Question 7.3.1: The 100-kg steel wheel in Figure 1 has a radius of 50 mm and...

Goal Find the ramp angle θ at which the steel wheel will begin to roll down the wooden ramp.
Given The mass and radius of the steel wheel and the coefficient of rolling resistance between the steel and the wood.
Assume The wheel will roll and not slide down the ramp.
Draw We draw the free-body diagram of the wheel (Figure 2a). Figure 2b shows an alternate free-body diagram, where the weight W of the wheel is drawn in terms of its components (W sinθ and W cosθ ).
Formulate Equations and Solve Now we find the value of R (the force of the ramp pushing on the wheel) by enforcing equilibrium:

\sum{F_{y}\left(\uparrow + \right)}= – W + R= 0\Rightarrow R =W                  (1)
\sum{M_{z @ A} } \left(\curvearrowleft + \right)=- W \sin \theta \left(d_{1}\right) + W \cos \theta \left(d_{2}\right)=0                    (2)

where d_{1} and d_{2} are shown in Figure 2b. For small wheel deformations d_{1} ≈ r and d_{2} = a (as shown in Figure 7.3.2c). Therefore (2) can be rewritten as

– W \sin \theta \left(r\right) + W \cos \theta \left(a\right)=0            (3)

Rearrange (3), to

\frac{\sin \theta}{\cos \theta} =\tan \theta =\frac{a}{r}=f_{r}             (4)

by definition, the coefficient of rolling resistance is f_{r}   = a/r. With f_{r} = 0.02 we solve (4) for θ to find θ= 1.14°

θ = 1.14°

At θ = 1.14°, the steel wheel will roll down the ramp at a constant speed. (Note: At angles greater than 1.14°, the wheel will accelerate down the ramp. Also notice that the answer is independent of the weight of the wheel.)
Check If the wheel–ramp interface had larger rolling resistance, the value of a would increase. Substituting a larger value of a into (4) would result in a larger angle θ . This makes intuitive sense—the larger the rolling resistance, the greater the force required to roll the wheel. In this case this force is the component of the wheel’s weight parallel to the ramp, and this component gets bigger with ramp angle.