Find interaction force between 3Kg and 4Kg block.

In the arrangement shown in figure pulley P can move whereas other two pulleys are fixed. All of them are light. String is light and inextensible. The coefficient of friction between 2 kg and 3 kg block is mu = 0.75 and that between 3 kg block and the table is mu = 0.5 . The system is released from rest (i) Find maximum value of mass M, so that the system does not move. Find friction force between 2 kg and 3 kg blocks in this case. (ii) If M = 4 kg, find the tension in the string attached to 2 kg block. (iii) If M = 4 kg and mu_(1) = 0.9 , find friction force between the two blocks, and acceleration of M. (iv) Find acceleration of M if m_(1) = 0.75, m_(2) = -0.9 and M = 4 kg.

Find the nomral force between 2 kg and 3 kg blocks.

Find the contact force between the 3kg and 3kg block as shown in figure.

Three blocks of masses 2kg,3kg and 5kg are placed in contact as shown in the diagram. A horizontal force of 30N is applied on 2kg block. Find the contact force between (a) 2kg and 3kg block and (b) 3kg and 5kg block.

Find out the contact force between 2kg & 3kg block placed on the incline plane as shown in figure.

When F = 2n, the frictional force between 10 kg block 5 kg block is

When F = 2n, the frictional force between 10 kg block 5 kg block is

Block B is placed on a smooth surface. Block A is placed on rough surface of bolck B with coefficient of friction 0.60 . The mass of A and B are 2 kg and 4 kg respectively. Find the frictional force between A and B ( in N )

A lift is moving upwards with an acceleration of 2m//sec^(2) . Inside the lift a 4kg block is kept on the floor. On the top of it , 3kg block is placed and again a 2kg block is kept on the 3kg . Calculator : (i) contact force between 2kg block and the 3kg block. (ii) contact force between 4kg block and floor of the lift. Draw the free body diagrams properly & take g=10m//sec^(2)

A lift is moving upwards with an acceleration of 2m//sec^(2) . Inside the lift a 4kg block is kept on the floor. On the top of it , 3kg block is placed and again a 2kg block is kept on the 3kg . Calculator : (i) contact force between 2kg block and the 3kg block. (ii) contact force between 4kg block and floor of the lift. Draw the free body diagrams properly & take g=10m//sec^(2)