In the figure the man hanging upside down is holding a partner who weighs 600 N.

Question

In the figure the man hanging upside down is holding a partner who weighs600N. Assume that the partner moves on a circle that has a radius of 6.50 m. At a swinging speed of3.35m/s, what force must the man apply to his partner in the straight-down position?
N

In the figure the man hanging upside down is holding a partner who weighs600N. Assume that the partner moves on a circle that has a radius of 6.50 m. At a swinging speed of3.35m/s, what force must the man apply to his partner in the straight-down position? N

Explanation / Answer

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=495/9.8=50.51kg v=3.6m/s r=6.5m F=((50.51kg)(3.6 m/s)^2)/6.5m=100.71N. 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 = 100.71N + 495N = 595.71N

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