posted by jiskha .
Workers who wash windows or paint the outside of buildings use an interesting contraption known as a painter's lift. This consists of a harness that the worker wears suspended by a rope. The rope runs through a pulley mounted on the roof of the building and back down to hang beside the worker. The worker simply pulls down on the hanging rope to raise herself up, and releases it to lower herself down (tieing the hanging rope to her harness keeps her at a constant height). What's neat is that the configuration also makes it easier for the worker to move up and down than if she was just hanging by a single rope. Let F1 be the force the worker exerts on the hanging rope in the painter's lift configuration to move upward at a constant speed. Let F2 be the force the worker would need to exert on a single rope to move upward at a constant speed. What is F1/F2? You can assume the rope itself doesn't have any significant mass.
The tension in the rope T pulls up on the worker's hand while the tension in the rope T also pulls up on the harness. Therefore the total force up on the worker is 2T. The force down is W, the weight. The acceleration is zero.
therefore T = W/2 and F1/F2 = 1/2
In other words the mechanical advantage is 2
This is the exact wording of one of this week's problems on brilliant dot o r g