Question

as shown in the figure, blocks of masses M/2,M and M/2 are connected through a light string as shown, pulleys are light and smooth. friction is only between block C and floor. system is released from rest. find acceleration of blocks A,B and C.
Correct answer a(A)=a(C)=(3/4)gsin(theta), a(B)=gsin(theta)
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Answer 1

Solution :

✴ Given :-

✏ A system of three blocks of mass M/2, M and M/2 is provided.

✏ Co-efficient of friction b/w block C and surface = tan$\theta$/2

✴ To Find :-

✏ Acceleration of all three blocks.

✴ Assumption :-

✏ All the pulleys are light and frictionless.

✴ Diagram :-

✏ Please see the attached image for better understanding.

✴ Calculation :-

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• For blocks A and C

$\implies\sf\: \dfrac{M}{2}g\sin\theta+\dfrac{M}{2}g\sin\theta-\mu(\dfrac{M}{2})g\cos\theta=( \dfrac{M}{2}+\dfrac{M}{2})a\\ \\ \implies\sf\:M[\dfrac{g\sin\theta}{2}+\dfrac{g\sin\theta}{2}-\dfrac{g\tan\theta\cos\theta}{4}]=Ma\\ \\ \implies\sf\:g\sin\theta-\dfrac{g\sin\theta}{4}=a\\ \\ \implies\: \boxed{\sf{\pink{\large{a_1=a_3=a=\dfrac{3g\sin\theta}{4}}}}}$

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• For block B

$\mapsto\sf\:Mg\sin\theta=Ma'\\ \\ \mapsto\: \boxed{\sf{\purple{\large{a'=a_2=g\sin\theta}}}}$

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☀ Note :-

▪ In this question, motion of B block is independent from motion of block A and block C.

✒ Nice question...

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Acceleration\:of\:block\:A=\frac{3g\:sin\:\theta}{4}}}}$

$\green{\tt{\therefore{Acceleration\:of\:block\:B=g\:sin\:\theta}}}$

$\green{\tt{\therefore{Acceleration\:of\:block\:C=\frac{3g\:sin\:\theta}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: block \: A( m_{1}) = \frac{M}{2} \\ \\ \tt: \implies Mass \: of \: block \: B( m_{2}) = M \\ \\ \tt: \implies Mass \: of \: block \: C( m_{3}) = \frac{M}{2} \\ \\ \tt: \implies Coefficient \: of \: friction \: between \: block \: and \: wedge( \mu) = \frac{tan \: \theta}{2} \\ \\ \tt : \implies angle \: inclined = \theta \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Acceleration \: of \: block \: A= ? \\ \\ \tt: \implies Acceleration \: of \: bock \: B= ? \\ \\ \tt: \implies Acceleration \: of \: bock \: C= ?$

• According to given question :

$\bold{For \: block \: A: } \\ \tt: \implies Downward \: force = \frac{M}{2} g \: sin \: \theta \\ \\ \tt: \implies Acceleration = \frac{Ma}{2} \\ \\ \bold{For \: block \:B: } \\ \tt: \implies Downward \: force = Mg \: sin \: \theta \\ \\ \tt: \implies Acceleration = {Ma} \\ \\ \bold{For \: block \: C: } \\ \tt: \implies Downward \: force = \frac{M}{2} g \: sin \: \theta \\ \\ \tt: \implies Acceleration = \frac{Ma}{2}$

$\bold{As \: we \: know \: that} \\ \tt: \implies Friction \: force = \mu m_{1}g \: cos \: \theta\\ \\ \tt: \implies Friction \: force = \frac{tan \: \theta}{2} \times \frac{M}{2} \times g \times cos \: \theta \\ \\ \tt: \implies Friction \: force = \frac{sin}{2 \: cos \: \theta} \times \frac{M}{2} \times g\times cos \: \theta \\ \\ \tt: \implies Friction \: force = \frac{Mg \: sin \: \theta}{4} \\ \\ \bold{For \: block \: A: } \\ \tt: \implies \frac{M}{2} g \: sin \: \theta = \frac{M}{2} a - - - - - (1) \\ \\ \bold{For \: block \: C : } \\ \tt: \implies m_{3} g \:sin \: \theta - fr = \frac{M}{2} a \\ \\ \tt: \implies \frac{Mg \: sin \: \theta}{2} - \frac{Mg \: sin \: \theta}{4} = \frac{M}{2} a - - - - - (2)$

$\text{Adding \: (1) \: and \: (2)} \\ \tt: \implies \frac{Mg \:sin \theta}{2} + \frac{Mg \: sin \theta}{2} - \frac{Mg \: sin \: \theta}{4} = \frac{Ma}{2} + \frac{Ma}{2} \\ \\ \tt: \implies \frac{2Mg \: sin \: \theta +2Mg \: sin \: \theta - Mg \: sin \: \theta}{4} = \frac{2Ma}{2} \\ \\ \tt: \implies \frac{3Mg \:sin \: \theta}{4} = Ma \\ \\ \tt: \implies a = \frac{3Mg \: sin \: \theta}{4M} \\ \\ \green{\tt: \implies a = \frac{3g \: sin \: \theta}{4} } \\ \\ \bold{For \: block \: B : } \\ \tt: \implies m_{2}g \: sin \: \theta = Ma \\ \\ \tt: \implies Mg \: sin \: \theta = Ma \\ \\ \tt: \implies a = \frac{Mg \: sin \: \theta}{m} \\ \\ \green{\tt: \implies a = {g \:sin \: \theta}} \\ \\ \green{\tt \therefore Acceleration \: of \: block \: A\: is \: \frac{3g \: sin \: \theta}{4} } \\ \\ \green{\tt \therefore Acceleration \: of \: block \:B \: is \: g \: sin \: \theta} \\ \\ \green{\tt \therefore Acceleration \: of \: block \: C \: is \: \frac{3g \: sin \: \theta}{4} }$