# Bicycle over curb

## Homework Statement

You are trying to raise a bicycle wheel of mass m and radius R up over a curb of height h. To do this, you apply a horizontal force F.

What is the least magnitude of the force F_vec that will succeed in raising the wheel onto the curb when the force is applied at the center of the wheel?

F = 0
torque_total = 0

## The Attempt at a Solution

I've tried for the last hours to figure out how to do this but I'm stuck. I've drawn a free body diagram with the horizontal force and also, the force the curb is exerting on the wheel as well? They must equal right..? I don't really have any idea how to start this problem other than that though. So frusterated sigh. :( Any help is appreciated, thanks.

Pic: http://img85.imageshack.us/img85/9453/yffigure1142xt0.jpg [Broken]

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Doc Al
Mentor
First identify all the forces exerted on the wheel. Then analyze the torques they exert about the pivot point. What's the pivot point as the wheel begins to climb the curb?

If youve already drawn a free body diagram, draw a torque diagram. At what distance does each force act over?

Thanks for the help.

I'm stuck on the forces. So far, I've identified gravity force, normal force from the ground, the force the curb is exerting on the wheel and the horizontal force. This is all right? But normal force and gravity force shouldn't affect the motion because they're in the y direction right? So it's only the horizontal force and curb force that matter....correct me if I'm wrong.

As for the pivot point and the distance each force act over, I'm confused there also. I think it should be where the wheel and the curb meet? But I'm not sure because that would make it very hard to find the torque for the force the curb is exerting on the wheel..? Unless there's something I'm missing...

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Doc Al
Mentor
I'm stuck on the forces. So far, I've identified gravity force, normal force from the ground, the force the curb is exerting on the wheel and the horizontal force. This is all right?
Good! Now consider the torque that they exert just as the wheel begins to climb the curb.

But normal force and gravity force shouldn't affect the motion because they're in the y direction right? So it's only the horizontal force and curb force that matter....correct me if I'm wrong.
Does gravity exert a torque about the pivot point? (What's the perpendicular distance?) What happens to the normal force as the wheel loses contact with the ground?

As for the pivot point and the distance each force act over, I'm confused there also. I think it should be where the wheel and the curb meet?
Good!
But I'm not sure because that would make it very hard to find the torque for the force the curb is exerting on the wheel..?
It actually makes finding the torque for the curb force very easy to find! (Hint: What's the distance in that case?)

The normal force should be zero because there's nothing pushing up anymore. For the torques, this is what I have so far:

Gravity - I'm confused. I'm thinking it should be mgR, because R is the distance from the gravity force (which should be at the center of mass right?) to the pivot point. I don't understand what you mean by perpendicular distance...

Normal force - 0

F of the curb - 0

Horizontal force - FR

Doc Al
Mentor
Torque is not just force*distance or FD--the angle the force makes makes a difference! One way to express torque is Force*perpendicular distance (also called "moment arm"); another way is Force*distance*sin(theta).

Look up the definition of torque in your text. Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/torq.html" [Broken]

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Ah erm, so here I go again.

Horizontal force - The perpendicular distance of F to the pivot line, is it (R - h)? So then that would make the torque = F(R - h)? I'm not sure about theta because it's not given in the problem. Unless there's some other logic that I'm missing again.

Gravity - It should still be mgR right? Because the perpendicular distance is just R and gravity force points downward so...

Am I wrong again?

Doc Al
Mentor
Horizontal force - The perpendicular distance of F to the pivot line, is it (R - h)? So then that would make the torque = F(R - h)?
Exactly right.
Gravity - It should still be mgR right? Because the perpendicular distance is just R and gravity force points downward so...
No. Since gravity acts down, the perpendicular distance is the horizontal distance between the line of gravity (through the center) and the curb corner. (That distance is less than R.) You'll have to figure that out. (That's the hard part--but not that hard.)

Thanks for all your help, really appreciate it. I got it now (cleared the torque confusion too). Phew.

Doc Al
Mentor
That's what I want to hear. Good work.