Question

In the figure the man hanging upside down is holding a partner who weighs 535 N. Assume that the partner moves on a circle that has a radius of 6.50 m. At a swinging speed of 3.75 m/s, what force must the man apply to his partner in the straight-down position?

Answer 1

The force at the bottom is mv^2/r + mg.

For mass, 535/9.8 = 54.6 kg

F = (54.6)(3.75)^2/6.5 + (535)
F = 653.1 N

Answer 2

In rotational motion, according to Newton's second law, F=mv^2/r. Therefore, the force required to keep the man's partner in circular motion is:

F=mv^2/r

m=535/9.8=54.59kg

v=3.75m/s

r=6.5m

F=((54.59kg)(3.7 m/s)^2)/6.5m=114.97N.

However, since his partner is in the straight-down position, the man also has to apply force to keep his partner from falling down due to gravity, so you have to add the partner's weight:

Force total = 114.97N + 535N = 649.97N

Answer 3