Equilibrium Formula Derivation for Torque and Lever Systems

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The discussion focuses on deriving the equilibrium formula for torque and lever systems, specifically the equation Xo = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3). The user is struggling to understand how to apply the definitions of torque and lever arms to arrive at this formula. They mention the relationship m1(x1 - x0) + m2(x2 - x0) + m3(x3 - x0) = 0 as a starting point. Additionally, they provide specific values for the masses and their respective distances from the balance point, Xo. The conversation aims to clarify the meanings of the variables and how to manipulate the equations to find Xo.
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I have this homework question that I an having a lot of trouble with. I am not sure if anyone could help, but here it is:

Using the definitions of torque and lever arm and the fact that the system is in equilibrium, derive this formula-
Xo= m1x1 + m2x2 + m3x3
... m1 + m2+ m3

So, I know all the different formulas for torque and the levers etc... but I can't figure out how to turn them into this formula?
 
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Mind telling what x_0,x_1,x_2,x_3 are?
All I can tell is that the equation can be written as such:
m_1(x_1-x_0)+m_2(x_2-x_0)+m_3(x_3-x_0)=0

Maybe that helps...
 
That does help, here are some of the given equations in the chapter before the problem:

T1=m1gl1 T2=m2gl2 l1=Xo-X1 l2=Xo-X2 l3=Xo-X3

The numbers in the equation:
Xo= 121.8(.3000) + 71.40(6500) + 145.4(.5005)/ 121.8 + 71.40 + 145.4


121.8=m1(hanging mass from bar on the left) 71.40=m2(hanging mass on the right)
145.4=m3(mass from the center of gravity) and we are looking for Xo-the balance point in between m1X1 and m3X3 wih m2X2 on the right of m3X3

Does that help or make it more confusing?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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