Determine the magnitude of frictional force and acceleration of the block in each of the following cases :

Explanation

a.`f_(max) = mu N = (0.2)(100) = 20 N`
Since `mg gt f_(max)` therefore, friction force is equal to `f_(max)` i.e.`f= f_(max) = 20 N`
b. `f_(max) = mu N= 0.2 xx 500 = 100 N`
Since `mg lt f_(max) ` therefore, friction force is equal to `mg ` The mean that `f = mg = 50 N`

c. `f_(max) = mu N = (mu 100 cos 30^(@)) = (0.2)(100)(sqrt(3))/(2) = 10 sqrt(3) N`
Since the vertical component of the external force `100 sin 30^(@) = 50 N` can balance the weight of the block , therefore `f = 0`