Assuming the masses of the pulley and the thread, as well as the friction, to be negligible, find the acceleration of the load $m_{1}$ relative to the elevator shaft

w_{t} = \frac{2 (m_{1} - m_{2}) g + 2 m_{2} w_{0}}{m_{1} + m_{2}}

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w_{t} = \frac{(m_{1} - m_{2}) g + m_{2} w_{0}}{m_{1} + m_{2}}

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w_{t} = \frac{2 (m_{1} - m_{2}) g + m_{2} w_{0}}{m_{1} + m_{2}}

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w_{t} = \frac{(m_{1} - m_{2}) g - 2 m_{2} w_{0}}{m_{1} + m_{2}}

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Solution

The correct option is D $w_{t} = \frac{(m_{1} - m_{2}) g - 2 m_{2} w_{0}}{m_{1} + m_{2}}$
Fom FBD,
for mass $m_{1} : m_{1} a = m_{1} (g + w_{0}) - T . . . (1)$
for mass $m_{2} : m_{2} a = T - m_{2} (g + w_{0}) . . . (2)$
$(1) + (2), a (m_{1} + m_{2}) = (m_{1} - m_{2}) (g + w_{0})$
or $a = \frac{(m_{1} - m_{2}) (g + w_{0})}{(m_{1} + m_{2})}$
As the acceleration of $m_{1}$ downward and the acceleration of elevator shaft upward.
So the acceleration of $m_{1}$ relative to elevator shaft is
$w_{t} = a - w_{0} = \frac{(m_{1} - m_{2}) (g + w_{0})}{(m_{1} + m_{2})} - w_{0}$
or $w_{t} = \frac{(m_{1} - m_{2}) g + m_{1} w_{0} - m_{2} w_{0} - m_{1} w_{0} - m_{2} w_{0}}{(m_{1} + m_{2})}$
$∴ w_{t} = \frac{(m_{1} - m_{2}) g - 2 m_{2} w_{0}}{(m_{1} + m_{2})}$

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Assuming the masses of the pulley and the thread, as well as the friction, to be negligible, find the acceleration of the load m1 relative to the elevator shaft

Assuming the masses of the pulley and the thread, as well as the friction, to be negligible, find the acceleration of the load $m_{1}$ relative to the elevator shaft