Question

A mass m is located at the origin; a second mass m is at x = d. A third mass m is above the first two so the three masses form an equilateral triangle. What is the net gravitational force on the third mass?

Answer 1

Answer:

√3 * Gm²/d²

Explanation:

m1 = m, m2= m, distance = d. hence:

F = Gm²/d²

Let the origin be A, the point x = d be B and the point above the first two is C.

The net force acting on the third mass (point C) $F_{net}$ = $F_A+F_B$

Let j represent the vertical component and i the horizontal component. Hence:

$F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}$

Answer 2

Answer:

• The net gravitational force on the third mass = $3^{0.5} * \frac{Gm^2}{d^2}$

Explanation:

For equilateral triangle,

$\theta = 30^o$

Force between masses,

$F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}$

Therefore,

$F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}$

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