# Tension equation

1. Mar 30, 2009

### msudawgs267

I have a question about this equation (link below). I am working on a tension problem and I don't understand how you go from Sum of F= Ft-mg to
v= [square root of] (Ft-mg)r /m

The progression from one to the other is confusing me. I don't see how can use (mg) in one part of the equation and in the next part it is (v)squared and the back to (mg) at the end.

http://img3.imageshack.us/img3/4830/equationi.jpg [Broken]

I hope my question was not written too unclear and any help would be amazing. Thanks!

Last edited by a moderator: May 4, 2017
2. Mar 31, 2009

### alphysicist

Hi msudawgs267,

I would not say the expression on the left is progression from one step to another. The starting point is Newton's law in the form:

$$\sum F = m a$$

and then they are saying that

$$\sum F \to F_T - mg$$

and

$$m a \to m\frac{v^2}{r} \ \ \ \mbox{(for centripetal acceleration)}$$

Setting these two new expressions equal to each other and solving for v gives the answer.