t v v v
v v kt^
v kt v v v v kt
t
v
v
v
v
2
2
0
2
1 0
0
0 0 0 0
0
0
a f
Section 5.5 The Fundamental Forces of Nature
P5.31 F Gm m r
×
= ×
− 1 2 − 2
11 2
e ja fa f
a f
N
P5.32 For two 70–kg persons, modeled as spheres,
F Gm m g (^) r
1 2 ×^ − ⋅ −
2
11 2
N m kg kg kg m
N
e 2 2 jb^ gb^ g
a f
P5.33 a MG R (^) E
b g
. m s (^2). m s (^2) toward the earth
P5.34 F k q q e r
= 1 2 = × +^ −^ = − × = ×
12
2
9 2
b g
e j
a fa f
a f
.. N (attractive). N downward
128 More Applications of Newton’s Laws
Additional Problems
P5.37 Applying Newton’s second law to each object gives:
(1) T 1 (^) = f (^) 1 + 2 m g b sin θ+ a g
(2) T 2 (^) − T 1 (^) = f (^) 2 + m g b sin θ+ a g
(3) T 2 (^) = M g b − a g
(a), (b) Equilibrium a a = 0 f and frictionless incline (^) b f (^) 1 = f 2 = (^0) g
Under these conditions, the equations reduce to
(1’) T 1 = 2 mg sin θ
(2’) T 2 − T 1 = mg sin θ
(3’) T 2 (^) = Mg
r r r
r
r r
r r r
r
r
r
FIG. P5.
Substituting (1’) and (3’) into equation (2’) then gives M = 3 m sin θ
so equation (3’) becomes T 2 = 3 mg sin θ
(c), (d) M = 6 m sin θ (double the value found above), and f 1 = f 2 = 0. With these conditions present,
the equations become T (^) 1 = 2 m g b sin θ+ a g, T (^) 2 − T (^) 1 = m g b sin θ+ a g and T 2 (^) = 6 m sin θb g − a g. Solved simultaneously, these yield
a = 1 + g^ sin 2 sin^ θθ , T 1 = 4 mg 11 ++ 2
F HG^
I KJ
sin θ sinsinθθ and T 2 = 6 mg 11 ++ 2
F HG^
I KJ
sin θ sinsinθθ
(e) Equilibrium a a = 0 f and impending motion up the incline so M = M max while
f 1 = 2 μ smg cos θand f 2 =μ smg cos θ, both directed down the incline. Under these
conditions, the equations become T 1 (^) = 2 mg (^) b sin θ +μ (^) s cosθg, T 2 (^) − T 1 = mg (^) bsin θ +μ (^) s cosθg, and T (^) 2 = M (^) max g , which yield M (^) max = 3 m (^) bsin θ +μ (^) s cosθg
(f) Equilibrium a a = 0 f and impending motion down the incline so M = M min , while
f 1 = 2 μ smg cos θand f 2 =μ smg cos θ, both directed up the incline. Under these conditions,
the equations are T 1 (^) = 2 mg (^) bsin θ −μ (^) s cosθg, T (^) 2 − T 1 = mg (^) bsin θ −μ (^) s cosθg, and T (^) 2 = M (^) min g , which yield M (^) min = 3 m (^) bsin θ −μ (^) s cosθg. When this expression gives a negative value, it corresponds physically to a mass M hanging from a cord over a pulley at the bottom end of the incline.
(g) T 2 ,max − T 2 ,min = M max g − M min g = 6 μ smg cosθ
Chapter 5 129
P5.38 For the system to start to move when released, the force tending to move m 2 down the incline,
m g 2 sin θ , must exceed the maximum friction force
which can retard the motion:
f f f n n f m g m g
s s s s
max =^ max +^ max=^ + = +
1 2 1 1 2 2 1 1 2 2
, , , , max , , cos
From Table 5.1, μ s , 1 = 0 610. (aluminum on steel) and
μ s , 2 = 0 530. (copper on steel). With
m 1 = 2 00. kg, m 2 = 6 00. kg, θ= 30 0. °,
r
r
FIG. P5.
the maximum friction force is found to be f max = 38 9. N. This exceeds the force tending to cause the system to move, m g 2 sin θ = 6 00. kg (^) e9 80. m s (^2) j sin 30 ° = 29 4. N. Hence,
the system will not start to move when released.
