Rotational Dynamics and torque

[Destrio]

What minimum force F applied horizontally at the axle of the wheel is necessary to raise the wheel over an obstacle of height h? Take r as the radius of the wheel and W as its weight.

We know torque is equal to the radius x perpendicular force
T = rFnet

I think I have to find angular acceleration, α.

T = rma
T = rmα
T = rm(αr)
T = α(mr^2)
ΣT = αΣ(mr^2)

am I on the right track so far?
How do I incorporate height into this, do I want to find angular acceleration in the y direction?

 

[learningphysics]

Do they give a picture? If not draw a picture... Draw a wheel standing against a block of height h.

the point where the block and the wheel touch... you want the torque about the point to be zero...

you have two forces involved for this torque... the weight and the horizontal applied force.

 

[ogb p]

Yes, you are on the right track. To incorporate the height into the equation, you can consider the torque required to lift the wheel over the obstacle. This torque would be equal to the weight of the wheel multiplied by the height of the obstacle, since the force required to lift the wheel is equal to its weight. So the equation would become:

T = rFnet + Wh

Since we want to find the minimum force required, we can set the net force equal to zero (assuming the wheel is just barely lifted over the obstacle). This gives us:

0 = rFmin + Wh

Solving for Fmin, we get:

Fmin = - Wh/r

So the minimum force required to raise the wheel over the obstacle would be equal to the weight of the wheel multiplied by the height of the obstacle, divided by the radius of the wheel. This makes intuitive sense, as a larger obstacle or smaller wheel radius would require a greater force to lift the wheel.