147

More Applications of Newton’s Laws

SOLUTIONS TO PROBLEMS Section 5.1 Forces of Friction P5.1 For equilibrium: f = F and n = Fg . Also, f = µn i.e.,

µ =fn=

FFg

µs =75.0 N

25.0 9.80( ) N= 0.306

and µk =

60.0 N25.0 9.80( ) N

= 0.245 .

FIG. P5.1

P5.2

Fy! = may : +n " mg = 0fs #µsn = µsmg

This maximum magnitude of static friction acts so long as the tires roll without skidding.

Fx! = max : " fs = ma The maximum acceleration is a = !µs g . The initial and final conditions are: xi = 0 ,

vi = 50.0 mi h = 22.4 m s , v f = 0 , v f2 = vi

2 + 2a x f ! xi( ) : !vi2 = !2µs gx f

(a) x f =

vi2

2µg

x f =

22.4 m s( )2

2 0.100( ) 9.80 m s2( )= 256 m

(b) x f =

vi2

2µg

x f =

22.4 m s( )2

2 0.600( ) 9.80 m s2( )= 42.7 m

148 More Applications of Newton’s Laws

P5.8 msuitcase = 20.0 kg , F = 35.0 N

Fx! = max : "20.0 N + F cos# = 0Fy! = may : +n + F sin# " Fg = 0

(a)

F cos! = 20.0 N

cos! =20.0 N35.0 N

= 0.571

! = 55.2°

FIG. P5.8

(b) n = Fg ! F sin" = 196 ! 35.0 0.821( )[ ] N

n = 167 N

P5.9 m = 3.00 kg , ! = 30.0° , x f = 2.00 m , t = 1.50 s , xi = 0 , vxi = 0

(a) x f =

12

at2 :

2.00 m =12

a 1.50 s( )2

a = 4.001.50( )2

= 1.78 m s2

FIG. P5.9

rF! =

rn +rf + mrg = mra :

Along x: 0 ! f + mg sin 30.0° = maf = m g sin 30.0° ! a( )

Along y: n + 0 ! mg cos 30.0° = 0n = mg cos 30.0°

(b) µk =

fn=

m g sin 30.0° ! a( )mg cos 30.0°

, µk = tan 30.0° !

ag cos 30.0°

= 0.368

(c) f = m g sin 30.0° ! a( ) , f = 3.00 9.80sin 30.0° ! 1.78( ) = 9.37 N (d) v f

2 = vi2 + 2a x f ! xi( ) where x f ! xi = 2.00 m

v f2 = 0 + 2 1.78( ) 2.00( ) = 7.11 m2 s2

v f = 7.11 m2 s2 = 2.67 m s

Chapter 5 149

P5.10 T ! fk = 5.00a (for 5.00 kg mass) 9.00g !T = 9.00a (for 9.00 kg mass) Adding these two equations gives:

9.00 9.80( ) ! 0.200 5.00( ) 9.80( ) = 14.0aa = 5.60 m s2

"T = 5.00 5.60( ) + 0.200 5.00( ) 9.80( )= 37.8 N

FIG. P5.10 P5.11 (a)

(b)

See Figure to the right.

68.0 !T ! µm2 g = m2a (Block #2)T ! µm1g = m1a (Block #1)

Adding,

68.0 ! µ m1 + m2( ) g = m1 + m2( ) a

a = 68.0m1 + m2( )

! µg = 1.29 m s2

T = m1a + µm1g = 27.2 N

FIG. P5.11

150 More Applications of Newton’s Laws

*P5.12 Let a represent the positive magnitude of the acceleration !aj of

m1 , of the acceleration !ai of m2 , and of the acceleration +aj of

m3 . Call T12 the tension in the left rope and T23 the tension in the cord on the right.

For m1 , Fy! = may +T12 ! m1g = !m1a For m2 , Fx! = max !T12 + µkn +T23 = !m2a and Fy! = may n ! m2 g = 0 for m3 , Fy! = may T23 ! m3 g = +m3a we have three simultaneous equations

!T12 + 39.2 N = 4.00 kg( ) a+T12 ! 0.350 9.80 N( ) !T23 = 1.00 kg( ) a

+T23 ! 19.6 N = 2.00 kg( ) a .

FIG. P5.12

(a) Add them up:

+39.2 N ! 3.43 N ! 19.6 N = 7.00 kg( ) a

a = 2.31 m s2 , down for m1 , left for m2 , and up for m3 .

(b) Now !T12 + 39.2 N = 4.00 kg( ) 2.31 m s2( )

T12 = 30.0 N and T23 ! 19.6 N = 2.00 kg( ) 2.31 m s2( )

T23 = 24.2 N .

