Question

Three masses m1 m2 and m3 are attached to a string pulley system as shown. all the three masses are held at rest and then released . if m3 remains at rest then prove that 4m1m2= m1m3 + m2m3

Answer 1

Two masses m1 and m2 are connected by an ideal string and passes over the pulley then we can write the equations for two masses as

$T - m_1g = m_1a$

$m_2g - T = m_2*a$

now add two equations to find acceleration

$m_2g - m_1g = (m_1 + m_2)*a$

now use the above value of acceleration to find the tension in the string

$T = m_1g + m_1*(\frac{m_2g - m_1g}{m_1 + m_2})$

now on solving above

$T = \frac{2m_1m_2g}{m_1+m_2}$

now then in the string connected to 3rd mass m3 will be given as

$T' = 2T$

$T' = \frac{4m_1m_2g}{m_1+m_2}$

now since mass m3 is at rest so

$m_3g = \frac{4m_1m_2g}{m_1+m_2}$

now rearranging the terms

$m_1m_3 + m_2m_3 = 4m_1m_2$

Answer 2

Explanation:

m1m3+m2m3=4m1m2

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