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Chapter 8: Problem 30

A house painter is standing on a uniform, horizontal platform that is held in equilibrium by two cables attached to supports on the roof. The painter has a mass of \(75 \mathrm{kg}\) and the mass of the platform is \(20.0 \mathrm{kg} .\) The distance from the left end of the platform to where the painter is standing is \(d=2.0 \mathrm{m}\) and the total length of the platform is $5.0 \mathrm{m}$ (a) How large is the force exerted by the left-hand cable on the platform? (b) How large is the force exerted by the right-hand cable?

Short Answer

Expert verified
Answer: The force exerted by the left-hand cable (cable A) is 49.05 N, and the force exerted by the right-hand cable (cable B) is 931.95 N.

Step by step solution

01

Write the force balance equation

For a system in equilibrium, the sum of the forces in any direction should be zero. Thus, we get: \(F_\text{A} + F_\text{B} = (75 + 20)g\)
02

Calculate the torque balance equation

For a system in rotational equilibrium, the sum of torque about any point should be zero. So, \(\tau_\text{A} - \tau_\text{platform} = 0\): \(F_\text{A}(2.0) - 20g \cdot 0.5 = 0\)
03

Solve for the force exerted by the left-hand cable (cable A)

Rearrange the torque balance equation to find the force exerted by cable A: \(F_\text{A} = \frac{20g \cdot 0.5}{2.0}\) Calculate the value: \(F_\text{A} = \frac{20 \cdot 9.81 \cdot 0.5}{2.0} = 49.05\,\text{N}\)
04

Solve for the force exerted by the right-hand cable (cable B)

Use the force balance equation to find the force exerted by cable B: \(F_\text{B} = (75 + 20)g - F_\text{A}\) Calculate the value: \(F_\text{B} = (75 + 20) \cdot 9.81 - 49.05 = 931.95\,\text{N}\) So, the force exerted by the left-hand cable (cable A) is \(49.05\,\text{N}\), and the force exerted by the right-hand cable (cable B) is \(931.95\,\text{N}\).

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