Gravitational Force of Three Identical Masses

AI Thread Summary
To find the net gravitational force on the mass at the origin from the other two identical masses, the gravitational force formula F_grav = GMm/R^2 is used, where G is the gravitational constant. The user calculated the forces from each mass but is confused about whether to use the total mass of 1650 or the individual mass of 550 for calculations. The calculated forces were -0.00202 for the mass at X1 and 0.00109 for the mass at X2. The user suspects an error in squaring the mass and seeks clarification on how to properly account for the masses in the net force calculation. Correctly applying the gravitational force formula with the individual mass of 550 will yield the accurate net force.
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Homework Statement



Three identical masses of 550 each are placed on the x axis. One mass is at X1= -10.0 , one is at the origin, and one is at X2= 43.0 .
What is the magnitude of the net gravitational force on the mass at the origin due to the other two masses?

G = 6.673 x 10-11

Homework Equations



I'm using just the regular force of grav.

F_grav= GMm/R^2

Net force:
F_net= F1-F2

and I tried Gm^2(1/X1^2-1/X2^2)

I'm getting logical numbers, but they're not right.

I calculated Force1 to be -0.00202
I calculated Force2 to be 0.00109

I think I may have squared the mass to get the these numbers...what do I do with the masses? there are three of them, do I use 1650? or 550?
 
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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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