A block of mass `8 kg` is kept on a rough floor. The coefficients of friction are `mu_(s) = 0.5` and `mu_(k) = 0.4`. Force `F_(1) = 15 N` and `F_(2) = 20 N` are applied on the block as shown.
(a) Find the magnitude and the direction of friction force .
(b) If `F_(1) = 30 N , F_(2) = 40 N`, find the acceleration of the block.

Explanation

(a) `N_(0) = 80 N`
`(f_(s))_(max) = mu_(s) N = 0.5 xx 80 = 40 N`
Resultant external force :
`F = sqrt(F_(1)^(2) + F_(2)^(2)) = sqrt((15)^(2) + (20)^(2)) = 25 N`
`tan theta = (15)/(20) = (3)/(4) rArr theta = 37^(@)`
Since `Flt (f_(s))_(max)`, the block is at rest. The friction force `f` will act opposite to the direction of resultant force.
`Friction = 25 N`
(b) Resultant force:
`F = sqrt(F_(1)^(2) + F_(2)^(2)) = sqrt((30)^(2) + (40)^(2)) = 50 N`
Now `F gt (f_(s))_(max)` , the block is in motion.
`f = f_(k) = mu_(k) N = 0.4 xx 80 = 32 N`
`F - f_(k) = ma`
`50 - 32 = 18 a`
`a = (18)/(8) = (9)/(4) = 2.25 m//s^(2)`