107

The Laws of Motion

SOLUTIONS TO PROBLEMS Section 4.3 Mass P4.2 Since the car is moving with constant speed and in a straight line, the resultant force on it must be

zero regardless of whether it is moving

(a) toward the right or (b) the left.

Section 4.4 Newton’s Second Law⎯The Particle Under a Net Force P4.3

m = 3.00 kgra = 2.00i + 5.00 j( ) m s2

rF = mra = 6.00i + 15.0 j( ) N!rF! = 6.00( )2 + 15.0( )2 N = 16.2 N

108 The Laws of Motion

P4.4 (a) !

rF =

rF1 +

rF2 = "9.00i + 3.00 j( ) N

Acceleration

ra = ax i + ay j = !rF

m=

"9.00i + 3.00 j( ) N2.00 kg

= "4.50i + 1.50 j( ) m s2

Velocity

rv f = vx i + vy j = rv i +

rat =rat

rv f = !4.50i + 1.50 j( )m s2( ) 10 s( ) = !45.0i + 15.0 j( ) m s

(b) The direction of motion makes angle θ with the x-direction.

! = tan"1 vy

vx

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'(= tan"1 "

15.0 m s45.0 m s

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! = "18.4° + 180° = 162° from + x-axis

(c) Displacement: x-displacement = x f ! xi = vxit +

12

axt2 =12!4.50 m s2( ) 10.0 s( )2 = !225 m

y-displacement = y f ! yi = vyit +

12

ayt2 =12+1.50 m s2( ) 10.0 s( )2 = +75.0 m

!

rr = "225i + 75.0 j( ) m

(d) Position:

rrf =

rri + !rr

rrf = !2.00i + 4.00 j( ) + !225i + 75.0 j( ) = !227i + 79.0 j( ) m

P4.7 (a)

rF! =

rF1 +

rF2 = 20.0i + 15.0 j( ) N

rF! = mra : 20.0i + 15.0 j = 5.00ra

ra = 4.00i + 3.00 j( ) m s2

or

a = 5.00 m s2 at ! = 36.9° (b)

F2x = 15.0cos 60.0° = 7.50 NF2y = 15.0sin 60.0° = 13.0 NrF2 = 7.50i + 13.0 j( ) NrF! =

rF1 +

rF2 = 27.5i + 13.0 j( ) N = m

ra = 5.00ra

ra = 5.50i + 2.60 j( ) m s2 = 6.08 m s2 at 25.3°

FIG. P4.7

Chapter 4 109

P4.8

rF! = mra reads

!2.00i + 2.00 j + 5.00i ! 3.00 j ! 45.0i( ) N = m 3.75 m s2( ) a

where a represents the direction of

ra

!42.0i ! 1.00 j( ) N = m 3.75 m s2( ) a

rF! = 42.0( )2 + 1.00( )2 N at

tan!1 1.00

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below the –x-axis

rF! = 42.0 N at 181° = m 3.75 m s2( ) a .

For the vectors to be equal, their magnitudes and their directions must be equal.

(a) ! a is at 181° counterclockwise from the x-axis

(b) m =

42.0 N3.75 m s2 = 11.2 kg

(d)

rv f =

rv i +rat = 0 + 3.75 m s2 at 181°( )10.0 s so

rv f = 37.5 m s at 181°

rv f = 37.5 m s cos181°i + 37.5 m s sin181° j so

rv f = !37.5i ! 0.893 j( ) m s

(c) rv f = 37.52 + 0.8932 m s = 37.5 m s

Section 4.5 The Gravitational Force and Weight P4.9 (a) Fg = mg = 120 lb = 4.448 N lb( ) 120 lb( ) = 534 N

(b) m =

Fg

g=

534 N9.80 m s2 = 54.5 kg

P4.11 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just

canceled out by a glass of tomato juice. By subtraction,

Fg( )p = mgp and

Fg( )C = mgC give

!Fg = m gp " gC( ) .

For a person whose mass is 88.7 kg, the change in weight is

!Fg = 88.7 kg 9.809 5 " 9.780 8( ) = 2.55 N . A precise balance scale, as in a doctor’s office, reads the same in different locations because it

compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference.

110 The Laws of Motion

P4.13 (a) F! = ma and v f2 = vi

2 + 2ax f or a =

v f2 ! vi

2

2x f.

Therefore,

F! = mv f

2 " vi2( )

2x f

F! = 9.11 # 10"31 kg7.00 # 105 m s2( )2 " 3.00 # 105 m s2( )2$

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2 0.050 0 m( )= 3.64 # 10"18 N .

(b) The weight of the electron is

Fg = mg = 9.11 ! 10"31 kg( ) 9.80 m s2( ) = 8.93 ! 10"30 N

The accelerating force is

4.08 ! 1011 times the weight of the electron. P4.14 We find acceleration:

rrf !rri =

rv it +12

rat2

4.20 mi ! 3.30 mj=0+ 12

ra 1.20 s( )2 = 0.720 s2rara = 5.83i ! 4.58 j( ) m s2 .

Now

rF! = mra becomes

rFg +

rF2 = mrarF2 = 2.80 kg 5.83i ! 4.58j( ) m s2 + 2.80 kg( ) 9.80 m s2( ) jrF2 = 16.3i + 14.6 j( ) N .

