Question

a man of mass m stands on a platform of equal mass m and pulls himself by two ropes passing over pulleys. If he pulls each rope with a force equal to half its weight, his upward acceleration will be?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{The}\mathrm{downward}\mathrm{force},\mathrm{f}=\mathrm{mg}+\mathrm{mg}=2\mathrm{mg} \\ \mathrm{Tension}\mathrm{in}\mathrm{the}\mathrm{string},\mathrm{T}\text{'}=\mathrm{T}+\mathrm{T}=2\mathrm{T} \\ \mathrm{Tension},\mathrm{T}\mathrm{applied}\mathrm{by}\mathrm{the}\mathrm{man}=\frac{1}{2}\left(2\mathrm{mg}\right)=\mathrm{mg} \\ \mathrm{Hence},\mathrm{total}\mathrm{force},\mathrm{f}\text{'}=2\times \mathrm{mg}=2\mathrm{mg} \\ \mathrm{So},\mathrm{the}\mathrm{net}\mathrm{force}\mathrm{on}\mathrm{the}\mathrm{system},\mathrm{F}=2\mathrm{mg}-2\mathrm{mg}=0 \\ \mathrm{Hence},\mathrm{acceleration}=0 \end{gathered}$$