The truck at B is used to hoist the cabinet up to the fourth floor of a building
ID: 1822900 • Letter: T
Question
The truck at B is used to hoist the cabinet up to the fourth floor of a building using the rope and pulley arrangement. The truck has a speed of vB = 2 ft/s, and is accelerating at aB = 0.6 ft/s2. Determine the acceleration of the cabinet when sA = 40 ft. Note that when sB = 0, sA = 0 (points A and B are coincident).
The truck at B is used to hoist the cabinet up to the fourth floor of a building using the rope and pulley arrangement. The truck has a speed of vB = 2 ft/s, and is accelerating at aB = 0.6 ft/s2. Determine the acceleration of the cabinet when sA = 40 ft. Note that when sB = 0, sA = 0 (points A and B are coincident).Explanation / Answer
when the the truck is at a distance x, let the cabinet is moved a distance y upwards.
Let the point at pulley = P
PA + PB is always constant
PA + PB = 100
50-y + (x^2 + 2500) = 100
(y + 50)^2 = x^2 + 2500
y^2 + 100y + 2500 = x^2 + 2500
y^2 + 100y = x^2, when y = 40, x = 74.833
2ydy/dt + 100dy/dt = 2xdx/dt
2*40*dy/dt + 100dy/dt = 2*74.83*2
dy/dt = 1.66
2yd^y/dt^2 + 2(dy/dt)^2 + 100d^2y/dt^2 = 2xd^2x/dt^2 + 2(dx/dt)^2
2*40*a + 2*1.66^2 + 100a = 2*78.33*0.6 + 2*2^2
a = 0.536 ft/s^2
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