Find minimum force to raise wheel

[vu10758]

What minimum force F applied horizontally at the axle of the wheel is necessary to raise the wheel of mass M and radius R over a step of height H. A picture for this is at

http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460005538


The answer key says that the correct answer is F =[mg*SQRT(2RH-H^2)]/r-h

I know that mg is gravity so the force of gravity plays a role here. I don't know the direction of friction because F is pointing right, but the object also rotates clockwise. Friction would resist the sliding but would contribute to the rotation.

I know torque is the cross product of R and F. I have no idea though why we have SQRT(2RH-H^2). I don't know what to do. Am I suppose to work with torque somehow?

 

[Pyrrhus]

Take moment about the contact point, and use a force with unknown direction for the reaction at the contact point.

 

[vu10758]

I am not sure about this...is the moment of inertia with respect to the contact point MR^2 + R^2 with the parallel axis theorem? I know that whatever force the wheel pushes against the contact point, the same force will push the wheel back. But I am not sure about how to find that force for the reaction.

 

[Pyrrhus]

By moment, i meant sum of torque, and consider Normal force = 0.

 

[vu10758]

With net torque, I have mg*R + F(R-H).

So F = mg*R/(R-H)

From the way it seems, I got the F(R-H) part right, but mg*R is still not quite right. HOw do I get SQRT(2RH-H^2)?

 

[Pyrrhus]

With net torque, I have mg*R + F(R-H).

So F = mg*R/(R-H)

From the way it seems, I got the F(R-H) part right, but mg*R is still not quite right. HOw do I get SQRT(2RH-H^2)?

Your lever arm for the weight is wrong, REMEMBER THE LEVER ARM IS PERPENDICULAR TO THE LINE OF ACTION OF THE FORCE.