## Engineering Mechanics: Statics & Dynamics (14th Edition)

$P=205N$
We can determine the required magnitude of $P$ as follows: We apply the principle of impulse and momentum in the y-direction: $mv_{y_1}+\Sigma \int_{t_1}^{t_2} F_ydt=mv_{y_2}$ We plug in the known values to obtain: $0+5+Psin30-(50)(90.81)(5)=0$ This simplifies to: $N=490.5-0.5P~~~~$eq(1) Now we apply the principle of impulse and momentum in the x-direction: $mv_{x_1}+\Sigma \int_{t_1}^{t_2}F_xdt=mv_{x_2}$ We plug in the known values to obtain: $0+(5)(0.2)N+5Pcos30=50(10)$ This simplifies to: $4.33P-N=500~~~~$eq(2) After solving eq(1) and eq(2), we obtain: $N=387.97N$ and $P=205N$