Question

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 52 kg, and her mass is 63 kg. The painter pulls downward on the rope with a force of 600.0 N, while she and the scaffold are hanging from the other end above the ground.
Required:
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?

Answer 1

Solution :

a). From Newtons second law,

F = ma

The total tension force is 2T.

∴ 2T - (m + M)g = (m+ M)a

Then

$a=\frac{2T-(m+M)g}{m+M}$

$a=\frac{2\times 600-(52+63)9.8}{52+63}$

   $=0.63 \ m/s^2$

b). From the person,

   F = ma

 T - Mg + N = Ma

or N = Ma + Mg - T

        = (63 x 9.8) + (52 x 9.8) - 600

        = 617.4 + 509.6 - 600

        = 527 N