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-

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

Thank you very much Sir!