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Question Number 21747 by Tinkutara last updated on 02/Oct/17

Three blocks of masses m_1 , m_2 and m_3 are connected as shown. All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of m_1 .

$$\mathrm{Three}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:{m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} \:\mathrm{and}\:{m}_{\mathrm{3}} \\ $$$$\mathrm{are}\:\mathrm{connected}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{All}\:\mathrm{the}\:\mathrm{surfaces} \\ $$$$\mathrm{are}\:\mathrm{frictionless}\:\mathrm{and}\:\mathrm{the}\:\mathrm{string}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{pulleys}\:\mathrm{are}\:\mathrm{light}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:{m}_{\mathrm{1}} . \\ $$

Commented by Tinkutara last updated on 02/Oct/17

Commented by Tinkutara last updated on 02/Oct/17

m_1 on horizontal surface. m_3 is below m_2 .

$${m}_{\mathrm{1}} \:\mathrm{on}\:\mathrm{horizontal}\:\mathrm{surface}. \\ $$$${m}_{\mathrm{3}} \:\mathrm{is}\:\mathrm{below}\:{m}_{\mathrm{2}} . \\ $$

Commented by mrW1 last updated on 03/Oct/17

m_3 g−T=m_3 a_3 T−m_2 g=m_2 a_2 2T=m_1 a_1 a_3 −2a_1 =a_2 (!) or a_2 −a_3 =−2a_1 T=((m_1 a_1 )/2) a_3 =g−(T/m_3 )=g−(m_1 /(2m_3 ))×a_1 a_2 =(T/m_2 )−g=(m_1 /(2m_2 ))×a_1 −g ⇒ (m_1 /(2m_2 ))×a_1 −g−(g−(m_1 /(2m_3 ))×a_1 )=−2a_1 (m_1 /2)((1/m_2 )+(1/m_3 ))×a_1 −2g=−2a_1 (m_1 /4)((1/m_2 )+(1/m_3 ))×a_1 −g=−a_1 ⇒a_1 =(g/(1+(m_1 /4)((1/m_2 )+(1/m_3 ))))

$$\mathrm{m}_{\mathrm{3}} \mathrm{g}−\mathrm{T}=\mathrm{m}_{\mathrm{3}} \mathrm{a}_{\mathrm{3}} \:\:\: \\ $$$$\mathrm{T}−\mathrm{m}_{\mathrm{2}} \mathrm{g}=\mathrm{m}_{\mathrm{2}} \mathrm{a}_{\mathrm{2}} \:\:\:\:\: \\ $$$$\mathrm{2T}=\mathrm{m}_{\mathrm{1}} \mathrm{a}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{a}_{\mathrm{3}} −\mathrm{2a}_{\mathrm{1}} =\mathrm{a}_{\mathrm{2}} \:\:\:\:\:\:\:\left(!\right) \\ $$$$\mathrm{or}\:\mathrm{a}_{\mathrm{2}} −\mathrm{a}_{\mathrm{3}} =−\mathrm{2a}_{\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{T}=\frac{\mathrm{m}_{\mathrm{1}} \mathrm{a}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{3}} =\mathrm{g}−\frac{\mathrm{T}}{\mathrm{m}_{\mathrm{3}} }=\mathrm{g}−\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{2m}_{\mathrm{3}} }×\mathrm{a}_{\mathrm{1}} \\ $$$$\mathrm{a}_{\mathrm{2}} =\frac{\mathrm{T}}{\mathrm{m}_{\mathrm{2}} }−\mathrm{g}=\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{2m}_{\mathrm{2}} }×\mathrm{a}_{\mathrm{1}} −\mathrm{g} \\ $$$$\Rightarrow\:\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{2m}_{\mathrm{2}} }×\mathrm{a}_{\mathrm{1}} −\mathrm{g}−\left(\mathrm{g}−\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{2m}_{\mathrm{3}} }×\mathrm{a}_{\mathrm{1}} \right)=−\mathrm{2a}_{\mathrm{1}} \\ $$$$\:\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{m}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{m}_{\mathrm{3}} }\right)×\mathrm{a}_{\mathrm{1}} −\mathrm{2g}=−\mathrm{2a}_{\mathrm{1}} \\ $$$$\:\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{m}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{m}_{\mathrm{3}} }\right)×\mathrm{a}_{\mathrm{1}} −\mathrm{g}=−\mathrm{a}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{g}}{\mathrm{1}+\frac{\mathrm{m}_{\mathrm{1}} }{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{m}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{m}_{\mathrm{3}} }\right)} \\ $$

Commented by Tinkutara last updated on 03/Oct/17

Thank you very much Sir!

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

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