What Force is Needed to Wheel Over an Obstacle?

[Litcyb]

Homework Statement

what magnitude of force applied horizontally at the axle of the wheel is necessary to wheel over an obstacle of height h = 0.108 m? The wheel's radius is r = 0.698 m and its mass is m = 1.71 kg.

Homework Equations

I know that in order to have a static equilibrium we must have the total amount of forces and the total amount of torques(at any point) equal zero.
ƩF=0
Ʃτ=0

The Attempt at a Solution

I have no idea how to solve this problem since we have a circular object. do we apply angular momentum? I am really confused on how to solve this problem! :( please help.

 

[Doc Al]

The torque exerted by the applied force must overcome the torque due to the weight of the object. Hint: Take the top of the [STRIKE]object[/STRIKE] obstacle as your pivot point.

(Edit: Corrected typo. I meant obstacle, but wrote object!)

 

[Litcyb]

Why are we making the pivot point at the top of the circle, wouldn't it make more sense to put it at the corner since we don't know the friction giving by the corner? because if we put the pivot point at the corner we would just need to calculate the torques with respect to the corner.

We have two forces acting at the center of mass ; gravity( Mg) and F(the force being exerted)
now, we would just calculate the lever arm for each force acting upon the object in respect to the pivot point(corner).

Then we calculate the lever arm of Mg and F, in which they have to be perpendicular to the axis of the pivot point.

Am I doing the right steps here?

 

[Doc Al]

Why are we making the pivot point at the top of the circle,

Who said anything about putting the pivot at the top of the circle? (Ah... when I said top of the "object" I meant to say top of the "obstacle". My bad! )

wouldn't it make more sense to put it at the corner since we don't know the friction giving by the corner?

Imagining the obstacle as a step with a corner (of the given height), then that is exactly where the pivot should be.

We have two forces acting at the center of mass ; gravity( Mg) and F(the force being exerted)
now, we would just calculate the lever arm for each force acting upon the object in respect to the pivot point(corner).

Sure.

Then we calculate the lever arm of Mg and F, in which they have to be perpendicular to the axis of the pivot point.

The lever arm for a particular force is the perpendicular distance between the line of the force and the pivot.

Am I doing the right steps here?

I think so.

 

[Litcyb]

Thank you so much for reassuring me. I got the correct answer. the lever arm for force mg is (2rh-h^2))^1/2 and the lever arm for force F is (r-h)

Thus, the sum of all torques Ʃτ= τF-τmg= 0 => τF=τmg
= F(h-r)=mg(2hr-h^2)^1/2
F= [mg(2hr-h^2)^1/2]/(r-h)

Thanks!