Posted by **micole** on Thursday, May 24, 2007 at 12:14pm.

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?

I started by using the equation 350N/cos15.8 = Weight in one rope (2) = 727.5. This answer was incorrect. I am not sure if I have the picture drawn wrong. Where exactly does the length of the ropes come in? Please help!

Max tension will occur at the bottome, and tension will equal weight plus centripetal force. Veloicty at the bottom is dependent on the height, when the swing started: here height was 2.75(1-cosTheta).

Tension= weight+ mv^2/r

and mv^2=2mgh, with h above.

## Answer This Question

## Related Questions

- physics - a child's swing is held up by two ropes tied to a tree branch that ...
- physics - A child's swing is held up by two ropes ties to a tree branch that ...
- Physics - A 50- kg child sits in a swing suspended with 3.8- m-long ropes. The ...
- Physics PLEASE HELP - A 80- kg child sits in a swing suspended with 2.6- m-long ...
- PHYSICS PLEASE HELP - A 80- kg child sits in a swing suspended with 2.6- m-long ...
- Physics - A 40 kg child sits in a swing suspended with 2.5 m long ropes. The ...
- PHYSICS - A 27.0 kg child on a 3.00 m long swing is released from rest when the ...
- physics - A 25.0-kg child on a 3.00-m-long swing is released from rest when the ...
- Last physics help question (my thoughts given) - A 160-N child sits on a light ...
- ap Physics - AP PROBLEM CHILD ON A SWING AN ADULT EXERTS A HORIZONTAL FORCE ON A...

More Related Questions