Find the force acting on a wheel through moments

Question

A wheel of mass 20 kg and radius 80 cm is pulled by a horizontal force F against a step of height 20cm. Determine the magnitude of the initial force F so that the wheel just turns over the step.

Answer 1

I am pretty sure there is more than one way to solve this.  Here is my suggestion:

 

Calculate the total change in the energy of the wheel, get the moment of force from the change in energy, and finally, get the force from the moment and the dimensions.

 

Change in energy:

 

Assuming the change in energy of the wheel is just enough for the bottom of the wheel to reach the top of the step, we only need to find the change in potential energy U.

 

The wheel rises a height h = 20 cm = 0.20 m

 

U_i = 0

 

U_f = mgh

 

ΔU = U_f - U_i = mgh - 0 = mgh

 

Since it rotates by an angle θ due to a moment M, the work performed on the wheel is

 

W = Mθ

 

The work and the change in potential energy must balance.

 

W = mgh

 

Mθ = mgh

 

M = mgh/θ  [θ must be in radians]

 

This may be hard to see without a diagram (which I cannot insert in this window), but

 

h = R - Rcosθ ⇒

 

(R - h)/R = cosθ = (0.80 - 0.20)/(0.80) = 0.75

 

θ = cos^-1(0.75) ≅ 0.7227 radians

 

M = mgh/θ = (20 kg)(9.8 m/s²)(0.20 m)/(0.7227) ≅ 54.24 Nm

 

So far, we have the moment.  We still need the force F.  Assuming F is applied through the center of mass of the wheel, find the moment about the contact point between the wheel and the top of the step.

 

M = Fd = F(R - h) = F(0.80 - 0.20)m = (0.60 m)F

 

(0.60 m )F = 54.24 Nm

 

F = 54.24/0.60 N = 90.4 N