The two blocks shown are originally at rest. Neglecting the masses of the pulley

Question

The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the incline, determine (a) the acceleration of each block, (b) the tension in the cable. Part 1 out of 2 aA = ft/s2 30 degree aB = ft/s2

Explanation / Answer

Probably easiest to find tension first: Pulleys cannot supply moment forces (neglecting moment of inertia), thus, tensions on either side must be the same. Force balance on A: T - mgsin(30) --> mg = 180 (lbs are force, not mass) --> Sum F = T - 90 a(A) = Sum F / m = (T - 90) / (180 / g) Force balance on B: Sum F = 360 - 2*T (moving down) a(B) = Sum F / m = (360 - 2*T) / (360 / g) Note that if the wire does not stretch, A should have twice the acceleration of B (for every inch A moves, B moves half an inch) Thus, (T - 90) / (180 / g) = 2*((360 - 2*T) / (360 / g)) (T - 90) / 180 = (720 - 4*T) / 360 T - 90 = (720 - 4*T) / 2 T - 90 = 360 - 2*T 3*T = 450 T = 150 lbs From this, we can find the acceleration of each part: a(A) = (T - 90) / (180 / g) = 60 / (180 / g) = g / 3 a(A) = 10.715 ft/(s^2) at 30 degrees incline (moving up) a(B) = (360 - 2*T) / (360 / g) = 60 / (360 / g) = g / 6 a(B) = 5.358 ft/(s^2) moving down Note that this satisfies that a(B) is half of a(A)

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