In the figure shown, there is no relative motion between the two blokcs. Force of friction acting on 1 kg block is

The frictional force acting on 1 kg block is

The frictional force acting on 1 kg block is

In the arrangement shown in figure, coefficient of friction between the two blocks is mu=1//2 . The force of friction acting between

Wedge is fixed on horizontal surface. Block A is pulled upward by applying a force F as shown in figure and there is no friction between the wedge and the block A while coefficient of friction between A and B is mu . If there is no relative motion between the block A and B then frictional force developed between A and B is

A force barF=hati+4hatj acts on block shown. The force of friction acting on the block is :

A force barF=hati+4hatj acts on block shown. The force of friction acting on the block is :

A force barF=hati+4hatj acts on block shown. The force of friction acting on the block is :

Consider the shown arrangement. The coefficient of friction between the two blocks is 0.5 . There is no friction between the 4 kg block and of 12 N is applied on the 2 kg block as shown in the figure, the acceleration of the 4 kg block would be

Block 'A' is placed ober the block 'B' as shown in the figure. Wedge is fixed and there is no friction. Between block 'B' and fixed inclined. Force of friction on the block 'A' is