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Question Number 23399 by Tinkutara last updated on 29/Oct/17

Two blocks of masses 2 kg and 3 kg are kept on a smooth inclined plane. A constant force of magnitude 20 N is applied on 2 kg block parallel to the inclined. The contact force between the two blocks is

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{3}\:\mathrm{kg} \\ $$$$\mathrm{are}\:\mathrm{kept}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{inclined}\:\mathrm{plane}. \\ $$$$\mathrm{A}\:\mathrm{constant}\:\mathrm{force}\:\mathrm{of}\:\mathrm{magnitude}\:\mathrm{20}\:\mathrm{N}\:\mathrm{is} \\ $$$$\mathrm{applied}\:\mathrm{on}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{inclined}.\:\mathrm{The}\:\mathrm{contact}\:\mathrm{force}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{blocks}\:\mathrm{is} \\ $$

Commented by Tinkutara last updated on 29/Oct/17

Commented by mrW1 last updated on 29/Oct/17

F_(12) =(M/(m+M))×F=(3/(2+3))×20=12 N

$$\mathrm{F}_{\mathrm{12}} =\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}×\mathrm{F}=\frac{\mathrm{3}}{\mathrm{2}+\mathrm{3}}×\mathrm{20}=\mathrm{12}\:\mathrm{N} \\ $$

Commented by Physics lover last updated on 29/Oct/17

Can you plz explin your method sir.

$${Can}\:{you}\:{plz}\:{explin}\:{your}\:{method} \\ $$$${sir}. \\ $$

Commented by mrW1 last updated on 30/Oct/17

If a total force F is applied on a total mass m, the partial force F_i which is applied on a part of mass (m_i ) is F_i =(m_i /m)×F in our example we can also do like this: acceleration of (m+M) is a=((F−(m+M)g×sin θ)/(m+M)) Ma=F_(12) −Mg×sin θ ⇒F_(12) =M(g×sin θ+a)=M[g×sin θ+((F−(m+M)g×sin θ)/(m+M))] =(M/(m+M))×F

$$\mathrm{If}\:\mathrm{a}\:\mathrm{total}\:\mathrm{force}\:\mathrm{F}\:\mathrm{is}\:\mathrm{applied}\:\mathrm{on}\:\mathrm{a}\:\mathrm{total} \\ $$$$\mathrm{mass}\:\mathrm{m},\:\mathrm{the}\:\mathrm{partial}\:\mathrm{force}\:\mathrm{F}_{\mathrm{i}} \:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{applied}\:\mathrm{on}\:\mathrm{a}\:\mathrm{part}\:\mathrm{of}\:\mathrm{mass}\:\left(\mathrm{m}_{\mathrm{i}} \right)\:\mathrm{is} \\ $$$$\mathrm{F}_{\mathrm{i}} =\frac{\mathrm{m}_{\mathrm{i}} }{\mathrm{m}}×\mathrm{F} \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{example}\:\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{do}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\left(\mathrm{m}+\mathrm{M}\right)\:\mathrm{is} \\ $$$$\mathrm{a}=\frac{\mathrm{F}−\left(\mathrm{m}+\mathrm{M}\right)\mathrm{g}×\mathrm{sin}\:\theta}{\mathrm{m}+\mathrm{M}} \\ $$$$ \\ $$$$\mathrm{Ma}=\mathrm{F}_{\mathrm{12}} −\mathrm{Mg}×\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{F}_{\mathrm{12}} =\mathrm{M}\left(\mathrm{g}×\mathrm{sin}\:\theta+\mathrm{a}\right)=\mathrm{M}\left[\mathrm{g}×\mathrm{sin}\:\theta+\frac{\mathrm{F}−\left(\mathrm{m}+\mathrm{M}\right)\mathrm{g}×\mathrm{sin}\:\theta}{\mathrm{m}+\mathrm{M}}\right] \\ $$$$=\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}×\mathrm{F} \\ $$

Answered by Physics lover last updated on 29/Oct/17

⇒a_(two bocks ) = ((5g∙Sin 37° − 20)/5) ⇒ a = ((5(10)(3/5) − 20)/5) ⇒ a = 2 m/s^(2 ) along the inclined downwards for 3kg block ⇒ ma = mg∙Sin 37 ° − F_(contact) ⇒ 3(2) = 3(10)(3/5) − F_(contact) ⇒ F_(contact ) = 12 N

$$ \\ $$$$\:\Rightarrow{a}_{{two}\:{bocks}\:} =\:\frac{\mathrm{5}{g}\centerdot{Sin}\:\mathrm{37}°\:−\:\mathrm{20}}{\mathrm{5}} \\ $$$$\:\Rightarrow\:\:{a}\:\:\:=\:\frac{\mathrm{5}\left(\mathrm{10}\right)\frac{\mathrm{3}}{\mathrm{5}}\:−\:\mathrm{20}}{\mathrm{5}}\:\: \\ $$$$\:\Rightarrow\:\:\:{a}\:\:\:=\:\mathrm{2}\:{m}/{s}^{\mathrm{2}\:} \:{along}\:{the}\:{inclined}\:{downwards} \\ $$$$ \\ $$$${for}\:\:\mathrm{3}{kg}\:{block} \\ $$$$\Rightarrow\:\:\:{ma}\:=\:{mg}\centerdot{Sin}\:\mathrm{37}\:°\:−\:{F}_{{contact}} \\ $$$$\Rightarrow\:\:\:\mathrm{3}\left(\mathrm{2}\right)\:=\:\mathrm{3}\left(\mathrm{10}\right)\frac{\mathrm{3}}{\mathrm{5}}\:−\:{F}_{{contact}} \\ $$$$\Rightarrow\:{F}_{{contact}\:} =\:\mathrm{12}\:{N} \\ $$$$ \\ $$

Commented by Tinkutara last updated on 29/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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