Gravitational Force between three forces

AI Thread Summary
The discussion centers on calculating the net gravitational force on a 20kg ball influenced by two 10kg balls positioned 20cm above it, with one ball 5cm to the right and the other 5cm to the left. The gravitational force formula used is F=GMm/R^2, with the calculated distance R being approximately 0.206m. The net force in the x-direction is zero due to symmetry, while the net force in the y-direction is calculated to be about 6.1E-7 N. There is a discrepancy noted, as the expected answer is 3E-7 N, leading to a suggestion that the textbook may be incorrect. The conversation emphasizes the importance of verifying calculations and the potential for errors in educational materials.
DRC12
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1. Homework Statement [/b]
There's a 20kg ball and two 10kg balls 20cm above it. One of the 10kg balls is 5cm to the right of center and the other is 5cm to the left of center. What is the net gravitational force on the 20kg ball

2. Homework Equations [/b]
F=GMm/R23. The Attempt at a Solution [/b]
M=20kg, m=10kg, G=6.67E-11 and R=√(.202+.052)=.206m
\SigmaFx=0; because the 10kg ball are equal distances away from the 20kg ball along the x-axis
\SigmaFy=2*(((6.67*10-11*10*20)/(.2062))*(.2/.206))=6.1*10-7
I know the answers supposed to be 3*10-7
 
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Your work looks correct to me.
 
All right thanks I'm just going to assume the books wrong it wouldn't be the first time
 
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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