Determine the acceleration of the masses w.r.t. lift and tension in the string if the whole system in moving vertically upwards with uniform acceleration $a_{0}$ . $(m_{1} > m_{2})$

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Solution

let the tension be $T$

$m_{1} (g + a_{0}) - T = m_{1} a$

$T - m_{2} (g + a_{0}) = m_{2} a$

add both the equation ,

$(m_{1} - m_{2}) (g + a_{0}) = a (m_{1} + m_{2})$

$a = \frac{(m_{1} - m_{2}) (g + a_{0})}{(m_{1} + m_{2})}$

acceleration of the blocks wrt the lift.

$T + \frac{m_{2} (m_{1} - m_{2}) (g + a_{0})}{(m_{1} + m_{2})} + m_{2} (g + a_{0})$ tension in the string.

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Determine the acceleration of the masses w.r.t. lift and tension in the string if the whole system in moving vertically upwards with uniform acceleration a0. (m1>m2)

let the tension be T m1(g+a0)−T=m1a T−m2(g+a0)=m2aadd both the equation , (m1−m2)(g+a0)=a(m1+m2) a=(m1−m2)(g+a0)(m1+m2) acceleration of the blocks wrt the lift. T+m2(m1−m2)(g+a0)(m1+m2)+m2(g+a0) tension in the string.

Determine the acceleration of the masses w.r.t. lift and tension in the string if the whole system in moving vertically upwards with uniform acceleration $a_{0}$ . $(m_{1} > m_{2})$