In Fig. 12-73, a uniform beam with a weight of $\mathbf{60}\mathbf{\text{\hspace{0.17em}N}}$and a length of $\mathbf{3}\mathbf{.2}\mathbf{\text{\hspace{0.17em}m}}$ is hinged at its lower end, and a horizontal force of magnitude 50 N acts at its upper end. The beam is held vertical by a cable that makes angle $\mathbf{\text{θ}}\mathbf{=}\mathbf{}\mathbf{25}\mathbf{\text{°}}$with the ground and is attached to the beam at height $\mathbf{\text{h}}\mathbf{=}\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$ . What are (a) the tension in the cable and (b) the force on the beam from the hinge in unit-vector notation?

$\begin{array}{l}W=60\text{\hspace{0.17em}N}\\ L=3.2\text{\hspace{0.17em}m}\\ F=50\text{\hspace{0.17em}N}\\ h=2.0\text{\hspace{0.17em}m}\end{array}$

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.

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

Free body diagram,

First, we have to resolve the tension alongthe X and Y-axis as the components $Tcos25°andTsin25°,$respectively.

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

$\begin{array}{l}F\times H=Tcos25\times h\\ 50\text{\hspace{0.17em}N}\times 3.2\text{\hspace{0.17em}m}=T\left(0.906\right)\times 2.0\text{\hspace{0.17em}m}\\ T=88.27\text{\hspace{0.17em}N}\end{array}$

By condition of equilibrium, we can say that

$\begin{array}{l}\sum F_x=0\\ \sum F_y=0\\ F_{px}=Tcos25°-F\\ F_{px}=79.99N-50N\\ F_{px}=29.99N\\ F_{px}~30\text{\hspace{0.17em}N}\\ F_{py}=Tsin25°+W\\ F_{py}=37.30N+60N\\ F_{py}=97.30N\\ F_{py}~97\text{\hspace{0.17em}N}\end{array}$

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

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