Answer
(a) $160N$
(b) $146.1N$
Work Step by Step
(a) We can determine the required friction as follows:
As the sum of forces in the y- direction is zero
$\Sigma F_y=0$
$\implies N-P_y-W=0$
$N-200\times 0.6-50\times 9.81=0$
$\implies N=610.5N$
As $F_{max}=\mu_s N$
$\implies F_{max}=0.30\times 610.5$
$\implies F_{max}=183.15N$
Now $P=\frac{4}{5}(200)=160N$
As the magnitude of the force $P$ is less than the maximum static friction $F_{max}$, thus, the crate will not move and hence the crate is in equilibrium. The required friction is $F=160N$
(b) We can find the required friction as follows:
$\Sigma F_y=0$
$\implies N-P_y-W=0$
$\implies N-400(0.6)-50(9.81)=0$
$\implies N=730.5N$
$F_{max}=\mu_s N$
$F_{max}=0.30(730.5)=219.15N$
As $P=\frac{4}{5}(400)=320N$
The magnitude of the force $P$ is greater than the value of maximum static friction, which shows that the crate is in motion and is not in equilibrium.
Now the expression for kinetic friction is given as
$F=\mu_k N$
$F=(0.20)(730.5)$
$\implies F=146.1N$