Request a call back

Solve:
Asked by Anish | 22 Oct, 2018, 11:56: AM
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   is acting on block-B towards left which is due to static friction force of block-A.
This reaction force is equal to 60N.
Since this reaction force is less than limiting friction, static friction fBC acting between block-B and block-C is 60 N

Reaction force   is acting on block-C towards left which is due to static friction force of block-B.
This reaction force is equal to 60N.
Since this reaction force 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
CBSE 11-science - Physics
Asked by sy123946 | 07 Apr, 2024, 04:23: PM
CBSE 11-science - Physics
Asked by derhebha955 | 03 Apr, 2024, 09:03: AM
CBSE 11-science - Physics
Asked by sumedhasingh238 | 29 Mar, 2024, 05:15: PM
CBSE 11-science - Physics
Asked by sumedhasingh238 | 28 Mar, 2024, 11:10: PM
CBSE 11-science - Physics
Asked by roshnibudhrani88 | 23 Mar, 2024, 05:52: PM
CBSE 11-science - Physics