Maximum weight.....?

Question

A person is to be released from rest on a swing pulled away from the vertical by an angle of 15.8°. The two frayed ropes of the swing are 2.75 m long, and will break if the tension in either of them exceeds 350 N.

What is the maximum weight the person can have and not break the ropes?

Answer 1

♠ when swinging you are moving along a circular path ±15.8° to and fro around equilibrium position – draw a picture.
thus your potential energy with respect to equilibrium E= P*h, where P=mg is your weight, m is your mass, gravity g=9.8 m/s^2, h= r-r*cos(15.8°) is your height above equilibrium, r=2.75 m.
♣ when released your speed v increases and attains max when you pass equilibrium position. Potential energy E becomes your kinetic energy; thus E=0.5m*v^2; therefore mg*h=0.5m*v^2, hence max speed squared is v^2=2gh;
♦ moving in a circular track a centripetal force F=m*v^2/r exerts your bottom; also take your weight into account; then P+F force must not exceed
350N = mg + m*v^2/r, hence v^2 =(350-mg)*r/m; or
(350-mg)*r/m = 2gh /see ♣/; or (350-mg)*r/m =2g*(r –r*cos15.8) /see ♠/, then 350-mg =2mg(1-cos15.8); or 350 = mg*(2 –2cos15.8 +1), hence
♥ your max weight mg =350/(3-2cos15.8) = 267 N =P;
note: r=2.75m is auxiliary datum.

Answer 2

2 ton