A house painter is standing on a uniform, horizontal platform that is held in eq

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 (as shown in the figure below ). The painter has a mass of 76.0 kg and the mass of the platform is 21.3 kg. The distance from the left end of the platform to where the painter is standing is d = 2.02 m and the total length of the platform is 4.86 m. How large is the force exerted by the left-hand cable on the platform? How large is the force exerted by the right-hand cable?

Explanation / Answer

Required for equilibrium: Sum of forces = 0 Sum of the torques (moments) = 0 Our forces DOWN are: Painter:76 kg (F = MA = 76 kg * 9.8 m/s^2 = 744.8 N) Platform: 21.3 kg (F =- MA = 21.3 * 9.8 m/s^2 = 208.74 N) Total:744.8 + 208.74= 953.54 N DOWN ===> Total forces by the rope UP must also = 953.54 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: 744.8 N * 2.02 m = 1504.49 N*m Platform CG is mid-way: 208.74 N * 4.86/2 m = 507.23 N*m Total: 1504.49+ 507.23= 2011.72 N*m ClockWise The RIGHT rope must counter that moment Right rope = X N * 4.86 m = 2011.72 = 2011.72 /4.86 = 413.93 N Then the LEFT rope must be:953.54 - 413.93 = 539.61 N

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