91Ó°ÊÓ

Mass, $\text{m}=53\text{ k²µ}$

Length of the beam,$\text{L}=5.0\text{″¾}$

$\mathrm{θ}=60°$

Using the condition for the static equilibrium, you can write the equation for forces for the given figure. In addition, the equation for the torque can be written about a pivot point. The equations are given below and can be solved for the unknown forces.

Condition of equilibrium,

$\begin{array}{l}\mathbf{Ï„}\mathbf{=}\mathbf{0}\\ \mathbf{\text{²Ñ´Ç³¾±ð²Ô³Ù´Ç´Ú´Ú´Ç°ù³¦±ð=±è±ð°ù±è±ð²Ô»å¾±³¦³Ü±ô²¹°ù»å¾±²õ³Ù²¹²Ô³¦±ð×¹ó´Ç°ù³¦±ð}}\end{array}$

As we see the diagram, we can say that

$\begin{array}{l}Tsin60°×L−mg×\left(\frac{L}{2}\right)=0\\ Tsin60°=mg×\frac{\left(\frac{L}{2}\right)}{L}\\ Tsin60°=\frac{mg}{2sin60}\\ Tsin60°=300\text{ N}\end{array}$

As the system is in equilibrium, we can say

$\begin{array}{l}∑{\text{F}}_{\text{x}}=0\\ ∑{\text{F}}_{\text{y}}=0\end{array}$

So,

$\begin{array}{c}{F}_{Px}=−Tcos\mathrm{θ}\\ =−\left(300\text{ }N\right)\left(0.5\right)\\ =−150\text{ N}\\ F_{Py}=Mg−Tsin60°\\ =\left(519.4\text{ â¶Ä‹}N\right)−(300\text{ }N×0.8660)\\ =260\text{ }N\end{array}$

So, we get

$F_P=(−150\text{ }N)\widehat{i}+\left(260\text{ }N\right)\widehat{j}$