Question

A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The painter is standing 2 m from the first rope and 8 m from the second rope. What is the tension in the first rope (in Newtons)

Answer 1

Explanation:

For equilibrium, $\sum M = 0$.

So,   $8 m \times mg - (10 m) T_{1}$ = 0

             $T_{1} = \frac{8 \times mg}{10}$

                        = $\frac{8 \times 90 \times 9.8}{10}$

                        = 705.6 N

Also, for equilibrium $\sum F_{y}$ = 0

              $T_{1} + T_{2} - mg$ = 0

or,         $T_{2} = mg - T_{1}$

                        = $90 \times 9.8 - 705.6$

                        = 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.