Question

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 340 kg, mB = 567 kg, and mC = 139 kg. Take the positive direction to be to the right. Find the net gravitational force, including sign, acting on (a) particle A, (b) particle B, and (c) particle C.

Answer 1

Formula of the gravitational force between two particles:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}$ is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

$F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle C on particle A is

$F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N$ to the right

So the net gravitational force on particle A is

$F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N$ to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

$F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N$ to the left

The gravitational force exerted by particle C on particle B is

$F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N$ to the right

So the net gravitational force on particle B is

$F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N$ to the right

(c) Particle C

The gravitational force exerted by particle A on particle C is

$F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N$ to the left

The gravitational force exerted by particle B on particle C is

$F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N$ to the left

So the net gravitational force on particle C is

$F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N$ to the left

Answer 2

Particle A,B, and C is  $\rm \bold{ 5.6 \times 10^-^5N}$,$\rm \bold{ 3.27 \times 10^-^5N}$, $\rm \bold{ -8.97 \times 10^-^5N}$ respectively. Negative sign represents left direction.

The gravitational force between two bodies

$\rm \bold { F= G\frac{m^1 m^2}{r^2} }$

Where,

G- gravitational constant = $\rm \bold{6.67\times 10^1^1 Nm^2kg^-^2 }$

$\rm \bold { {m^1 m^2} }$ = mass of bodies

r - is distance between them

Net gravitational force on Particle A

$\rm \bold{F_a = F_b+ F_c}$

The gravitational force exerted by particle B on particle A is $\rm \bold{ 5.14 \times 10^-^5N}$ to the right .

The gravitational force exerted by particle C on particle A is $\rm \bold{ 5.6 \times 10^-^6N}$ to the right.

Hence, net Gravitational force on A is $\rm \bold{ 5.6 \times 10^-^5N}$

Net gravitational force on Particle B

$\rm \bold{F_b = F_a+ F_c}$

The gravitational force exerted by particle A on particle B is $\rm \bold{ -5.14 \times 10^-^5N}$ on to the left.

The gravitational force exerted by particle C on particle B is $\rm \bold{ 8.41 \times 10^-^5}$ to the right.

Hence net  gravitational force on particle B will be $\rm \bold{ 3.27 \times 10^-^5N}$

Net gravitational force on Particle C is

$\rm \bold{F_c = F_a+ F_b}$

The gravitational force exerted by particle A on particle C is $\rm \bold{ -5.6 \times 10^-^6N}$ to the left.

The gravitational force exerted by particle B on particle C is $\rm \bold{ -8.41 \times 10^-^5N}$ to the left.

Hence,  net gravitational force on particle C is $\rm \bold{ -8.97 \times 10^-^5N}$ to the left.

Therefore we can conclude that particle A,B, and C is  $\rm \bold{ 5.6 \times 10^-^5N}$,$\rm \bold{ 3.27 \times 10^-^5N}$, $\rm \bold{ -8.97 \times 10^-^5N}$ respectively.

To know more about Gravitational force, refer to the link:

https://brainly.com/question/12528243?referrer=searchResults