Question: A scaffold of mass 60 Kg and length 5.0 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 80 kg stands at a point1.5 mfrom one end. What is the tension in (a) the nearer cable and (b) the farther cable?

The mass of the scaffold, M = 60 kg

The length of the scaffold, I = 5.0 m

The mass of the window washer, M = 80 Kg

The distance of the window washer from one end of the scaffold, I₁=1.5 m

Using the free body diagram and condition for equilibrium, you can write the equation for torque at point B. Using this, you can find the tension . You can balance the forces in upward and downward directions as the system is at equilibrium. From this, you can get the tension . The formulas used are given below.

Newton’s second law:

$F_{net}=ma$

At equilibrium,

$F_{net}=0$

The free body diagram:

For the equilibrium, torque at point B should be balanced,

$T_{1}l=mg{l}_2+Mg\left(\frac{l}{2}\right)$

So, the tension in nearer cable is,

$$\begin{gathered} T_1=\frac{mg{l}_2+Mg\left(\frac{l}{2}\right)}{l} \\ =\frac{80\left(9.8\right)\left(3.5\right)+\left(60\right)\left(9.8\right)\left(2.5\right)}{5} \\ =8.4\times {10}^2\text{N} \end{gathered}$$

Therefore, the tension in the nearer cable is$8.4\times {10}^2\text{N}$.

At equilibrium,

$$\begin{gathered} F_{net}=0 \\ \text{Upwardforces}\text{=}\text{Downwardforces} \\ {T}_1+T_2=mg+Mg \\ {T}_1+T_2=\left(m+M\right)g \\ {T}_1+T_2=\left(60+80\right)\left(9.8\right) \\ {T}_1+T_2=1.4\times {10}^3\text{N} \\ {T}_2=1.4\times {10}^3-8.4\times {10}^2 \\ =5.3\times {10}^2\text{N} \end{gathered}$$

Therefore, the tension in the farther cable is $5.3\times {10}^2\text{N}.$