# Force around a curve

1. Feb 21, 2006

### finlejb

I have this problem as a part of my online phyics homework... it's the first question and I can't figure it out so it doesn't bode well for the rest of the assignment, so I might have to be back a lot before it's due Friday. Any hints on how I should go about solving it?

A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed, 14.7 m/s, the total force on the driver has magnitude 125 N. What is the total vector force on the driver if the speed is 18.9 m/s instead?

I've got that F=ma=m(v^2/r), but I don't have mass or radius, and I don't see any ways of rewriting that formula. I thought about a=change in velocity/change in time, but I don't have time so that doesn't help much either.

By the way, are there any ways to write formulas and such to make them easier to read than how I've been doing it?

Thanks for any help.

2. Feb 21, 2006

### gulsen

Well, you don't have mass or radius. But, it turns out that you don't need to know them either!
$$125 = \frac{m(14.7)^2}{r}$$
and you get $$m/r = 125/{14.7^2}$$
So, for a new speed, your fictous force turns out to be $$F = (125/{14.7^2}) 18.9^2$$

3. Feb 21, 2006

### finlejb

I don't quite understand that. I think my problem, though, is in the actual math part, not the concepts...

I've got that F=mass*centripetal acceleration, and you rearranged that to get m/r = F/v^2, but I don't really understand that last step. And how do you type the formulas in the different font?

Thanks for the help though.

4. Feb 21, 2006

### gulsen

The last step way simply $$F = (m/r) v^2$$, I just got m/r from the first equation, and plugged it into the second one. To put it another way,

$$F_1 = (m/r) v_1^2$$
$$F_2 = (m/r) v_2^2$$

where $$F_1 = 125$$, $$v_1=14.7$$ and $$v_2=18.9$$.
Now, if you divide each equation, $$m/r$$s will cancel and you get

$$\frac{F_1}{F_2} = \frac{v_1^2}{v_2^2}$$