Question

Three identical very small 50-kg masses are held at the corners of an equilateral triangle, 0.30 m on each side. if one of the masses is released, what is its initial acceleration if the only forces acting on it are the gravitational forces due to the other two masses

Answer 1

Gravitational force between any two masses is given by

$F = \frac{Gm_1m_2}{r^2}$

$F = \frac{6.67*10^-11 * 50 * 50}{0.30^2}$

$F = 1.853 * 10^{-6} N$

Now net force on it due to two masses is given by

$F_{net} = 2Fcos(\frac{\theta}{2})$

here $\theta = 60 degree$

$F_{net} = 2*1.853 * 10^{-6}cos(30)$

$F_{net} = 3.21 * 10^{-6} N$

Now acceleration is given by

F= ma

$a = \frac{3.21*10^{-6}}{50}$

$a = 6.42*10^{-8} m/s^2$

Answer 2

If one of the masses is released, its initial acceleration is about 6.4 × 10⁻⁸ Newton

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Further explanation

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

$\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }$

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

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Given:

mass of the object = m = 50 kg

distance between the object = R = 0.30 m

Asked:

initial acceleration = a = ?

Solution:

Firstly , let's find the gravitational force between 2 objects as follows:

$F = G \frac{m_1m_2}{R^2}$

$F = 6.67 \times 10^{-11} \times \frac{ 50 (50)}{0.30^2}$

$F \approx 1.85 \times 10^{-6} \texttt{ Newton}$

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Next, we will calculate the resultant vector of these gravitational forces acting on a object:

$(\Sigma F)^2 = (F_1)^2 + (F_2)^2 + 2F_1F_2\cos \theta$

$(\Sigma F)^2 = F^2 + F^2 + 2F(F) \cos 60^o$

$(\Sigma F)^2 = 2F^2 + 2F^2(\frac{1}{2})$

$(\Sigma F)^2 = 3F^2$

$\Sigma F = \sqrt{3F^2}$

$\Sigma F = F\sqrt{3}$

$\Sigma F \approx 1.85 \times 10^{-6} (\sqrt{3})$

$\Sigma F \approx 3.2 \times 10^{-6} \texttt{ Newton}$

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Finally, we could calculate the initial acceleration of the object by using 2nd-Newton's Law of Motion as follows:

$a = \frac{\Sigma F}{m}$

$a \approx \frac{3.2 \times 10^{-6}}{50}$

$a \approx 6.4 \times 10^{-8} \texttt{ m/s}^2$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields