Figure-shows-an-arrangement-of-blocks-pulley-and-strings-Strings-and-pulley-are-massless-and-frictionless-The-relation-between-acceleration-of-the-blocks-as-shown-in-the-figure-is-

Question Number 21145 by Tinkutara last updated on 14/Sep/17

$$\mathrm{Figure}\:\mathrm{shows}\:\mathrm{an}\:\mathrm{arrangement}\:\mathrm{of}\:\mathrm{blocks}, \\ $$$$\mathrm{pulley}\:\mathrm{and}\:\mathrm{strings}.\:\mathrm{Strings}\:\mathrm{and}\:\mathrm{pulley} \\ $$$$\mathrm{are}\:\mathrm{massless}\:\mathrm{and}\:\mathrm{frictionless}.\:\mathrm{The} \\ $$$$\mathrm{relation}\:\mathrm{between}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{blocks}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{is} \\ $$

Commented by Tinkutara last updated on 14/Sep/17

Commented by mrW1 last updated on 14/Sep/17

T_2 =m_2 (g+a_2 ) m_1 (g−a_1 )=6T_2 =6m_2 (g+a_2 ) ((6m_2 )/m_1 )(g+a_2 )=g−a_1 ((6m_2 )/m_1 )a_2 +a_1 =(1−((6m_2 )/m_1 ))g ⇒6m_2 a_2 +m_1 a_1 =(m_1 −6m_2 )g

$$\mathrm{T}_{\mathrm{2}} =\mathrm{m}_{\mathrm{2}} \left(\mathrm{g}+\mathrm{a}_{\mathrm{2}} \right) \\ $$$$\mathrm{m}_{\mathrm{1}} \left(\mathrm{g}−\mathrm{a}_{\mathrm{1}} \right)=\mathrm{6T}_{\mathrm{2}} =\mathrm{6m}_{\mathrm{2}} \left(\mathrm{g}+\mathrm{a}_{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{6m}_{\mathrm{2}} }{\mathrm{m}_{\mathrm{1}} }\left(\mathrm{g}+\mathrm{a}_{\mathrm{2}} \right)=\mathrm{g}−\mathrm{a}_{\mathrm{1}} \\ $$$$\frac{\mathrm{6m}_{\mathrm{2}} }{\mathrm{m}_{\mathrm{1}} }\mathrm{a}_{\mathrm{2}} +\mathrm{a}_{\mathrm{1}} =\left(\mathrm{1}−\frac{\mathrm{6m}_{\mathrm{2}} }{\mathrm{m}_{\mathrm{1}} }\right)\mathrm{g} \\ $$$$\Rightarrow\mathrm{6m}_{\mathrm{2}} \mathrm{a}_{\mathrm{2}} +\mathrm{m}_{\mathrm{1}} \mathrm{a}_{\mathrm{1}} =\left(\mathrm{m}_{\mathrm{1}} −\mathrm{6m}_{\mathrm{2}} \right)\mathrm{g} \\ $$

Commented by Tinkutara last updated on 15/Sep/17

$$\mathrm{How}\:\mathrm{a}_{\mathrm{2}} =\mathrm{6a}_{\mathrm{1}} ? \\ $$

Commented by Tinkutara last updated on 14/Sep/17

Answer is a direct relation between a_1 and a_2 .

$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{a}\:\mathrm{direct}\:\mathrm{relation}\:\mathrm{between} \\ $$$${a}_{\mathrm{1}} \:\mathrm{and}\:{a}_{\mathrm{2}} . \\ $$

Commented by mrW1 last updated on 15/Sep/17

a_2 =6a_1 ⇒(36m_2 +m_1 )a_1 =(m_1 −6m_2 )g ⇒a_1 =((m_1 −6m_2 )/(36m_2 +m_1 ))×g

$$\mathrm{a}_{\mathrm{2}} =\mathrm{6a}_{\mathrm{1}} \\ $$$$\Rightarrow\left(\mathrm{36m}_{\mathrm{2}} +\mathrm{m}_{\mathrm{1}} \right)\mathrm{a}_{\mathrm{1}} =\left(\mathrm{m}_{\mathrm{1}} −\mathrm{6m}_{\mathrm{2}} \right)\mathrm{g} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{1}} =\frac{\mathrm{m}_{\mathrm{1}} −\mathrm{6m}_{\mathrm{2}} }{\mathrm{36m}_{\mathrm{2}} +\mathrm{m}_{\mathrm{1}} }×\mathrm{g} \\ $$

Commented by ajfour last updated on 15/Sep/17

for the larger block to come down by x the smaller has to rise by 6x. so x_2 =6x_1 ⇒ v_2 =6v_1 ⇒ a_2 =6a_1 .

$${for}\:{the}\:{larger}\:{block}\:{to}\:{come}\:{down} \\ $$$${by}\:{x}\:{the}\:{smaller}\:{has}\:{to}\:{rise}\:{by}\:\mathrm{6}{x}. \\ $$$${so}\:{x}_{\mathrm{2}} =\mathrm{6}{x}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\:{v}_{\mathrm{2}} =\mathrm{6}{v}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\:\:{a}_{\mathrm{2}} =\mathrm{6}{a}_{\mathrm{1}} \:. \\ $$

Commented by Tinkutara last updated on 15/Sep/17

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

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