# Homework Help: Force on the pilot

1. Oct 15, 2009

### KillerZ

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

The plane is travelling at a constant speed of 800 ft/s along the curve y = 20(10^-6)x^2 + 5000, where x and y are in feet. If the pilot has a weight of 180 lb, determine the normal and tangential components of the force the seat exerts on the pilot when the plane is at its lowest point.

2. Relevant equations

$$\sum F = ma$$

$$\rho = \frac{[1 + (dy/dx)^{2}]^{3/2}}{|d^{2}y/dx^{2}|}$$

3. The attempt at a solution

FBD:

$$\uparrow\sum F_{n} = ma_{n}$$

$$N - 180lb = (180lb/32.2ft/s^{2})((800ft/s)^{2}/\rho)$$

$$\rightarrow\sum F_{t} = ma_{t}$$

$$0 = (180lb/32.2ft/s^{2})a_{t}$$

$$a_{t} = 0$$

$$\rho = \frac{[1 + (4x10^{-4}x)^{2}]^{3/2}}{|4x10^{-4}|}$$

x = 0

$$\rho = 2500 ft$$

$$N - 180lb = (180lb/32.2ft/s^{2})((800ft/s)^{2}/2500ft)$$

$$N = 1611.06 lb$$

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2. Oct 15, 2009

### Andrew Mason

Find the algebraic solution and plug in the numbers. Otherwise it is very hard to follow your reasoning.

The force on the seat is the weight of the pilot + the centripetal force. I think that is what you have shown in your diagram.

To find the centripetal force you have to find the radius of curvature. Since this is a parabolic curve, you have to find the radius of the circle that approximates the parabola at the vertex. You seem to have worked that out to be 2500 ft. I don't think that is right.

I think the radius of curvature at the vertex is the distance from the parabola to the focus. The focus is p where x^2 = 4py. Since in this case, 1/4p = 20 x 10^-6 = 1/50,000, the radius, p, should be 50,000/4.

AM

3. Oct 15, 2009

### KillerZ

thanks for the help.

I checked the back of the book and the answer is F_n = 323 lb so I looked over my work I made a mistake on the p its 25000 ft not 2500 ft I messed up my derivative of y = 20(10^-6)x^2.

4. Oct 15, 2009

### Andrew Mason

Ok, so the radius of curvature of the circle at the vertex is 2p not p. I stand corrected.

AM