Question

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?

Short answer

a) Tension in the cable, $\text{T}=88.27\text{\hspace{0.17em}N}$.

b) Force on the beam from the hinge$F=\left(30N\right)\widehat{i}+\left(97N\right)\widehat{j}$, .

Step-by-step solution

Understanding the given information

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

Concept and formula used in the given question

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.

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

(a) Calculation for the tension in the cable

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,

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

(b) Calculation for the force on the beam from the hinge in unit-vector notation

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:

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