II$.$ Problems

Two forces are applied at the origin as shown on the figure. The forces are known: $P=365lb$ and $Q=220lb.$

a$.(8$ points$)$ Use Parallelogram Law and Triangle Rule to draw a force parallelogram and triangle.

b$.(4$ points$)$ Determine the resultant $R$ of the two forces $P$ and $Q.$

c$.(4$ points$)$ Determine the angle $$ the resultant makes with the positive x axis.

TUA

$ta{n}^{-1}(\frac{-223\mathrm{.}55}{172\mathrm{.}34})=-52\mathrm{.}379$

$R=\sqrt[\phantom{2}]{Q^2+P2-2(P)()cos}$

$P_x=(-\frac{4}{5})365+(\frac{3}{5})365$

$P=-292$hat$(i)+219$hat$(j)$

$Q=-220sin40i-270cos40$hat$(j)$

$=-141.$hat$(i)-168\mathrm{.}53$hat$(j)$

$R+$vec$(P)+$vec$(C)=>(-292-141\mathrm{.}4)hat(i)+(219-16*53)hat(j)$

$R=-433\mathrm{.}4i+50\mathrm{.}47$hat$(j)$

$n=436\mathrm{.}36$

$ta{n}^{-1}(\frac{1036}{?})$

$=-\frac{5}{2\mathrm{.}37}$