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What is the magnitude of the net gravitational force on the mass at the origin due to the other two masses?
Three identical masses of 640 kg each are placed on the x axis. One mass is at x1 = -10.0 cm, one is at the origin, and one is at x2 = 36.0 cm. Take the gravitational constant to be G = 6.67×10−11 N * m^2/kg^2.
Can someone solve for this and show me how they arrived at the answer? Thank you.
2 Answers
- Captain MephistoLv 71 decade agoFavorite Answer
Gravitational force = G*M*m/r^2
Since the mass on the left is closer to the center mass, the net force will point towards the left.
Since all the masses are the same I will call it m and in the above equation we have M = m.
G = 6.67x10^(-11) M m^2 / kg^2
m = 640 kg
x1 = 10 cm = 0.1 m
x2 = 60 cm = 0.6 m
Fnet = G*m^2[1/x1^2 - 1/x2^2] = G*m^2[100 - 100/36] = G*m^2[3500/36]
Fnet = 1.891x10^(-3) N and it is directed towards the mass on the left
- Anonymous4 years ago
Use the Newtonian gravitation equation: F = G * m1 * m2 / r^2. F1 = G * 470 kg * 470 kg / (-0.13 m)^2 = 8.7 x 10^-4 N. this is directed leftward, so supply it a unfavourable sign: -8.7 x 10^-4 N. F2 = G * 470 kg * 470 kg / (+0.30 m)^2 = a million.6 x 10^-4 N. this is directed rightward, so supply it a good sign: a million.6 x 10^-4 N. upload the two: F1+F2 = (-8.7 + a million.6) x 10^-4 N = -7.a million x 10^-4 N. this is, 0.71millinewtons, directed leftward, is the consequent.