# Tension problem

1. Oct 18, 2007

### dnt

1. The problem statement, all variables and given/known data

a man swings from a rope which is 10 feet long. his mass is 80 kg and at the bottom of his swing, he is going 8 m/s. The max tension the rope can take without breaking is 2000 N. Does he make it?

2. Relevant equations

Fc = mv^2/r

3. The attempt at a solution

first i had to convert the feet to meters (3.05m). then i think the way to solve it is to first figure out the centripetal force generated by swinging - this would be (80 kg)(8)^2/(3.05 m) = 1678 N.

does this mean hes ok because the tension in the string doesnt exceed the 2000 N limit or do I need to include his weight in the problem?

2. Oct 18, 2007

### Robdog

You already have his weight in the problem with 80kg

3. Oct 19, 2007

### dnt

i konw but how do i put it into the calculations? so far all i figured out was the centripetal force?

how does the centripetal force, tension and his weight relate to each other at the bottom (when hes going 8 m/s)?

4. Oct 19, 2007

### dnt

anyone?

5. Oct 20, 2007

### dnt

another bump - still confused.

at the bottom of his swing, how does the centripetal force, tension force and his weight relate to each other?

i know weight points up and centripetal force points towards the middle (therefore up), but tension? i think it points up as well so would the equation be:

T + Fc - W = ma?

6. Oct 21, 2007

### steven10137

10 feet = 3.05m
and $$T_{max}=2000N$$

At the bottom;
$$\frac{mv^2}{{r}} = F_{T} - mg$$
hence
$$F_{T} = \frac{mv^2}{{r}} + mg$$

(If you need more clarification; http://dev.physicslab.org/Document.aspx?doctype=3&filename=OscillatoryMotion_VerticalCircles.xml)

Using this equation, you should be able to simply plug the values in and get something in the league of $$2.46x10^3 N$$