Answer
$T_1=842.8 N$
Work Step by Step
We can find the tension $T_1$ in the nearer cable (i.e left cable) as follows:
$x=1.5 m$ distance between one end and window washer
$y=2.5 m$ mid point of scaffold
$z=5 m$ length of the scaffold
$W=mg=(80)(9.8)=784 N$ weight of the window washer
$W_s=mg=(60)(9.8)=588 N$ weight of the scaffold
According to second condition of equilibrium, the sum of all counter clockwise torques must be zero.
$\Sigma \tau=0$
$\implies W_s(z-y)+W(z-x)-T_1\times z=0$
$T_1=\frac{W_s(z-y)+W(z-x)}{z}$
Plugging in the known values, we find:
$T_1=\frac{588(5-2.5)+784(5-1.5)}{5}$
$T_1=842.8 N$
