# Doubts with levers

I'm studying for a test of static rigid body and I'm having doubts on how to solve problems involving levers with weight. If I have, for example, a lever 10 kilograms and 3 meters long and one support point at 1 meter from one end, how much force must be applied to the other side to achieve equilibrium?

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@Ron Maimon's answer is the most elegant and @ja72's is the most general... but there's a dirty "trick" you can do with levers: treat them as ideal leavers with an extra weight concentrated at the centre of mass. This works as long as you don't need to take bending into account.

If the density is $\rho(\ell)$ as a function of horizontal position $\ell$ then the torque is $$\tau = \int \mathrm{d}\ell\ \rho g \ell = M g \frac{\int \mathrm{d}\ell\ \rho \ell}{\int \mathrm{d}\ell\ \rho} = M g \ell_{\text{CM}}$$ where $\ell_{\text{CM}}$ is the centre of mass and $M$ is the total mass.

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For any statics problems you have these steps:

1. Create Free Body Diagrams (FBD) of all your bodies (parts) including any connection forces between bodies (one or two components as needed).
2. For each body write three equations

a) Sum of forces along x-axis equals zero

b) Sum of forces along y-axis equals zero

c) Sum of all moments taken about about any point equals zero

3. Count your equations and number of unknown force components and check that they match

4. Systematically solve the problem by eliminating the unknowns one by one, or compose the system as a matrix equation.

So if you follow these steps, you will do fine. See example in accepted answer to this question.

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The best way to do this is not to do a formal solution as ja72 says (although that works), but to use the principle of virtual work. Imagine the lever is displaced by an amount $\epsilon$. Then the work done by the two weights (their weight times their displacement) must be equal in order for the system to be in equilibrium. All masses which are distributed can be considered as pushing at the center of mass.

for your example, the 10kg lever can be imagined to be concentrated at the 1.5 m center, which is .5 m away from the fulcrum, which means that you need 5kg on the very end on the other side to balance it.

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