Question

Body A in the figure weighs 97 N, and body B weighs 80 N. The coefficients of friction between A and the incline are μs = 0.57 and μk = 0.23. Angle θ is 28°. Let the positive direction of an x axis be up the incline. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

Answer 1

Let's start with some basics:
A's normal force: Fn = 97.0N * cos28º = 85.65N
→ so max static friction Fs = 0.57 * 85.65 N = 48.82 N
→ and max kinetic friction Fk = 0.23 * 85.65N = 19.7 N
A's weight component downslope Fg = 97N * sin28º = 45.54 N

a) if A is initially at rest, observe that there is a 80N force pulling A upslope and a gravitational force of 45.54 N pulling A downslope. The difference is about 34N, and this is not enough to overcome the static friction (= 49N). Therefore a = 0.

b) If A is moving up the incline, the friction force points downslope. For A,
Fnet = ma = 97N/9.8m/s² * a = 9.9a = T - Fk - Fg = T - 19.7N - 45.54N = T - 65.2N
→ so T = 9.9a + 65.2
For B, Fnet = ma = 80N/9.8m/s² * a = 8.2kg * a = mg - T = 80 N - T
→ so T = 80 - 8.2a
Since T = T,
9.9a + 65.2= 80 - 8.2a
18.1a = 14.8

a = 0.82 m/s²
that is, downslope

c) If A is moving down the incline, then the friction force points upslope. That means
Fnet = ma = 97N/9.8m/s² * a = 9.9a = T + Fk - Fg = T + 19.7N - 45.54N = T - 25.8N
→ so T = 9.9a + 25.8
For B, still true that
→ T = 80 - 8.2a
Since T = T,
9.9a + 25.8 = 80 - 8.2a
18.1a = 54.16
a = 3m/s²,

Hope this helps you.

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