What is the minimum acceleration with which bar A (figure) should be shifted horizontally to keep bodies 1 and 2 stationary relative to the bar? The masses of the bodies are equal and the coefficient of friction between the bar and the bodies is equal to k. The masses of the pulley and the threads are negligible, the friction in the pulley is absent.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
w min = g (1 – k) / (1 + k)
Sol. If acceleration in bar is zero, then the body (1) will slip on bar rightward and the body (2) moves
downward. To prevent slipping, net force on each body should be zero in the frame of bar (non-inertial
reference frame)
In fig. A, T = mw + kmg ..........(i)
In fig. B, T + kN = mg..........(ii)
N = mw .........(iii)
From (ii) and (iii), we get
T + kmw = mg
T = mg – kmw ............(iv)
From (i) and (iv), we get
w = 
............(v)
Since, relative acceleration of body with respect to bar (w rel ) is zero: So, the value of w in eqn.(v) is
minimum value of w.
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