Physics, 07.07.2019 16:30 kiasia24

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.

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Answer from: ciel8809

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

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 from: caylah6101

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:

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Answer from: Quest

It may possibly be c i think |-/

Answer from: Quest

Ithink the answer is c but then again i'm probably wrong

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