Formula of the gravitational force between two particles:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}[/tex] 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
[tex]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[/tex] to the right
The gravitational force exerted by particle C on particle A is
[tex]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[/tex] to the right
So the net gravitational force on particle A is
[tex]F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N[/tex] to the right
(b) Particle B
The gravitational force exerted by particle A on particle B is
[tex]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[/tex] to the left
The gravitational force exerted by particle C on particle B is
[tex]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[/tex] to the right
So the net gravitational force on particle B is
[tex]F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N[/tex] to the right
(c) Particle C
The gravitational force exerted by particle A on particle C is
[tex]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[/tex] to the left
The gravitational force exerted by particle B on particle C is
[tex]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[/tex] to the left
So the net gravitational force on particle C is
[tex]F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N[/tex] to the left
Particle A,B, and C is [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex],[tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex], [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] respectively. Negative sign represents left direction.
The gravitational force between two bodies
[tex]\rm \bold { F= G\frac{m^1 m^2}{r^2} }[/tex]
Where,
G- gravitational constant = [tex]\rm \bold{6.67\times 10^1^1 Nm^2kg^-^2 }[/tex]
[tex]\rm \bold { {m^1 m^2} }[/tex] = mass of bodies
r - is distance between them
Net gravitational force on Particle A
[tex]\rm \bold{F_a = F_b+ F_c}[/tex]
The gravitational force exerted by particle B on particle A is [tex]\rm \bold{ 5.14 \times 10^-^5N}[/tex] to the right .
The gravitational force exerted by particle C on particle A is [tex]\rm \bold{ 5.6 \times 10^-^6N}[/tex] to the right.
Hence, net Gravitational force on A is [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex]
Net gravitational force on Particle B
[tex]\rm \bold{F_b = F_a+ F_c}[/tex]
The gravitational force exerted by particle A on particle B is [tex]\rm \bold{ -5.14 \times 10^-^5N}[/tex] on to the left.
The gravitational force exerted by particle C on particle B is [tex]\rm \bold{ 8.41 \times 10^-^5}[/tex] to the right.
Hence net gravitational force on particle B will be [tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex]
Net gravitational force on Particle C is
[tex]\rm \bold{F_c = F_a+ F_b}[/tex]
The gravitational force exerted by particle A on particle C is [tex]\rm \bold{ -5.6 \times 10^-^6N}[/tex] to the left.
The gravitational force exerted by particle B on particle C is [tex]\rm \bold{ -8.41 \times 10^-^5N}[/tex] to the left.
Hence, net gravitational force on particle C is [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] to the left.
Therefore we can conclude that particle A,B, and C is [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex],[tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex], [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] respectively.
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