`m=20 kg, mu_(s)=0.5`, find friction on 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.

A block of mass m is kept on the floor of a freely falling lift .During the free fall of the lift, the block is pulled horizontally with a force of F = 5 N, mu_(s) = 0.1 The friction force on the block will be

A block of mass m is kept on the floor of a freely falling lift .During the free fall of the lift, the block is pulled horizontally with a force of F = 5 N, mu_(s) = 0.1 The friction force on the block will be

On horizontal surface of a truck a block of mass 1 kg is placed (mu = 0.5) and truck is moving with acceleration 5 ms^-2 , then frictional force on block will be

A force of 200N is applied as shown in the figure on block of 3 kg. The coefficient of friction between the block and the wall is 0.3. Find the friction acting on the block. (in N) {Take g = 10 m//s^(2)

In the figure shown, if friction coefficient of block 1 kg and 2kg with inclined plane is mu_(1) = 0.5 and mu_(2) = 0.4 respectively, then

In the figure shown, if friction coefficient of block 1 kg and 2kg with inclined plane is mu_(1) = 0.5 and mu_(2) = 0.4 respectively, then

A block of mass M is sliding down the plane. Coefficient of static friction is mu_(s) and kinetic friction is mu_(k) . Then friction force acting on the block is :