Answers for Problem Set 13

Chapter 13

#15. [My answers agree with the text.]

The diagram shows the forces on the board.

(a) T₂ = M g ( L - d ) / d = 1160 newtons, down

(b) T₁ = M g L / d = 1740 newtons, up

(c) Left is stretched and right is compressed [remember Newton's Third Law]

#20.

cos (45^o ) = sin (45^o) = 1/sqrt(2)

Normal force by bottom of container on sphere, N₃ = 2 W
Normal forces on spheres by each other, N₁ = sqrt(2) W
Normal force by surface of container on either sphere, N₂ = W

#24. Proof.

#26. [My answers agree with an earlier edition of the text.]

(a) T = M g (2 W - h) sqrt( L² + W²) / ( 2 L W ) = 409 newtons

(b) F_{H x} = M g (2 W - h) / ( 2 L ) = 245 newtons right

(c) F_{H y} = M g h / ( 2 W ) = 163 newtons up

F = W [ sqrt( 2 r h - h² ) ] / [ r - h ]

#41. [My answers agree with the text]

(a) F₁ = 2 N_E y / sqrt ( L² - y² ) = 47 lbs

(b) N_A = W (L - x / 2 ) / L = 120 lbs

(c) N_E = W x / (2 L ) = 72 lbs

(a) a = - mu_k g = - 3.9 m /sec²

(b)

N_R/2 = (M g / 2 ) ( x - mu_k h ) / w = 2000 newtons on each back wheel;
N_F/2 = ( M g / 2 ) (w - x - mu_k h) / w = 3500 newtons on each front wheel

(c)

F_R/2 = 790 newtons on each back wheel;
F_F/2 = 1410 newtons on each front wheel