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

$83.33lb$
We can find the required forced $P$ as follows: $P-\mu mg=0$ We plug in the known values to obtain: $P-(0.4)(250)=0$ $\implies P=100$ We know that $\Sigma M_A=0$ $\implies P\times 4.5-250\times 1.5=0$ $P=83.33lb$