A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of the painter who wants to lift the scaffold. She plans to pull downward on the loose end of the rope, thinking that the scaffold will then rise vertically with her along for the ride. The scaffold has a mass of 47 kg, and her mass is 64 kg. The painter pulls downward on the rope with a force of 640.0 N, while she and the scaffold are hanging from the other end above the ground. (a) What is the net acceleration on the system consisting of the painter and the scaffold? (b) What is the magnitude of the normal force exerted on the painter by the scaffold?

Question

Physics

Lilly Dawson

11 July, 12:29

A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of the painter who wants to lift the scaffold. She plans to pull downward on the loose end of the rope, thinking that the scaffold will then rise vertically with her along for the ride. The scaffold has a mass of 47 kg, and her mass is 64 kg. The painter pulls downward on the rope with a force of 640.0 N, while she and the scaffold are hanging from the other end above the ground. (a) What is the net acceleration on the system consisting of the painter and the scaffold? (b) What is the magnitude of the normal force exerted on the painter by the scaffold?

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Rafael Ferrell · Answer 1

Applying Newton's second law of motion;

let m = mass of scaffold M = her mass Tension = downward force = 640N g = 9.81m/s^2 Summation (F) = ma

hence total tension considering both ends of the scaffold;

= 2T, therefore, 2T - (m + M) g = a (m + M)

Making a the subject of the relation;

a = 2T - (m + M) g / (m + M)

plugging the values into the equation;

a = 2 x 640 - (47 + 64) x 9.81 / (47 + 64)

net acceleration = 1.72m/s^2

b) the magnitude of the normal force exerted on the painter by the scaffold; Summation (F) = ma

-Mg + T = N

N = T - mg

plugging the values;

N = - 64 x9.81 + 640

N = 12.16N