Laws of motion ( constraint motion problem )
There are three blocks A, B and C of masses M, m1 and m2 respectively kept on a smooth horizontal surface. Block B is placed to the left of block A and The position of block C is between the block A and B. There are two pulleys of which one is fitted with block A and other is fitted with block B on their vertical sides , horizontally. One inextensible string of negligible mass is tied up from the body of block B to the pulley of block A , then from the pulley of block A to the pulley of block B and then from pulley of block B to the body of block C. A force F is applied on the block A to its right side direction. What will be acceleration on block C and show the free body diagram .
Laws of motion ( constraint motion problem )
There are three blocks A, B and C of masses M, m1 and m2 respectively kept on a smooth horizontal surface. Block B is placed to the left of block A and The position of block C is between the block A and B. There are two pulleys of which one is fitted with block A and other is fitted with block B on their vertical sides , horizontally. One inextensible string of negligible mass is tied up from the body of block B to the pulley of block A , then from the pulley of block A to the pulley of block B and then from pulley of block B to the body of block C. A force F is applied on the block A to its right side direction. What will be acceleration on block C and show the free body diagram .
Asked by sudhanshubhushanroy | 12th Jul, 2017, 11:49: AM
The FBD of the three blocks is as shown above.
The following steps are to be followed in order to solve the query:
Step 1) Draw the FBD as shown above.
Step 2) Consider block A: Balance its forces. F = 2T. Get the tension from this equation. You will get T = MaA/2.
Step 3) Consider block C: As the block A is pushed towards the right, the string will tend to pull block B towards the right. Due to this, the string attached to block B via block A gets pushed to the right.
So, the string attached to C gets pulled towards the left. Hence, C moves towards the left. Balance the forces on C. m2aC = T. Substitute T. Find aC. You should get aC = MaA/2m2.

Answered by Romal Bhansali | 13th Jul, 2017, 05:51: PM
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