91Ó°ÊÓ

Mass of scaffold 1, $m_1=50\text{kg}$

Mass of lower scaffold 2, $m_2=30\text{kg}$

Length, $L_2=2.00\text{m}$

Mass of nail box, $m=20.0\text{kg}$

d = 0.500 m

Length of scaffold 1, $L_1=3.00\text{m}$

Tension on right, $T_R=196\text{N}$

Tension on left,$T_R=196\text{N}$

Applying equilibrium conditions to both scaffolds, you can get the equations in terms of the unknown tensionsT_R and T_L . By plugging in the known values, you can find the magnitudes of tensions. The equations used are given below.

Static Equilibrium conditions:

$$\begin{gathered} ∑F_x=0 \\ ∑F_y=0 \\ ∑\mathrm{τ}=0 \end{gathered}$$

FBD of scaffold .

Taking torque at left point.

.$\left(T_R×L_2\right)-\left(mgd\right)-\left(m_{2}g×\left(\frac{L}{2}\right)\right)=0$

Hence,

$$\begin{gathered} T_R=196\text{N} \\ ∑F_y=0 \\ {T}_L+T_R-mg-m_{2}g=0 \\ {T}_L+196-20×9.8-30×9.8=0 \\ {T}_L=294\text{N} \end{gathered}$$

FBD of scaffold :

Considering the pivot at left end of scaffold :
$$\begin{gathered} ∑\mathrm{τ}=0 \\ \left(T×L_1\right)-\left(T_L×d\right)-\left(m_{1}g×\left(\frac{L_1}{2}\right)\right)-T_R\left(L_1-d\right)=0 \\ \left(T×3\right)-\left(294×0.5\right)-\left(50×9.8×1.5\right)-196×2.5=0 \\ T=457\text{N} \end{gathered}$$

Hence, the tension in the indicated cable is 457 N.