# Rotating a lever — small end has more force but less speed?

Suppose I have a massless lever with two masses at the ends and a fixed pivot as depicted below. Assume no gravity in this scenario.

Further, let's say object $$B$$ is closer to the pivot, so $$r. Now let us suppose a force $$\vec{F}_{1}$$ perpendicular to the lever and of constant magnitude is applied to object $$A$$ for some time to supply a torque. By the law of the lever, a force of magnitude $$F_{2} = F_{1}\frac{R}{r}$$ is applied to object $$B$$. Since $$r < R$$, it must be that the force applied to $$B$$ is of greater magnitude than that of the force on $$A$$. However, both objects are attached to the same lever, so they must have the same angular velocity and in particular $$v_{B} = v_{A}\frac{r}{R}.$$ Thus the speed of $$B$$ must be less than the speed of $$A$$.

It seems that $$B$$ received a greater force, yet it is moving with less speed. This seems like an apparent contradiction. Is there a way to resolve this?

I am aware that there is centripetal force that needs to be considered in a thorough analysis. I am assuming the pivot keeps the entire system in place as it rotates.

There are two errors here.

First, $$F_{2} = F_{1}\frac{R}{r}$$ is about the balancing of two torques in static equilibrium. That is, the usually case where this holds is where $$F_1 R$$ is one torque, and $$F_2 r$$ is the other, and if there's no net torque then they cancel each other out and are equal in value, which produces the equation. But this doesn't generally hold in all circumstances. In fact, the fact that your system would speed up shows that there's definitely a net torque and the equation doesn't hold.

Second, scaling the velocities with the forces, is intuitive, but incorrect. In fact, they usually scale in the other direction, i.e. inversely.

To see the inverse scaling consider the work put into lifting a car with a lever. Then the work is the force*distance, so you move the small force (your muscles) a large distance, which moves the large force (the car) a small distance. But you can also think of the velocities and the power (work/time), to see that the small force end of the lever moves fast, and the large force end of the lever moves slowly. That is, inversely related: the large force side moves slowly.

• For the first mistake, I think you are right, but I'm not sure the second mistake is a mistake. I'm not sure what you meant by "scaling with the forces", but the point is that they scaled inversely wrt radius: $v_{B}/r = \omega = v_{A}/R \implies v_{B} = v_{A}\cdot r/R$. Jun 14 at 3:18
• What I mean by "scaling with the forces" is related to where you say, "It seems that B received a greater force, yet it is moving with less speed. This seems like an apparent contradiction." That is, you seem to think the velocity should scale in direct proportion to the force. But it doesn't in the normal lever case, in fact, there the velocity scales in inverse proportion to the force (to conserve work/energy/power/etc). Jun 14 at 3:26
• I see. Those do seem like silly mistakes. Jun 16 at 1:56

What accelerates object A is not the force $$F_1$$ that you applied to it, it's the net force that it experiences.

$$F_1$$ is one component of the net force, but there is also a reaction force from the lever that you need to consider.

Rather than calculate this reaction force, the usual way to solve the problem is to find the moment of inertia of the two joined masses, and the torque you are applying at object A, and from those calculate the angular acceleration of the lever and masses.

If you must know the net force on object A, you can calculate it from its angular acceleration, lever arm, and mass.