The drawing shows three particles far away from any other...

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are $m_{\mathrm{A}}=363 \mathrm{~kg}_{0} m_{\mathrm{B}}=517 \mathrm{~kg}_{o}$ and $m_{\mathrm{C}}=154 \mathrm{~kg} .$ Find the magnitude and direction of the net gravitational force acting on (a) particle A, (b) particle $\mathrm{B},$ and (c) particle $\mathrm{C}$.

Key Concepts

Newton's Law of Universal Gravitation

This is the fundamental law that describes the gravitational force between two point masses. It states that every two masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This law provides the mathematical framework to calculate the magnitude of the gravitational force in any two-body scenario.

Superposition Principle

In gravitational problems involving more than two masses, the net gravitational force acting on an object is the vector sum of the individual forces exerted by each of the other masses. This principle allows us to analyze complex systems by breaking them down to pairwise interactions and then summing those forces to find the overall effect.

Vector Addition of Forces

Because gravitational forces are vector quantities, they have both magnitude and direction. When calculating the net force, it is essential to consider the direction of each gravitational force and perform vector addition, typically by resolving the forces into their components, to accurately determine both the magnitude and the direction of the resultant force.

Transcript

00:01 For this equation, i have three particles set at the distance is shown.

00:04 Mass of particle a is 363 kilograms.

00:08 The mass of particle b is 517 kilograms.

00:16 The mass of particle c is 154 kilograms.

00:23 We're asked to find the magnitude and direction of the net forces acting on it.

00:28 For this, we will need the universal gravitational constant g, which equals 6 .673 times 10 to the negative 11.

00:35 Newton meter squared per kilogram squared.

00:43 And the formula of force equals the universal gravitational constant times the mass of 1 times the mass of 2 divided by r squared.

00:56 For particle a, we'll find the force of beyond a, which equals g times m .a times mb over the distance between a and b squared.

01:15 Which equals 6 .673 times 10 to negative 11 times 363 times 517 divided by 0 .5 squared, which equals 50 .09 times 10 to the negative 6 newtons.

01:47 Next we'll find the force of c on day, which equals g times the mass of a times the mass of c divided by the distance between a and c squared.

02:02 Which equals 6 .673 times 10 to the negative 11 times 363 times 154 divided by 0 .5 plus 0 .25 squared, which equals 6 .632 times 10 to the negative 6 newtons.

02:32 Now to find the net force, we add the forces of b and a and the forces of c and a together, which equal 50 .09 plus 6.

02:47 0 .632 times 10 to the negative 6, which gives us a total net force of 56 .722 times 10 to the negative 6 newtons in the positive x direction...

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