Question

A rope thrown over a pulley has a ladder with a man A no one of its ends and a counterbalancing mass M on its other end. The man whose mass is m, climbs upwards by $\overrightarrow{\Delta }r$ relative to the ladder and then stops. Ignoring the masses of the pulley and the rope, as well as the friction in the pulley axis, find the displacement of the centre of mass of this system

Answer 1

$Letdisplacementofcounterweightbe\Delta ythereforethedisplacementofladderwillbe(-\Delta y).Displacementofman\Delta y_m=\Delta y_{m,l}+\Delta y_l,y_{m,l}=displacementofmanw.r.t.ladder=\Delta r\to y_m=\Delta r-\Delta yHencedisplacementofCOMis=\frac{M_{mass}{\Delta r}_{mass}+M_{ladder}{\Delta r}_{ladder}+M_{man}{\Delta r}_{man}}{M_{mass}+M_{ladder}+M_{man}}=\frac{m\Delta r}{2M}$