Will the rope break when the man swings?

[dnt]

Homework Statement

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?

Homework Equations

Fc = mv^2/r

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 he's ok because the tension in the string doesn't exceed the 2000 N limit or do I need to include his weight in the problem?

 

[Robdog]

You already have his weight in the problem with 80kg

 

[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 he's going 8 m/s)?

 

[dnt]

anyone?

 

[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?

 

[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)

Now you have your mass, radius and velocity given to you.
Using this equation, you should be able to simply plug the values in and get something in the league of $$2.46x10^3 N$$

This is more than the maximum tension of 2000N, meaning that the rope will collapse.

Hope this helps
Steven