Potential Energy: Suspended Person, Force of mg(3cosθ-2cosβ)

[mirandasatterley]

A person with mass,m, is suspended by 2 ropes (paralell) both of length l. The person begins on a platform in the air. The person steps off the platform, starting from rest with the rope at an angle beta, with respect the the verticle. Air resistance is negligable.
a) When the ropes make an angle theta with the verticle, the person must exert a force of: mg(3 cos(theta) -2 cos(beta)).
b) Find the angle beta where the force needed to hang on at the bottom of the swing is twice the persons weight.

Any help is appreciated. Thanks

 

[PhanthomJay]

A person with mass,m, is suspended by 2 ropes (paralell) both of length l. The person begins on a platform in the air. The person steps off the platform, starting from rest with the rope at an angle beta, with respect the the verticle. Air resistance is negligable.
a) When the ropes make an angle theta with the verticle, the person must exert a force of: mg(3 cos(theta) -2 cos(beta)).
b) Find the angle beta where the force needed to hang on at the bottom of the swing is twice the persons weight.

Any help is appreciated. Thanks

You've got to use conservation of mechanical energy principle whereby initial PE + initial KE = final PE + Final KE. Initial KE is 0, so that's a help. For initial PE, you need to draw a diagram and see how the beta angle and radius of the arc detrmine the height h. It's a bit messy but you've got to plug and chug through it. Same for final PE at angle theta. Final KE you are trying to solve, solve for V as function of R,m,theta , and beta. Then do a FBD at the angle theta, T-mgcos theta = mv^2/r. Solve for T, the required force the person must exert.
Part b is similar, except this time you know that T = 2mg.