Question

A uniform beam is $\mathbf{5}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$long and has a mass of $\mathbf{53}\mathbf{\text{\hspace{0.17em}kg}}$. In Fig. 12-74, the beam is supported in a horizontal position by a hinge and a cable, with angle$\mathit{\theta }\mathbf{=}\mathbf{60}\mathbf{°}$. In unit-vector notation, what is the force on the beam rom the hinge?

Short answer

The force on the beam in unit vector notation,$F_P=(-150N)\widehat{i}+\left(260N\right)\widehat{j}$ .

Step-by-step solution

Understanding the given information

Mass, $\text{m}=53\text{\hspace{0.17em}kg}$

Length of the beam,$\text{L}=5.0\text{\hspace{0.17em}m}$

$\theta =60°$

Concept and formula used in the given question

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,

$\begin{array}{l}\mathbf{\tau }\mathbf{=}\mathbf{0}\\ \mathbf{\text{Momentofforce=perpendiculardistance×Force}}\end{array}$

Calculation for theforce on the beam from the hinge

As we see the diagram, we can say that

$\begin{array}{l}Tsin60°\times L-mg\times \left(\frac{L}{2}\right)=0\\ Tsin60°=mg\times \frac{\left(\frac{L}{2}\right)}{L}\\ Tsin60°=\frac{mg}{2sin60}\\ Tsin60°=300\text{\hspace{0.17em}N}\end{array}$

As the system is in equilibrium, we can say

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

So,

$\begin{array}{c}{F}_{Px}=-Tcos\theta \\ =-\left(300N\right)\left(0.5\right)\\ =-150\text{\hspace{0.17em}N}\\ F_{Py}=Mg-Tsin60°\\ =\left(519.4\text{\hspace{0.17em}}N\right)-(300N\times 0.8660)\\ =260N\end{array}$

So, we get

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