Question Number 20346 by Tinkutara last updated on 25/Aug/17
$$\mathrm{To}\:\mathrm{paint}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{building},\:\mathrm{painter} \\ $$$$\mathrm{normally}\:\mathrm{hoists}\:\mathrm{himself}\:\mathrm{up}\:\mathrm{by}\:\mathrm{pulling}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{rope}\:{A}\:\mathrm{as}\:\mathrm{in}\:\mathrm{figure}.\:\mathrm{The}\:\mathrm{painter}\:\mathrm{and} \\ $$$$\mathrm{platform}\:\mathrm{together}\:\mathrm{weigh}\:\mathrm{200}\:\mathrm{N}.\:\mathrm{The} \\ $$$$\mathrm{rope}\:{B}\:\mathrm{can}\:\mathrm{withstand}\:\mathrm{300}\:\mathrm{N}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{painter}. \\ $$
Commented by Tinkutara last updated on 25/Aug/17
Commented by ajfour last updated on 25/Aug/17
$${let}\:{mass}\:{of}\:{painter}={M} \\ $$$${mass}\:{of}\:{platform}={m} \\ $$$${tension}\:{in}\:{rope}\:{A}\:={T}=\frac{\mathrm{1}}{\mathrm{2}}{T}_{{B}} \\ $$$${T}={T}_{{A}} =\mathrm{150}{N} \\ $$$$\mathrm{2}{T}−\left({M}+{m}\right){g}=\left({M}+{m}\right){a} \\ $$$${a}_{{max}} =\frac{\mathrm{2}{T}}{\left({M}+{m}\right)}−{g} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{300}}{\left(\mathrm{200}/{g}\right)}−{g}\:=\:\frac{{g}}{\mathrm{2}}\:. \\ $$
Commented by Tinkutara last updated on 25/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$