Section 4.6 Newton’s Third Law P4.17 (a) 15.0 lb up

(b) 5.00 lb up

(c) 0

Chapter 4 111

Section 4.7 Applications of Newton’s Laws P4.22 (a) Isolate either mass

T ! mg = ma = 0T = mg .

The scale reads the tension T, so

T = mg = 5.00 kg 9.80 m s2( ) = 49.0 N . (b) Isolate the pulley

rT2 + 2

rT1 = 0T2 = 2 T1 = 2mg = 98.0 N .

(c)

rF! =

rn +rT + mrg = 0

Take the component along the incline

rnx +

rTx + m

rgx = 0 or

0 +T ! mg sin 30.0° = 0

T = mg sin 30.0° = mg2

=5.00 9.80( )

2= 24.5 N .

FIG. P4.22(a)

FIG. P4.22(b)

FIG. P4.22(c)

112 The Laws of Motion

P4.24 (a) First construct a free body diagram for the

5 kg mass as shown in the Figure 4.24a. Since the mass is in equilibrium, we can require T3 ! 49 N = 0 or T3 = 49 N . Next, construct a free body diagram for the knot as shown in Figure 4.24a. Again, since the system is moving at constant velocity, a = 0 and applying Newton’s second law in component form gives

Fx! = T2 cos 50° "T1 cos 40° = 0

Fy! = T2 sin 50° +T1 sin 40° " 49 N = 0

Solving the above equations

simultaneously for T1 and T2 gives

T1 = 31.5 N and T2 = 37.5 N and above

we found T3 = 49.0 N . (b) Proceed as in part (a) and construct a free

body diagram for the mass and for the knot as shown in Figure 4.24b. Applying Newton’s second law in each case (for a constant-velocity system) we find:

T3 ! 98 N = 0T2 !T1 cos 60° = 0T1 sin 60° !T3 = 0

Solving this set of equations we find:

T1 = 113 N T2 = 56.6 N and T3 = 98.0 N

FIG. P4.24(a)

FIG. P4.24(b)

Chapter 4 113 P4.26 The two forces acting on the block are the normal force, n, and the

weight, mg. If the block is considered to be a point mass and the x-axis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle θ is the angle of inclination of the plane. Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have

Fy! = n " mg cos# = 0 : n = mg cos!

Fx! = "mg sin# = ma : a = !g sin"

FIG. P4.26 (a) When ! = 15.0°

a = !2.54 m s2 (b) Starting from rest

v f2 = vi

2 + 2a x f ! xi( ) = 2ax f

v f = 2ax f = 2 !2.54 m s2( ) !2.00 m( ) = 3.18 m s

P4.27 Choose a coordinate system with i East and j North.

rF! = mra = 1.00 kg 10.0 m s2( ) at 30.0°

5.00 N( ) j +rF1 = 10.0 N( )!30.0° = 5.00 N( ) j + 8.66 N( ) i

!F1 = 8.66 N East( )

FIG. P4.27 P4.28 First, consider the block moving along the horizontal. The only

force in the direction of movement is T. Thus, Fx! = ma

T = 5 kg( ) a (1) Next consider the block that moves vertically. The forces on it

are the tension T and its weight, 88.2 N. We have Fy! = ma

88.2 N !T = 9 kg( ) a (2)

FIG. P4.28

Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be

added to give 88.2 N = 14 kg( ) a . Then

a = 6.30 m s2 and T = 31.5 N .

114 The Laws of Motion

P4.30 m1 = 2.00 kg , m2 = 6.00 kg , ! = 55.0° (a) Fx! = m2 g sin" #T = m2a and

T ! m1g = m1a

a = m2 g sin" ! m1gm1 + m2

= 3.57 m s2

(b) T = m1 a + g( ) = 26.7 N

FIG. P4.30

(c) Since vi = 0 , v f = at = 3.57 m s2( ) 2.00 s( ) = 7.14 m s .

Additional Problems P4.44 (a) Following the in-chapter Example about a block on a frictionless incline, we have

a = g sin! = 9.80 m s2( )sin 30.0°

a = 4.90 m s2

(b) The block slides distance x on the incline, with sin 30.0° = 0.500 m

x

x = 1.00 m : v f2 = vi

2 + 2a x f ! xi( ) = 0 + 2 4.90 m s2( ) 1.00 m( )

v f = 3.13 m s after time ts =

2x f

v f=

2 1.00 m( )3.13 m s

= 0.639 s .

(c) Now in free fall y f ! yi = vyit +

12

ayt2 :

!2.00 = !3.13 m s( )sin 30.0°t ! 12

9.80 m s2( )t2

4.90 m s2( )t2 + 1.56 m s( )t ! 2.00 m = 0

t =!1.56 m s ± 1.56 m s( )2 ! 4 4.90 m s2( ) !2.00 m( )

9.80 m s2

Only one root is physical

t = 0.499 s

x f = vxt = 3.13 m s( )cos 30.0°[ ] 0.499 s( ) = 1.35 m

(d) total time = ts + t = 0.639 s + 0.499 s = 1.14 s (e) The mass of the block makes no difference.

Chapter 4 115 P4.47 F! = ma For m1 : T = m1a For m2 : T ! m2 g = 0 Eliminating T,

a = m2 g

m1

For all 3 blocks:

FIG. P4.47

F = M + m1 + m2( ) a = M + m1 + m2( ) m2 g

m1

!

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