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Please show all steps. Thank you.
Constants Part A Four identical masses of mass 5.00 kg each are placed at the corners of a square
whose side lengths are 3.00 m What is the magnitude of the net gravitational force on one of the
masses, due to the other three? Express your answer with the appropriate units aal
Solution
The expression for the gravitational force of attraction between the two masses in Newtons is -
F = G m?m?/r²
where,
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
#1, at +x = 3?2 meter distance
F1 = G(5)² / (3?2)² = 1.39*G
angle is 0º
#2 at distance 1 meter
F2 = G(5)² / (3)² = 2.78*G
angle is 45º
#3 at distance 1 meter
same as F2 but 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 = 1.39*G + 1.414*2.78*G = 5.32*G = 5.32*6.674*10^-11 = 3.55 x 10^-10 N