Question

A shop sign weighing 215 N hangs from the end of a uniform 155-N beam, as shown in Fig. 9–58. Find the tension in the supporting wire (at 35.0°) and the horizontal and vertical forces exerted by the hinge on the beam at the wall. [Hint: First, draw a free-body diagram

Short answer

The tension in the supporting wire is 642.2 N.

The horizontal and vertical components of the forces by the hinge are $526.1\mathrm{N}$ and $1.6\mathrm{N}$, respectively.

Step-by-step solution

Concepts

In equilibrium, the net force in the x and y directions should be zero, and the torque about any point is zero.For this problem, you can calculate the tension in the wire using the condition for the zero torque about the hinge.

Given data

The weight of the shop sign is $W_1=215\mathrm{N}$.

The weight of the beam is $W_2=155\mathrm{N}$.

The angle of the wire is $\theta ={35.0}^{\circ }$.

The length of the beam is $L_1=1.70\mathrm{m}$.

The supported wire is attached at a distance of $L_2=1.35\mathrm{m}$ from the hinge.

You can assume that the mass of the beam is in the middle of the beam.

Let $F_{\mathrm{x}}$ and $F_{\mathrm{y}}$ be the horizontal and vertical components of the forces exerted by the hinge on the beam.

Calculation of the tension

The torque about the hinge is

$\begin{array}{c}T\sin \theta \times L_2-W_1{L}_1-W_2\frac{L_1}{2}=0\\ T\sin \theta \times L_2=\left(W_1+\frac{W_2}{2}\right)L_1\\ T\times \sin {35.0}^{\circ }\times 1.35\mathrm{m}=\left(215\mathrm{N}+\frac{155\mathrm{N}}{2}\right)1.70\mathrm{m}\\ T=642.2\mathrm{N}\end{array}$.

Hence, the tension in the supporting wire is 642.2 N.

Calculation of the force by the hinge

Now, the condition for the horizontal forces at equilibrium is

$\begin{array}{c}{F}_{\mathrm{x}}=T\cos \theta \\ =\left(642.2\mathrm{N}\right)\times \cos {35.0}^{\circ }\\ =526.1\mathrm{N}\end{array}$.

Now, the condition for the vertical forces at equilibrium is

$\begin{array}{c}{F}_{\mathrm{y}}+T\sin \theta =W_1+W_2\\ F_{\mathrm{y}}+\left[\left(642.2\mathrm{N}\right)\times \sin {35.0}^{\circ }\right]=215\mathrm{N}+155\mathrm{N}\\ F_{\mathrm{y}}=1.6\mathrm{N}\end{array}$.

Hence, the horizontal and vertical components of the forces by the hinge are $526.1\mathrm{N}$ and $1.6\mathrm{N}$, respectively.