Question

find net gravitional force on the object of the mass​

Answer 1

Answer:

5gm^2/r^2

Explanation:

Answer 2

Answer:

Gravitational Force:-

Gravitational force acts on masses due to the gravitational attraction between them. This force is a function of the masses and the distance that separate them. When multiple masses are interacting together, the total gravitational force on an object is equal to the sum of the gravitational forces acting on it between it and other masses.

Answer and Explanation: -

To calculate the force between two masses, we use the equation:

{eq}F = \dfrac{Gm_1m_2}{r^2}{/eq}, where:

1.  G is the gravitational constant, {eq}(6.67\times 10^{-11}\frac{m^3}{kg\cdot s^2}){/eq}

2. {eq}m_1 \text{ and }m_2{/eq} are the masses of the objects in question

3. r is the distance between the mass centers

The direction of the force will always point toward the larger of the two masses. Let's find the forces between A and B, and between A and C. If we take forces to the right as "positive" direction and forces to the let as "negative" direction, we can find the direction of the net force at the same time as we find the magnitude. We will answer both parts of the question together since they go hand in hand.

{eq}F_{A-B} = \dfrac{Gm_Am_B}{r_{A-B}^2}\\ F_{A-B} = \dfrac{(6.67\times 10^{-11}\frac{m^3}{kg\cdot s^2})(5\:kg)(15\:kg)}{(0.40\:m)^2}\\ F_{A-B} = 3.13 \times 10^{-8}\:N\\{/eq}

Between forces A and B, the force will point toward mass B (positive) due to it being the larger mass. Let's follow the same calculations between masses A and C.

{eq}F_{A-C} = \dfrac{Gm_Am_C}{r_{A-C}^2}\\ F_{A-C} = \dfrac{(6.67\times 10^{-11}\frac{m^3}{kg\cdot s^2})(5\:kg)(10\:kg)}{(0.10\:m)^2}\\ F_{A-C} = -3.34 \times 10^{-7}\:N\\{/eq}

Because C is the larger of the two masses, the force points to C (negative). Now let's add the two forces together to find the total force acting on mass A.

{eq}\Sigma F = F_{A-B} + F_{A-C}\\ \Sigma F = 3.13 \times 10^{-8}\:N - 3.34 \times 10^{-7}\:N\\ \Sigma F = -3.03 \times 10^{-7}\:N{/eq}

Because the final answer is negative, this means the force points to mass C. So the final magnitude and direction is {eq}3.03 \times 10^{-7}\:N {/eq} toward mass C.

Explanation: