Question

A
house painter is standing on a uniform horizontal platform that is
held in equilibrium by two cables attached to supports on the roof
of the building. The painter has a mass of 79.2 kg and the mass of
the platform is 18.0 kg. The distance from the left end of the
platform to where the painter is standing is d = 2.0 m and the
total length of the platform is 4.70 m.

How large is the force exerted by the right-hand cable?

Answer #1

Required for equilibrium:

Sum of forces = 0

Sum of the torques (moments) = 0

Our forces DOWN are:

Painter: 79.2 kg (F = MA = 79.2 kg * 9.8 m/s^2 = 776.16 N)

Platform: 18 kg (F =- MA = 18 * 9.8 m/s^2 = 176.4 N)

Total: 776.16 + 176.4 = 952.56 N DOWN ===> Total forces by the rope UP must also = 952.56 N

We take moments about the LEFT END ===> We can ignore the tension in the LEFT rope for the moment since it has no lever arm.

Painter Moment: 776.16 N * 2 m = 1552.32 N*m

Platform CG is mid-way: 176.4 N * 4.7/2 m = 414.54 N*m

Total: 1552.32 + 414.54 = 1966.86 N*m ClockWise

The RIGHT rope must counter that moment

Right rope = X N * 4.7 m = 1966.86

=> X = **418.5
N**

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