A block of mass `4kg` pressed againest a rought wall by two perpendicular horizontal forces `F_(1) and F_(2)`as shown in figure Coefficient of static friction between the block and wall is `0.6` and that of kinetic friction is `0.5`

For `F_(1) = 300 N and F_(2) = 100 N` find the direction and magnitude of friction force acting on the block

Options

A. `180 N` vertically upward
B. `40 N` vertically upward
C. `107.7 N` making an angle of `tan^(-1)((2)/(5))` will the horizontal the upward direction
D. `91.6 N` making an angle of `tan^(-1)((2)/(5))` will the horizontal the upward direction

Answer

c

Explanation

The force acting on the block are `F_(1) ,F_(2) mg` normal contact force and friction as the component of resultant force along the surface acting no body is not along vertyical direction and direction of the friction force is ether force is either opposite to the motion of block (direction of acceleration of block)it is moving or opposite to component of resultant force along the surface if a is not moving
`N_(1)= 300 N`
so `f_(L)= mN_(1)= 0.6 xx 300 = 180 N`
Resultant of `4g` and `F_(2)` is `107.7 N` making an angle `tan^(-1)((2)/(5))`with the horizontal as force applied `F_(2) = 100 N` is less then `f_(D)` the block does not move and friction is static in nature
`f = 107.7N` making an angle of `tan^(-1)((2)/(5))`with the horizontal in