Question

Four identical masses of mass 600 kg each are placed at the corners of a square whose side lengths are 10.0 cm .

a) What is the magnitude of the net gravitational force on one of the masses, due to the other three?

b) What is the direction of the net gravitational force on one of the masses, due to the other three?:

1- toward the center of the square

2- outward the center of the square

Answer 1

Gravitational attraction in newtons
F = G m?m?/r²
G = 6.674e-11 m³/kgs²
m? and m? are the masses of the two objects in kg
r is the distance in meters between their centers
10 cm = 0.1 m

#1, at +x = 0.1?2 distance
F1 = G(900)² / (0.1?2)² = G(810000) / (0.02)
angle is 0º

#2 at distance 0.10
F2 = G(900)² / (0.10)² = G(810000) / (0.01)
angle is 45º

#3 at distance 0.10
F3 = G(900)² / (0.10)² = G(810000) / (0.01)
angle is –45º

Y components of 2 and 3 cancel, so all we have to do is add the x components
F = F1 + F2cos45 + F3cos45
F = F1 + 1.414 F2
F = G(810000)[ (1/0.02) + (1.414/0.01) ]
F = G(810000)(50+ 141.4)
F = 0.01034 newtons

Force will ne toward the center of the square.