Evaluate the motion of system, Mechanical Engineering

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.

 

887_Evaluate the motion of system.png

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

1084_Evaluate the motion of system1.png

Figure

C + µARcos60° - Rcos30° = 0                                                                      ...(iv)

By putting all the values we get

R = 751.85N                                                                                                  ...(v)

∑V = 0;

µ ARsin60° + Rsin60° - W = 0

By solving,

W = 538.7N

 

...(vi)

Thus weight of block

A = 538.7N

.......ANS

 

 

Posted Date: 10/18/2012 2:20:35 AM | Location : United States







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