Chapter 9, Problem 18P

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 35.0⁰)and the horizontal and vertical forces exerted by the hinge on the beam at the wall.[ Hint. First draw a free-body diagram.]

To determine

The tension in the supporting wire and the horizontal and vertical forces exerted by the hinge on the beam at the wall.

Answer to Problem 18P

Solution:

The tension in the supporting wire is $642.19\text{N}$ and the horizontal and vertical forces exerted by the hinge on the beam at the wall are $526.05$ and $1.65\text{N}$

Explanation of Solution

Given:

The shop sign weighing 245 N is supported by a uniform 155 N.

Formula used:

The beam is in equilibrium, to calculate the tension in wire and forces use the condition of equilibrium.

  ${\displaystyle \sum \tau }=r_i\times F_i=0$

Here, $F_i$ is the applied force and $r_i$ represents its corresponding distance from the rotational axis.

Calculation:

  

From figure

  $\begin{array}{c}\sum \tau =(F_T\sin \theta )l_2-\frac{m_{1}g{l}_1}{2}-m_{2}g{l}_1=0\\ \sum \tau =\frac{1\times m_{1}gl}{2}+\frac{m_{2}g{l}_1}{l_2\sin \theta }\\ =\frac{1\times (155)\times (1.70)}{2}+\frac{(215)(1.70)}{1.35(\sin 35°)}\\ =603.8\text{N}\end{array}$

  $\begin{array}{c}\sum F_x=F_{Hx}-F_T\cos \theta =0\\ F_{Hx}=F_T\cos \theta =526.05\\ \sum F_y=F_{Hy}-F_T\sin \theta -m_{1}g-m_{2}g=0\\ F_{Hy}=m_{1}g+m_{2}g-F_T\sin \theta \\ =155+215-603.8(\sin 35)\\ =23.67\text{N}\end{array}$

Conclusion:

Hence, the tension in the supporting wire is $603.8\text{N}$ and the horizontal and vertical forces exerted by the hinge on the beam at the wall are $526.05$ and $23.67\text{N}$ .