# Pilot in a circular arc

1. Dec 20, 2012

### Infamous911

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

A pilot of mass 50 kg comes out of a vertical dive in a circular arc such that her upward acceleration is 6g.

(a) What is the magnitude of the force exerted by the airplane seat on the pilot at the bottom of the arc?

(b) If the speed of the plane is 390 km/h, what is the radius of the circular arc?

2. Relevant equations

Fc = (mv^2)/r
(v^2/r)=a

3. The attempt at a solution

(a)
Fc = 50(6)(9.8)
Fc = 2940

(b)
2940 = (50(360^2))/r
r=2586.735

Apparently these answers are wrong and I have no idea why.

2. Dec 20, 2012

### HallsofIvy

Staff Emeritus
There would be 1 g, or 9.8 times mass, force even if the pilot were flying in a straight line. If the upward acceleration is 6g, then the force will be 50(7)(9.8), adding that original 1g.

Where did "360" come from? Did you mean to use 390?

3. Dec 20, 2012

### Staff: Mentor

The first answer looks right to me. Are you sure it's wrong?

In (b), you first have a typo for the speed, and second, you have not converted the speed from km/hr to m/s. You need to work in a consistent set of units, usually the SI/mks system.

4. Dec 20, 2012

### Staff: Mentor

Ah, good point about adding in the extra 1g for level flight, Halls. I thought about it but reasoned incorrectly.

5. Dec 20, 2012

### Infamous911

Ok so I have solved (a) with a solution of 3430 N, thanks.

I'm still having problems with (b) though.

390 km/h = 108.333 m/s

3430=(50(108.333^2))/r
r=171.079

This isn't the correct answer apparently (I put in my answer into an online form and it tells me if it's right or wrong).

6. Dec 20, 2012

### Infamous911

Nevermind, I figured out that it should instead be just:

a=v^2/r
6(9.8)=108.333^2/r
r=199.592

7. Dec 20, 2012

### Staff: Mentor

You're Infamous! Good work.