Three 55.5kg masses are at the corners of an equilateral triangle and located in

Question

Three 55.5kg masses are at the corners of an equilateral triangle and located in space far from any other masses.

(a) If the sides of the triangle are 14.0m long, find the magnitude of the net force F1 exerted on each of the three masses. Enter your answer with four significant figures.

(b) If the distance between the masses is increased by a factor of 17, how will the new net force compare to the initial net force? Express your answer as the ratio F2F1; simplify your answer as far as possible.

Please explain the answer. Thank you in advance

Explanation / Answer

force between two bodies is F = Gm1 m2/r^2

F = 6.67 *10^-11 * 55.5 *55.5/(14^2)

F = 1.05 *10^-9 N

due to equilateral triangle net force acts at 60 deg

Fnet^2 = F1^2 + F2^2 + 2F1 F2 cos 60

Fnet^2 = (1.05*10^-9)^2 + (1.05*10^-9)^2 + (2 * 1.05*10^-9 * 1.05*10^-9* cos 60)

Fnet = 1.81 *10^-9 N

-------------------------

if distance is increased by 17, force decreases by 17^2 = 289 times

so Fnew - 1.81 *10^-9/289

Fnew = 6.26 *10^-11 N

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