Question

Three balls A, B and C are kept in a straight line. The separation between A and C is 1 m, and B is placed at the midpoint between them. The masses of A, B, C are 100 g, 200 g and 300 g respectively. Find the net gravitational force on (a) A, (b) B, and (c) C​

Answer 1

Explanation:

Heya!!

Here is your Answer!!

Mass of A = 100 g= 0.1 kg.

Mass of B= 200g= 0.2 kg.

Mass of C =300 g=0.3 kg.

Distance between AC = 1m

Distance between AB= 1/2 m

Distance between BC=1/2 m

Since B is midpoint.

We know that by Newton's Law of Gravitation that:

F=GMm÷(R)^2

Therefore:

1.The net gravitational force on A:

F(AB)+F (AC)

G×0.1×0.2÷ (1/2)^2 + G ×0.1× 0.3÷ (1)^2

G×0.20

6.67×10^-11×0.20

=1.334×10^-11 N.

2.The net gravitational force on B:

F(BA)+F(BC)

=21.44 ×10^-11 N.

3.The net gravitational force on C:

F(AC) + F (BC)

=1.8009×10^-11 N.

Hope it helps you!!

Answer 2

$\Large\bf{\underline{\bf{\color{indigo}GiVeN,}}}$

$\bf\pink{We\: have,}$

• Three balls A, B & C are kept in a straight line.

• Separation between Ball A & C is 1 m.

$\longmapsto\:\:\bf\blue{AC\:=\:1\:m}$

• Ball B is placed at the mid point between ball A & C.

$\longmapsto\:\:\bf\purple{AB\:=\:BC=\:\dfrac{1}{2}\:m\:=\:0.5\:m}$

• Mass of ball A is 100 g.

$\longmapsto\:\:\bf\orange{m_A\:=\:100\:g\:=\:0.1\:kg}$

• Mass of ball B is 200 g.

$\longmapsto\:\:\bf\green{m_B\:=\:200\:g\:=\:0.2\:kg}$

• Mass of ball C is 300 g.

$\longmapsto\:\:\bf\red{m_C\:=\:300\:g\:=\:0.3\:kg}$

$\Large\bf{\underline{\bf{\color{cyan}To\:FiNd,}}}$

1. The net gravitational force on A.
2. The net gravitational force on B.
3. The net gravitational force on C.

$\Large\bf{\underline{\bf{\color{lime}CaLcUlAtIoN,}}}$

$\bf\red{Using\: the\: following\: relation,}$

$\pink\bigstar\:\:{\underline{\green{\boxed{\bf{\color{peru}F\:=\:G\:\dfrac{m_1\:m_2}{r^2}\:}}}}}$

$\bf\purple{Where,}$

• F is the gravitational force acting on body.

• G is the gravitational acceleration, i.e. 6.67 × 10⁻¹¹ m/s².

• m₁ & m₂ are the masses, where one mass acts force on other mass.

• r is the distance between two masses.

❥︎ See the attachment free body diagram for direction of forces act on other.

$\bf\pink{According\: to\: the\: diagram,}$

• $\bf{F_{AB}}$ is the force on A due to B.

• $\bf{F_{AC}}$ is the force on A due to C.

• $\bf{F_{BC}}$ is the force on B due to C.

• $\bf{F_{BA}}$ is the force on B due to A.

• $\bf{F_{CA}}$ is the force on C due to A.

• $\bf{F_{CB}}$ is the force on C due to B.

Cᴀsᴇ - 1 ;-

☯︎ Force acting on ball A is "(Force on A due to B + Force on A due to C)".

