Answers for Problem Set 6

Chapter 6

27.

(a) F = mu_k m g / [ sin(theta) - mu_k cos(theta) ]

(b) theta = arc tan (mu_s )

36.

Agrees with a past edition of the text:

(a) T = m₁ m₂ g (mu₁ - mu₂) cos (theta) / (m₁ + m₂)
        = 1.05 N tension;

(b) a = [ g / (m₁ + m₂ ) ] [ (m₁ + m₂ ) sin (theta) - (mu₁ m₁ + mu₂ m₂)cos(theta) ]
        = 3.62 m/sec² down the plane;

(c) If m₁ and m₂ (and their coefficients of friction) are interchanged, the acceleration is of course unchanged but the rod shifts from being under tension to being under compression.

43.

- k v = m dv/dt

v = v₀ exp [ - k t / m ]

t = - (m/k) ln [ v_f / v₀ ]

70.

(b) T_l = [ T_u sin (theta) - m g ] / [ sin (theta) ] = 9 N

(c) F_net = [ T_u + T_l ] cos ( theta ) = 38 N

(d) v = sqrt [ d ( 2 T_u sin ( theta ) - m g ] cos ( theta ) / sqrt [ m sin (theta ) ] = 6.5 m/s

IX

F_friction = M g - 5 M a

X.

The acceleration of the skier is zero. Make sure you understand why.

XI.

tan (theta) = 4 pi² x / ( g T² )

y = 2 pi² x² / ( g T² ) - C