91Ó°ÊÓ

$\begin{array}{l}W=60\text{ N}\\ L=3.2\text{″¾}\\ F=50\text{ N}\\ h=2.0\text{″¾}\end{array}$

Youcan resolve the tension along vertical and horizontal directions. Using the equilibrium condition, you can find the tension in the cable. The formulas are given below.

$\begin{array}{l}\mathbf{τ}\mathbf{=}\mathbf{r}\mathbf{×}\mathbf{F}\\ \mathbf{F}\mathbf{=}\sqrt{{\mathbf{F}}_{\mathbf{p}\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{F}}_{\mathbf{p}\mathbf{y}}^{\mathbf{2}}}\end{array}$

Free body diagram,

First, we have to resolve the tension alongthe X and Y-axis as the components $Tcos25°\text{ }and\text{ }Tsin25°,$respectively.

By the condition of equilibrium, the moment of force we can write from the figure is,

$\begin{array}{l}F×H=Tcos25×h\\ 50\text{ N}×3.2\text{″¾}=T\left(0.906\right)×2.0\text{″¾}\\ T=88.27\text{ N}\end{array}$

By condition of equilibrium, we can say that

$\begin{array}{l}∑F_x=0\\ ∑F_y=0\\ F_{px}=Tcos25°−F\\ F_{px}=79.99\text{ }N−50\text{ }N\\ F_{px}=29.99N\\ F_{px}~30\text{ N}\\ F_{py}=Tsin25°+W\\ F_{py}=37.30\text{ }N+60\text{ }N\\ F_{py}=97.30\text{ }N\\ F_{py}~97\text{ N}\end{array}$

These are the components of force on the hinge. You can write them in unit vector notation like so:

$F=\left(30\text{ }N\right)\widehat{i}+\left(97\text{ }N\right)\widehat{j}$