# Frictional force on bend.

1. May 24, 2010

### fcb

1. The problem statement, all variables and given/known data
A certain car experiances a limiting maximum frictional force equal to 75% of its weight. What is the smallest radius of bend that it can move around on a level road at 20ms-1

2. Relevant equations

F=mv2/r

3. The attempt at a solution
Couldnt get past last stage.

2. May 24, 2010

### Staff: Mentor

3. May 24, 2010

### Staff: Mentor

That's Newton's 2nd law applied to circular motion to give you the centripetal force. In this case, what is providing that force? (What is F equal to?) Plug in what you know about F and you'll be able to solve for v.

4. May 24, 2010

### fcb

All i know about 'F' is that it is equal to 75% of the cars weight. I dont know the cars mass, How am i able to solve for F? I am lost in my own little world.

5. May 24, 2010

### Staff: Mentor

Good!
Call the car's mass 'm'. How would you express F in terms of m?

6. May 24, 2010

### fcb

Would it be F=.75 x 'm'

7. May 24, 2010

### Staff: Mentor

Almost. Given the mass, how do you calculate the weight?

8. May 24, 2010

### fcb

0.75=1x202/r

0.75=400/r

r=$$\sqrt{}533$$

=23.06

9. May 24, 2010

### fcb

multiply it by acceleration due to gravity which is 9.8ms-2

10. May 24, 2010

### Staff: Mentor

Right. W = mg, where g = 9.8 m/s^2.

So revise your expression for F and plug it into the centripetal force formula.

11. May 24, 2010

### fcb

Scrap post #8. Its screwed.

12. May 24, 2010

### fcb

7.35=400/r
400/7.35 = 54.42

= 54.42

13. May 24, 2010

### Staff: Mentor

Good! That radius will have units of m.

Here's how I'd write it:

F = mv^2/r

.75 mg = mv^2/r

The mass cancels (so you don't need to know the mass after all):
.75 g = v^2/r

so: r = v^2/(.75 g)