# [Physics] Minimum required to lift a roller up a step

homework-and-exercisesnewtonian-mechanics

Consider a roller of diameter $$0.82\ \mathrm m$$ which is to be raised up a step measuring $$0.32\ \mathrm m$$. Find the minimum force required to do so. The roller weighs $$500\ \mathrm N$$.

The roller could be considered as a point mass $$F$$, and its body could be considered a lever with $$F_\text{applied}$$ acting at $$G$$ and $$F_\text{perpendicular}$$ acting perpendicular to diameter $$GC$$.

Let $$\angle FCH = \theta \implies \angle IFG = \theta$$.

$$F_\text{perpendicular} = mg\sin\theta$$.

After that's its a simple application of torque and moments. I'm not quite sure if this resolution of vectors into components is quite the right approach. At least, the value given in my textbook doesn't correspond to the one I got.

EDIT: In response to the comment below, I decided to post my workings:

$$\sin\theta = \frac{FH}{FC} = \frac{0.09\ \mathrm m}{0.41\ \mathrm m} = 0.2195 \implies F_\text{perpendicular} = 500\ \mathrm N\cdot 0.2195 = 109.756\ \mathrm N$$

The anticlockwise torque equals $$\tau_\text{anti-clockwise} = 109.756\ \mathrm N\cdot 0.41\ \mathrm m = 45\ \mathrm{Nm}$$

This has to equal the clockise torque $$\tau_\text{clockwise} = F_\text{applied}\cdot 0.82\ \mathrm m$$

$$\implies F_\text{applied} = \frac{45\ \mathrm{Nm}}{0.82\ \mathrm m} = \boxed{54.878\ \mathrm N}$$

According to the book, the answer is $$110\ \mathrm N$$. Maybe the book is incorrect?

I am afraid I have some problems with your solution. If your force is applied at G, it seems that the roller will slip at the point where it touches the upper corner of the step (C), as the force with which the step acts on the roller passes through the center of the roller (if there is no friction), so there is nonzero torque with respect to the center of the roller. Therefore, the force should be applied at the center of the roller, the shoulder of the lever will be half of what you have, therefore, the force will be twice as large as what you have, i.e., 110N - the answer from the book. However, there seems to be another problem with your solution: I would say you should multiply by $\cos \theta$, not by $\sin \theta$, to get $F_{perpendicular}$, so I have my doubts about the answer in the book as well.