# Moment on door vehicle

1. Nov 20, 2009

### bluelava0207

Validity of Einstein's Elevator Experiment

I solved the problem below using Einstein's http://en.wikipedia.org/wiki/Equivalence_principle" [Broken], or "elevator experiment," and I easily find an answer. I used traditional dynamics analysis and I get a different answer! Where is the mistake in my solution(s) or assumptions? I'm pretty good with dynamics, but I'm having trouble resolving this. Any help greatly appreciated!

A door is hinged to a vehicle at pt. O and is open at angle $$\theta$$. The vehicle's acceleration is a$$_{}o$$. I need to find the moment on the door caused by the vehicle's acceleration.

Answer from Solution 1: M = m*a$$_{}o$$*r*sin($$\theta$$)
Answer from Solution 2: M = a$$_{}o$$*r*sin($$\theta$$)/(1/m+r^2/I)

a$$_{}o$$ is the acceleration of the vehicle in the x direction.
$$\theta$$ is the angle of the door away from the x axis towards the y axis.
m is the mass of the door.
r is the distance from the hinge to the door's center of mass.
I is the door's moment of inertia.

SOLUTION 1
----------

Based on my interpretation of Einstein's "elevator experiment" or "equivalence principle," instead of saying that the vehicle is accelerating, I can say that the vehicle is stationary with the door being exposed to a gravity field in the -x direction with magnitude a$$_{}o$$. This makes it a very easy pendulum problem.

Force is applied at center of gravity
F = m*a$$_{}o$$

Moment is force times moment arm
M = F*r*sin($$\theta$$)
M = m*a$$_{}o$$*r*sin($$\theta$$)

SOLUTION 2
----------

I know that the vehicle pushes on the door at pt. O (but I don't know magnitude or direction):

F = force vector applied to pt. O

F has magnitude f in the $$\theta$$F direction. To see what this force does to the door, I move F to the center of gravity (pt. cg). The moment caused by moving this vector is:

M = R x F

R is the vector from pt. cg to pt. O. "x" is the cross product.

Now I know how the body will move for a given force vector.

A_CG = F/m
ALPHA = M/I

ALPHA is the radial acceleration vector, and A_CG is the acceleration vector of pt. cg.

I want to figure out what force vector will cause the door to move such that the hinge point accelerates in the x direction with magnitude a$$_{}o$$. I can then calculate the moment.

To find acceleration of pt. O:

A_O = A_CG + OMEGA x ( OMEGA x R ) + ALPHA x R

My vectors ([x,y,z]' indicates a column vector):

OMEGA = omega*[0,0,1]'
R = r*[-cos($$\theta$$),-sin($$\theta$$),0]'
F = f*[cos($$\theta$$F),sin($$\theta$$F),0]'
ALPHA = M/I = 1/I * R X F

From here out I'll use:

c$$\theta$$ = cos($$\theta$$), st = sin($$\theta$$)
c$$\theta$$f = cos($$\theta$$F), stf = sin($$\theta$$F)

My x direction and y direction equations come out to:

a$$_{}o$$ = f/m*c$$\theta$$f + omega^2*r*c$$\theta$$ - r^2*f/I*s$$\theta$$*(c$$\theta$$*s$$\theta$$f-s$$\theta$$*c$$\theta$$f)
0 = f/m*s$$\theta$$f + omega^2*r*s$$\theta$$ + r^2*f/I*c$$\theta$$*(c$$\theta$$*s$$\theta$$f-s$$\theta$$*c$$\theta$$f)

I can combine the two equations to get rid of the omega^2*r terms. Divide the first equation by c$$\theta$$ then subtract the second equation divided by s$$\theta$$.

a$$_{}o$$/c$$\theta$$ = f/m*(c$$\theta$$f/c$$\theta$$-s$$\theta$$f/s$$\theta$$) + r^2*f/I*(c$$\theta$$*s$$\theta$$f-s$$\theta$$*c$$\theta$$f)*(-s$$\theta$$/c$$\theta$$-c$$\theta$$/s$$\theta$$)

Multiply both sides by st*ct

a$$_{}o$$*s$$\theta$$ = f/m*(s$$\theta$$*c$$\theta$$f-c$$\theta$$*s$$\theta$$f) + r^2*f/I*(s$$\theta$$*c$$\theta$$f-c$$\theta$$*s$$\theta$$f)
a$$_{}o$$*s$$\theta$$ = f*(s$$\theta$$*c$$\theta$$f-c$$\theta$$*s$$\theta$$f)*(1/m+r^2/I)

f = a$$_{}o$$*s$$\theta$$/((s$$\theta$$*c$$\theta$$f-c$$\theta$$*s$$\theta$$f)*(1/m+r^2/I))
M = R x F = r*f*(s$$\theta$$*c$$\theta$$f-c$$\theta$$*s$$\theta$$f) (magnitude of M)

M = a$$_{}o$$*r*sin($$\theta$$)/(1/m+r^2/I)

Last edited by a moderator: May 4, 2017