Chapter 5 151 P5.13 (Case 1, impending upward motion) Setting

Fx! = 0 : P cos 50.0° " n = 0fs , max = µsn : fs , max = µsP cos 50.0°

= 0.250 0.643( )P = 0.161P

Setting

Fy! = 0 : P sin 50.0° " 0.161P " 3.00 9.80( ) = 0

Pmax = 48.6 N

(Case 2, impending downward motion) As in Case 1,

fs , max = 0.161P Setting

Fy! = 0 : P sin 50.0° + 0.161P " 3.00 9.80( ) = 0

Pmin = 31.7 N

FIG. P5.13

Section 5.2 Newton’s Second Law Applied to a Particle in Uniform Circular Motion

P5.16 (a) F =

m v2

r=

9.11 ! 10"31 kg( ) 2.20 ! 106 m s( )2

0.530 ! 10"10 m= 8.32 ! 10"8 N inward

(b) a = v2

r=

2.20 ! 106 m s( )2

0.530 ! 10"10 m= 9.13 ! 1022 m s2 inward

P5.19 T cos 5.00° = mg = 80.0 kg( ) 9.80 m s2( )

(a) T = 787 N : rT = 68.6 N( ) i + 784 N( ) j

(b) T sin 5.00° = mac : ac = 0.857 m s2 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.

FIG. P5.19

152 More Applications of Newton’s Laws

Section 5.3 Nonuniform Circular Motion P5.21 Let the tension at the lowest point be T.

F! = ma : T " mg = mac =mv2

r

T = m g + v2

r#

$%&

'(

T = 85.0 kg( ) 9.80 m s2 +8.00 m s( )2

10.0 m)

*++

,

-..= 1.38 kN > 1 000 N

He doesn t make it across the river because the vine breaks.

FIG. P5.21

P5.23

Fy! =mv2

r= mg + n

But n = 0 at this minimum speed condition, so

mv2

r= mg ! v = gr = 9.80 m s2( ) 1.00 m( ) = 3.13 m s .

FIG. P5.23

Section 5.5 The Fundamental Forces of Nature

P5.31 F =

Gm1m2r2 =

6.672 ! 10"11( ) 2( ) 2( )

0.30( )2= 2.97 ! 10"9 N

P5.32 For two 70–kg persons, modeled as spheres,

Fg =

Gm1m2r2 =

6.67 ! 10"11 N #m2 kg2( ) 70 kg( ) 70 kg( )

2 m( )2~ 10"7 N

Chapter 5 153

Additional Problems P5.44 (a)

FIG. P5.44 f1 and n1 appear in both diagrams as action-reaction pairs

(b) 5.00 kg: !Fx = ma : n1 = m1g = 5.00 9.80( ) = 49.0 N f1 !T = 0

T = f1 = µmg = 0.200 5.00( ) 9.80( ) = 9.80 N 10.0 kg: !Fx = ma : 45.0 ! f1 ! f2 = 10.0a

!Fy = 0 : n2 ! n1 ! 98.0 = 0

f2 = µn2 = µ n1 + 98.0( ) = 0.20 49.0 + 98.0( ) = 29.4 N

45.0 ! 9.80 ! 29.4 = 10.0a a = 0.580 m s2

154 More Applications of Newton’s Laws

*P5.45 (a) If the car is about to slip down the incline, f is directed up

the incline.

Fy! = n cos" + f sin" # mg = 0 where f = µsn gives

n =

mgcos! 1 + µs tan!( )

and f = µsmg

cos! 1 + µs tan!( ).

Then,

Fx! = n sin" # f cos" = mvmin

2

R yields

vmin =Rg tan! " µs( )

1 + µs tan!.

When the car is about to slip up the incline, f is directed

down the incline. Then, Fy! = n cos" # f sin" # mg = 0

with f = µsn yields

n =

mgcos! 1 " µs tan!( )

and f = µsmg

cos! 1 " µs tan!( ).

In this case,

Fx! = n sin" + f cos" = mvmax

2

R, which gives

vmax =Rg tan! + µs( )

1 " µs tan!.

(b) If vmin =

Rg tan! " µs( )1 + µs tan!

= 0 , then µs = tan! .

(c)

vmin =

100 m( ) 9.80 m s2( ) tan10.0° ! 0.100( )1 + 0.100( ) tan10.0°

= 8.57 m s

vmax =

100 m( ) 9.80 m s2( ) tan10.0° + 0.100( )1 ! 0.100( ) tan10.0°

= 16.6 m s

FIG. 5.45