$:\implies\:\:\bf{F_A\:=\:G\:\dfrac{m_A\:m_B}{(AB)^2}\:+\:G\:\dfrac{m_A\:m_C}{(AC)^2}}$

$:\implies\:\:\bf{F_A\:=\:G\:\Big[\dfrac{m_A\:m_B}{(AB)^2}\:+\:\dfrac{m_A\:m_C}{(AC)^2}\Big]}$

$:\implies\:\:\bf{F_A\:=\:6.67\times{10^{-11}}\:\Big[\dfrac{0.1\times{0.2}}{0.5^2}\:+\:\dfrac{0.1\times{0.3}}{1^2}\Big]}$

$:\implies\:\:\bf{F_A\:=\:6.67\times{10^{-11}}\:\Big[\dfrac{0.02}{0.25}\:+\:\dfrac{0.03}{1}\Big]}$

$:\implies\:\:\bf{F_A\:=\:6.67\times{10^{-11}}\:(0.08\:+\:0.03)}$

$:\implies\:\:\bf{F_A\:=\:6.67\times{10^{-11}}\times{0.11}}$

$:\implies\:\:\bf\blue{F_A\:=\:7.34\times{10^{-12}}\:N}$

$\Large\bold\therefore$ ❶ The gravitational force on ball A is 7.34 × 10⁻¹² N.

Cᴀsᴇ - 2 ;-

☯︎ Force acting on ball B is "(Force on B due to C - Force on B due to A)".

[NOTE → Forces are in opposite direction.]

$:\implies\:\:\bf{F_B\:=\:G\:\dfrac{m_B\:m_C}{(BC)^2}\:-\:G\:\dfrac{m_B\:m_A}{(AB)^2}}$

$:\implies\:\:\bf{F_B\:=\:G\:\Big[\dfrac{m_B\:m_C}{(BC)^2}\:-\:\dfrac{m_B\:m_A}{(AB)^2}\Big]}$

$:\implies\:\:\bf{F_B\:=\:6.67\times{10^{-11}}\:\Big[\dfrac{0.2\times{0.3}}{0.5^2}\:-\:\dfrac{0.2\times{0.1}}{0.5^2}\Big]}$

$:\implies\:\:\bf{F_B\:=\:6.67\times{10^{-11}}\:\Big[\dfrac{0.06}{0.25}\:-\:\dfrac{0.02}{0.25}\Big]}$

$:\implies\:\:\bf{F_B\:=\:6.67\times{10^{-11}}\:(0.24\:-\:0.08)}$

$:\implies\:\:\bf{F_B\:=\:6.67\times{10^{-11}}\times{0.16}}$

$:\implies\:\:\bf\orange{F_B\:=\:1.067\times{10^{-11}}\:N}$

$\Large\bold\therefore$ ❷ The gravitational force on ball B is 1.067 × 10⁻¹¹ N.

Cᴀsᴇ - 3 ;-

☯︎ Force acting on ball C is "(Force on C due to A + Force on C due to B)".

$:\implies\:\:\bf{F_C\:=\:G\:\dfrac{m_C\:m_A}{(AC)^2}\:+\:G\:\dfrac{m_C\:m_B}{(BC)^2}}$

$:\implies\:\:\bf{F_C\:=\:G\:\Big[\dfrac{m_C\:m_A}{(AC)^2}\:+\:\dfrac{m_C\:m_B}{(BC)^2}\Big]}$

$:\implies\:\:\bf{F_C\:=\:6.67\times{10^{-11}}\:\Big[\dfrac{0.3\times{0.1}}{1^2}\:+\:\dfrac{0.3\times{0.2}}{0.5^2}\Big]}$

$:\implies\:\:\bf{F_C\:=\:6.67\times{10^{-11}}\:\Big[\dfrac{0.03}{1}\:+\:\dfrac{0.06}{0.25}\Big]}$

$:\implies\:\:\bf{F_C\:=\:6.67\times{10^{-11}}\:(0.03\:+\:0.24)}$

$:\implies\:\:\bf{F_C\:=\:6.67\times{10^{-11}}\times{0.27}}$

$:\implies\:\:\bf\green{F_C\:=\:1.8009\times{10^{-11}}\:N}$

$\Large\bold\therefore$ ❸ The gravitational force on ball C is 1.8009 × 10⁻¹¹ N.