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Evaluate the motion of system:
Q : Block A and B are connected by a rigid horizontally bar planed at each end are placed on the inclined planes as shown in the figure given below. The weight of block B is 300N. Find the limiting values of weight of the block A to just start motion of system.
Sol.: Let Wa be weight of block A. Consider free body diagram of B. As shown in the figure. And Assume that AB be the Axis of reference.
∑V = 0;
Rsin45° - µBRcos45° - 300 = 0
On solving, R = 606.09N ...(i)
∑H = 0;
C - RCos45° - µBRsin45° = 0 ...(ii)
Putting value of R, we get
C = 557.14N ...(iii)
Where C is reaction imparted by rod.
Consider free body diagram of block A as shown in the figure
Figure
C + µARcos60° - Rcos30° = 0 ...(iv)
By putting all the values we get
R = 751.85N ...(v)
µ ARsin60° + Rsin60° - W = 0
By solving,
W = 538.7N
...(vi)
Thus weight of block
A = 538.7N
.......ANS
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