A cubical block is experiencing three forces `F_(1) = 20 N` acts at angle `45^(@)`with horizontal and lies in diagram plane of the cube `F_(2) = 30 N` acts along y- axis and `F_(3) = 40 N` acts in x- direction as shown in fig find the friction force acting on the block if it is at rest

Answer

`50 N` From x- axis

Explanation

As shown in figure `f_(1)` is having three component one along vertical`((F_(1))/(sqrt2))` . After along force `F_(2) ((F)/(sqrt2)xx (1)/(sqrt(2)))` and one againest `F_(3) ((F)/(sqrt2)xx (1)/(sqrt(2)))` .Here the net horizontal force acting on the block is given by
`F_("net") = sqrt((F_(2) + F// sqrt(4))^(2) +(F_(3) - F//sqrt(4))^(2))`
`= sqrt((30 + 10)^(2) +(40 - 10)^(2) = 50 N)`
As it is given that block is in static equalibruim this sum of horizontal force acting on it is mass be zero sum of given horizontal force is `50 N` and is in a direction `theta = tan^(-1)((4)/(3))= 53` from the direction of force `F_(3)` we can static that friction so as in keeep the block in static equlibrium