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


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

$Three blocks of masses m_{1}, m_{2} and m_{3}$ $are connected as shown . All the surfaces$ $are frictionless and the string and the$ $pulleys are light . Find the acceleration$ $of m_{1} .$
Commented by Tinkutara last updated on 02/Oct/17

Commented by Tinkutara last updated on 02/Oct/17

$m_{1} on horizontal surface .$ $m_{3} is below m_{2} .$
Commented by mrW1 last updated on 03/Oct/17

$m_{3} g - T = m_{3} a_{3}$ $T - m_{2} g = m_{2} a_{2}$ $2 T = m_{1} a_{1}$ $a_{3} - {2 a}_{1} = a_{2} (!)$ $or a_{2} - a_{3} = - {2 a}_{1}$ $T = \frac{m_{1} a_{1}}{2}$ $a_{3} = g - \frac{T}{m_{3}} = g - \frac{m_{1}}{{2 m}_{3}} \times a_{1}$ $a_{2} = \frac{T}{m_{2}} - g = \frac{m_{1}}{{2 m}_{2}} \times a_{1} - g$ $\Rightarrow \frac{m_{1}}{{2 m}_{2}} \times a_{1} - g - (g - \frac{m_{1}}{{2 m}_{3}} \times a_{1}) = - {2 a}_{1}$ $\frac{m_{1}}{2} (\frac{1}{m_{2}} + \frac{1}{m_{3}}) \times a_{1} - 2 g = - {2 a}_{1}$ $\frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}}) \times a_{1} - g = - a_{1}$ $\Rightarrow a_{1} = \frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}$
Commented by Tinkutara last updated on 03/Oct/17

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