Question

✪ Pʜʏsɪᴄs Qᴜᴇsᴛɪᴏɴ ✪

➪ Figure shows three blocks in contact and kept on a smooth horizontal surface. What is ratio of the force exerted by block A on B to that of B on C.
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Answer 1

QUESTION :

Figure shows three blocks in contact and kept on a smooth horizontal surface. What is the ratio of force exerted by block A on B to that of B on C.

ANSWER:

• Ratio of force exerted by block A on B to that of B on C = 3 : 1.

GIVEN:

• Three blocks in contact and kept on a smooth horizontal surface.

TO FIND:

• Ratio of force exerted by block A on B to that of B on C.

EXPLANATION:

Force exerted on the system = 16 N.

$\boxed{\bold{\large{\gray{Force = ma}}}}$

F = 16 N

m = 5 + 2 + 1

m = 8 kg

16 = 8a

a = 2 m/s²

∴ Acceleration of the system = 2 m/s²

By free body diagram,

$\sf F - F_{BA} = m_A \times a$

F = 16 N

$\sf m_A = 5\ kg$

a = 2 m/s²

$\sf 16 - F_{BA} = 5(2)$

$\sf - F_{BA} = 10 - 16$

$\sf F_{BA} = 6 \ N$

$\sf |F_{BA}|=|F_{AB}|$

$\sf F_{AB} = 6\ N$

By free body diagram,

$\sf F_{AB} - F_{CB} = m_B\times a$

$\sf F_{AB} = 6 \ N$

$\sf m_B = 2 \ kg$

a = 2 m/s²

$\sf 6 - F_{CB} =2(2)$

$\sf - F_{CB} =4 - 6$

$\sf F_{CB} =2\ N$

$\sf |F_{CB}| =|F_{BC}| =2\ N$

$\sf\dfrac{F_{AB}}{F_{BC}} = \dfrac{6}{2}$

$\sf\dfrac{F_{AB}}{F_{BC}} = \dfrac{3}{1}$

$\sf F_{AB}:F_{BC}= 3:1$

Hence the ratio of force exerted by block A on B to that of B on C = 3 : 1.

Answer 2

$\mathcal{\gray{\underbrace{\blue{GIVEN:-}}}}$

• Three blocks in contact and kept on a smooth horizontal surface .

• Force acts on Block A is 16N .

$\mathcal{\gray{\underbrace{\blue{TO\: FIND:-}}}}$

• The ratio of the force exerted by block A on B to that of B on C .

$\mathcal{\gray{\underbrace{\blue{SOLUTION:-}}}}$

✍️ See the attachment figure .

⚡Total mass of the three block = (5 + 2 + 1)kg

=> Total mass of the three block = 8kg

⚡And force acting on the block A is 16N .

Hence,

$\orange\bigstar\:\rm{\pink{\boxed{\purple{Acceleration\:=\:\dfrac{Force}{Mass}\:}}}}$

$\rm{\implies\:Acceleration\:(a)\:=\:\dfrac{16}{8}\:}$

$\rm{\green{\implies\:Acceleration\:=\:2\:m/s^2\:}}$

Let,

⚡ Force exerted by block A on B is $\rm{\underline{\red{\bold{F_{AB}}}}}$

⚡ And force exerted by B on C is $\rm{\underline{\red{\bold{F_{BC}}}}}$ .

Also,

⚡ Force exerted by block B on A is $\rm{\underline{\red{\bold{F_{BA}}}}}$

⚡And force exerted by C on B is $\rm{\underline{\red{\bold{F_{CB}}}}}$ .

✍️ We have know that,

$\mathcal{\green{F_{AB}\:=\:F_{BA}\:}}$ And $\mathcal{\green{F_{BC}\:=\:F_{CB}\:}}$

Now, apply Newton's second law,

$\rm{F\:-\:F_{BA}\:=\:m_A\times{a}}$

$\rm{\implies\:16\:-\:F_{BA}\:=\:5\times{2}}$

$\rm{\implies\:F_{BA}\:=\:16\:-\:10}$

$\rm{\implies\:F_{BA}\:=\:6\:N}$

$\rm{\boxed{\pink{F_{BA}\:=\:F_{AB}\:=\:6\:N\:}}}$

Again,

$\rm{F_{AB}\:-\:F_{CB}\:=\:m_B\times{a}\:}$

$\rm{\implies\:6\:-\:F_{CB}\:=\:2\times{2}\:}$

$\rm{\implies\:F_{CB}\:=\:6\:-\:4\:}$

$\rm{\implies\:F_{CB}\:=\:2\:N}$

$\rm{\boxed{\pink{F_{CB}\:=\:F_{BC}\:=\:2\:N\:}}}$

$\red\therefore$ $\rm{\green{\dfrac{F_{AB}}{F_{BC}}\:=\:\dfrac{6}{2}\:}}$

$\rm{\green{\implies\:\dfrac{F_{AB}}{F_{BC}}\:=\:\dfrac{3}{1}\:}}$

$\rm\red{\therefore}$$<font color=baby>$ The ratio of force exerted by block A on B to that of B on C is “3:1” .