Finding unknown mass using a lever system

AI Thread Summary
In the lever system problem, the goal is to find the unknown mass m2 using known masses m1 and m3 while maintaining equilibrium. The equations of static equilibrium were set up, involving the lever's mass (M) and the distance of the center of mass (x) from the pivot. The key insight is that the torque generated by the lever's mass can be ignored when calculating the relationship between the known and unknown masses, simplifying the equations. The final expression for m2 is derived as m2 = sqrt(m1 * m3), demonstrating that the center of mass does not need to be included in the final calculations. Understanding this allows for a clearer path to solving the problem without complicating the torque considerations.
Oldblood
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Homework Statement


We have a lever system. The end of the left side is a distance w1 from the pivot. The end of the right side is a distance w2 from the pivot. When no masses are attatched the lever is touching the ground on the left side. We place a known mass called m1 on the far left side and an unknown mass m2 on the far right side. The system is now in equilibrium. Then we remove the masses and place the mass m2 on the far left side and a know mass m3 on the right side. The system is again in equilibrium. Find an expression for the unknown mass m2 expressed by the known masses m1 and m3.


Homework Equations


Equation for static equilibrium.


The Attempt at a Solution


m1g*w1+Mgx=m2g*w2

m2g*w1+Mgx=m3g*w2

M is the mass of the lever and x is the distance the center of mass is from the pivot. Subtracting the frist equation from the 2nd let's me get rid of Mgx but I can't seem to get rid of w1 and w2. The answer is supoused to be m2=sqrt (m1*m3). I've notice that this is the answer you get if you ignore Mgx from the equations, so what I don't understand is why you can do that. Mgx is generating a torque and I don't understand why you can just say that m1g*w1=m2g*w2 and m2g*w1=m3g*w2

Any help is appreciated
 
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Since you are accounting for the individual weights, it would be redundant (and incorrect) to also account for the CM.
 
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