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CBSE Class 11-science Answered

Solve:

Asked by Anish | 22 Oct, 2018, 11:56: AM
Expert Answer
Free body diagram of all the blocks are given in figure. For block-A the limiting friction is μN = 0.3×300 = 90N
For block-B the limiting friction is μN = 0.2×400 = 80N,  For block-C the limiting friction is μN = 0.1×600 = 60N
 
Let the applied force on block-A acting on left side is 60N.
Since this is less than limiting friction,  static friction fAB between block-A and block-B is 60 N.
 
Reaction force begin mathsize 12px style f space apostrophe subscript A B end subscript end style  is acting on block-B towards left which is due to static friction force of block-A.
This reaction force begin mathsize 12px style f space apostrophe subscript A B end subscript end style is equal to 60N. 
Since this reaction force begin mathsize 12px style f space apostrophe subscript A B end subscript end style is less than limiting friction, static friction fBC acting between block-B and block-C is 60 N
 
Reaction force begin mathsize 12px style f space apostrophe subscript B C end subscript end style  is acting on block-C towards left which is due to static friction force of block-B.
This reaction force begin mathsize 12px style f space apostrophe subscript B C end subscript end style is equal to 60N. 
Since this reaction force begin mathsize 12px style f space apostrophe subscript B C end subscript end style is now equal to limiting friction fCG , block-C will start move
Block-A and block-B will not move for the applied force 60N on block-A,
because for both blocks, the force acting towards left is less than limiting friction.
 
Conclusion :- if 60N force is applied on block-A towards left, only block-C start moving,
Answered by Thiyagarajan K | 23 Oct, 2018, 09:14: AM
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