F_{t} = 0 lbForce on Pilot at Lowest Point of Plane

[KillerZ]

Homework Statement

The plane is traveling 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.

Homework Equations

\sum F = ma

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

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

 

[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

 

[KillerZ]

Homework Statement

The plane is traveling 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.

Homework Equations

\sum F = ma

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

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^{-5}x)^{2}]^{3/2}}{|4x10^{-5}|}
x = 0

\rho = 25000 ft

N - 180lb = (180lb/32.2ft/s^{2})((800ft/s)^{2}/25000ft)

N = 323 lb = F_{n}
F_{t} = 0

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.

 

[Andrew Mason]

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

AM