The friction forces increase in magnitude until the total friction force retarding the motion, f = f (^) 1 + f 2 , equals the force tending to set the system in motion. That is, until
f = m g 2 sin θ =29 4. N.
P5.39 (a) The crate is in equilibrium, just before it starts to move. Let the normal force acting on it be n and the friction force, fs. Resolving vertically:
n = Fg + P sin θ. Horizontally: P cos θ = f s. But, f s ≤ μ sn i.e.,
P cos θ ≤ μ s c F (^) g + P sin θh or P acos θ − μ (^) s sinθ f ≤μ s Fg. Divide by cos θ : P a 1 − μ (^) s tan θ f ≤μ (^) s Fg secθ. Then
P
s Fg s
minimum =^ −
μ θ μ θ
sec 1 tan
r r
r
r
FIG. P5.
(b) P =
. sec . tan
N θ
θ (^) bdeg g 0.00 15.0 30.0 45.0 60. P N a f 40.0^ 46.4^ 60.1^ 94.3^260
If the angle were 68 2. ° or more, the expression for P would go to infinity and motion would become impossible.
Chapter 5 131
P5.42 (a) See Figure (a) to the right.
(b) See Figure (b) to the right.
(c) For the pin,
F ma C C
∑ y =^ y −^ =
: cos
cos
N
N
For the foot,
m g r = (^) b36 4. kg (^) ge9 8. m s (^2) j= 357 N
r
r (^) r r
r
FIG. P5.42(a) FIG. P5.42(b)
F ma n C n
y y B B
∑ =^ +^ −^ =
: cos .
357 N
(d) For the foot with motion impending,
F ma f C n C C n
x x s s s B s s s B
s s s
∑ =^ +^ −^ =
: sin sin sin cos^ sin^ tan.
N
N
b g
(e) The maximum coefficient is
μ s = tan θ s = tan 50 2. ° = 1 20..
132 More Applications of Newton’s Laws
P5.43 (a) First, draw a free-body diagram, (top figure) of the top block. Since a (^) y = 0 , n 1 = 19 6. N. And f (^) k = μ (^) kn 1 = 0 300 19 6. a. N f =5 88. N. Σ F (^) x = maT
10 0. N − 5 88. N = (^) b 2 00. kgg a (^) T
or a (^) T = 2 06. m s 2 (for top block). Now draw a free- body diagram (middle figure) of the bottom block and observe that Σ Fx = MaB gives f = 5 88. N =b8 00. kgg a (^) B or a (^) B = 0 735. m s 2 (for the bottom block). In time t , the distance each block moves (starting from rest) is d (^) T = 1 a tT 2
(^2) and d (^) B = 1 a tB 2
(^2). For the top block to reach
the right edge of the bottom block, (see bottom figure) it is necessary that d (^) T = d (^) B + L or
1 2 2 06^
2 0 735^ 3 00
e.^ m s^2 j t^2^^ =^ e.^ m s^2 j t^2 +. m
which gives: t = 2 13. s.
(b) From above, d (^) B = 1 = 2 e0 735.^ m s^2 ja2 13^.^ s^ f^2 1 67. m.
r r
r r
r
r r
FIG. P5.
P5.44 (a)
r r
r r
r r
r r
FIG. P5.
f (^) 1 and n 1 appear in both diagrams as action-reaction pairs
(b) 5.00 kg: Σ Fx = ma : n (^) 1 = m g 1 = 5 00 9 80. a. f =49 0. N f (^) 1 − T = 0 T = f (^) 1 = μ mg = 0 200 5 00. a. fa 9 80. f = 9 80. N
10.0 kg: Σ Fx = ma : 45 0. − f 1 (^) − f (^) 2 =10 0. a Σ Fy =0: n (^) 2 − n 1 − 98 0. = 0
continued on next page
