Question: In Fig.12-34, a uniform beam of weight 500 Nand length 3.00 m is suspended horizontally. On the left it is hinged to a wall; on the right it is supported by a cable bolted to the wall at distance Dabove the beam. The least tension that will snap the cable is 1200 N. (a) What value of D corresponds to that tension? (b) To prevent the cable from snapping, should Dbe increased or decreased from that value?

$$\begin{gathered} W=500\text{N} \\ L=3.0\text{m} \\ T=1200\text{N} \end{gathered}$$

Using the concept of static equilibrium, you can write the equation for torque in terms of forces and their distances from the pivot point. Solving these equations, you can find the unknown distance. The formula used is given below.

Static Equilibrium conditions:

$$\begin{gathered} \sum F_x=0 \\ \sum F_y=0 \\ \sum \tau =0 \end{gathered}$$

Applying equilibrium condition for torque:

$$\begin{gathered} \sum \tau =0 \\ T\times L\times \cos \left(\theta \right)-Mg\times \left(\frac{L}{2}\right)=0 \\ (1200\times 3\times \cos \left(\theta \right))-\left(500\times 1.5\right)=0 \\ \theta ={78}^0 \end{gathered}$$

From triangle geometry:

$$\begin{gathered} \tan \left(\theta \right)=\frac{L}{D} \\ \tan \left({78}^0\right)=\frac{3}{D} \\ D=0.64\text{m} \end{gathered}$$

Hence, D 0.64 m

Increasing the distance d would increase the angle. The higher the angle, the smaller the tension would be. This is as the weight of the beam would be supported by the vertical component of the tension which is$T\sin \theta$ . Hence to prevent the cable from snapping, we have to increase the distance D.