Three masses that are 10 kg each are located at the corners of an equilateral triangle,...

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Three masses that are 10 kg each are located at the corners of an equilateral triangle,...

Three masses that are 10 kg each are located at the corners of an equilateral triangle, with side length of 0.55 m. What is the magnitude of the total force acting on one mass due to the other two masses?

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Answer 1

Answer #1

Say we have an upright equilateral triangle, and we want the force on the topmost particle.

Note that the force of interaction on each 2 masses is

F = G m1 m2 / s^2 = 2.2049*10^-8 N

The force along the horizontal cancels, and only the vertical component matters. The vertical component of this force is

Fy = 1.9095*10^-8 N

Two such forces act, thus,

Fnet = 3.82*10^-8 N [ANSWER]

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Answer 2

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