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olasank [31]
4 years ago
10

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 pai

nter 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)
Physics
1 answer:
RSB [31]4 years ago
8 0

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.

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