Question

Question Part Submissions Used Body A in the figure weighs 108 N, and body B weighs 40 N. The coefficients of friction between A and the incline are μs = 0.56 and μk = 0.25. Angle θ is 40°. Let the positive direction of an x axis be up the incline. Answer the following questions in unit-vector notation. a.) Find the acceleration of A if A is initially at rest. b.) Find the acceleration of A if A is moving up the incline. c.) Find the acceleration of A if A is moving down the incline.

Answer 1

given

weight of body A, Wa = 108 N

weoight of body 2, Wb = 40 N

coefficient of friction between A and incline, ks = 0.56, k = 0.25

theta = 40 deg

a. if A is initially at rest

then let A have downward acceleration

then tension in the rope is T

Wa*sin(theta) - T - k*N = Wa*a/g

T - Wb = Wb*a/g

hence

Wa*sin(theta) - kN - Wb = (Wa _ Wb)a/g

now, N = Wa*cos(theta)

hence

Wasin(theta) - kWa*cos(theta) - Wb = (Wa + Wb)a/g

now, if we substitude k with ks we get -ve acceleration

hence this means if the object is initially ate rest the static friction is enough to balance and hence the object has 0 aceleration

b. if A is moving up the incline

friciton is downwards

Wa*sin(theta) + kWa*cos(theta) - Wb = (Wa + Wb)a/g

hence

a = 3.3211 m/s/s ( downwards, along the incline)

c. if A is sliding downwards

friciton acts upwards

Wasin(theta) - kWacos(theta) - wB = (Wa + Wb)a/g

hence

a = 0.57917854 m/s/s ( downwards)