# Pushing a Wheel Over a Curb

1. Jul 1, 2012

### Bennigan88

So I made an observation that I would really like anyone who is good with mechanics to look over to see if I got the math right and didn't miss anything obvious. It's the latest post in my blog "leapingleibniz.wordpress.com" [Broken] and it involves the derivation of an equation for how much force is required to push a wheelbarrow over a curb based only on angles and the force of gravity. Specifically I am wondering if I screwed up by ignoring the normal force. My reasoning for doing so is that I wanted to find the minimum force required to maintain static equilibrium with only the applied force causing torque up over the curb without the normal force contributing, and this would the minimum force required to push the wheel up over the curb. The normal force won't apply any torque once the wheel starts moving up off the ground, but this is still bugging me... anyway, any observations would be greatly appreciated!

Last edited by a moderator: May 6, 2017
2. Jul 2, 2012

### haruspex

I think you should at least acknowledge the normal force since without it these equations are simply wrong:
$\sum F_x = F\cos\gamma$
$\sum F_y = F\sin\gamma + f_g$
But since you then take moments about the point of contact, the normal force makes no contribution.
This equation is wrong:
$\sum \tau = rF \sin\gamma + rf_g \sin\phi - rF\cos\gamma\sin\phi$
But it comes out corrected in the following line. Should also explain why total torque should be set to 0.
Wrt the $\phi - \gamma$ term, you can see that in the geometry. Extend the line of F down and drop a perpendicular to it from the point of contact. The angle the line of F makes to the radius of contact is $\phi - \gamma$.
Of course, anyone with experience of negotiating kerbs with barrows will pull them up, not push them.

3. Jul 2, 2012

### Bennigan88

Ok, I fixed that line and explained why torque is set to zero. Man, this post was pretty disorganized.