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Question Number 23748 by mrW1 last updated on 05/Nov/17

Commented by mrW1 last updated on 05/Nov/17

$$\mathrm{The}\mathrm{same}\mathrm{question}\mathrm{as}\mathrm{Q}23707,$$$$\mathrm{but}\mathrm{the}\mathrm{pulley}\mathrm{C}\mathrm{has}\mathrm{a}\mathrm{mass}1\mathrm{kg}\mathrm{and}$$$$\mathrm{radius}5\mathrm{cm}.$$

Answered by ajfour last updated on 05/Nov/17

$$Tensiononleft\left(Aside\right)=T_1$$$$T_1-mg=2mA...\left(i\right)$$$$F-T_1-T_2-mg=mA....\left(ii\right)$$$$(T_2-T_1)r=\left(\frac{{mr}^2}{2}\right)\left(\frac{A}{r}\right)...\left(iii\right)$$$$\Rightarrow T_2=T_1+\frac{mA}{2}$$$$usingin\left(ii\right):$$$$F-2T_1=\frac{3mA}{2}+mg.....\left(iv\right)$$$$BlockAisliftedwhen{T}_1=mg$$$$so,F=3mg(A=0tillthen)$$$$\Rightarrow 20t=30or\mathit{t}=\frac{3}{2}s$$$$From\left(iv\right)and\left(i\right):$$$$F-2(mg+2mA)=\frac{3mA}{2}+mg$$$$or20t-30=\frac{11A}{2}$$$$A=\frac{40}{11}(t-\frac{3}{2})$$$$thisisvalidtill{T}_2⩽2mg$$$$Eliminating{T}_{1}from\left(i\right)and\left(iii\right)$$$$T_1-mg=2mA;T_2=T_1+\frac{mA}{2}+mg$$$$T_2=mg+2mA+\frac{mA}{2}$$$$whenblockBislifted,$$$$T_2=2mg\Rightarrow \frac{5mA}{2}=mg$$$$A=\frac{2g}{5}or\frac{40}{11}(t-\frac{3}{2})=\frac{2g}{5}$$$$t=\frac{11}{10}+\frac{3}{2}=2.6s$$$$s_p={\int }_{3/2}^{2.6}{v}_{p}dt$$$$v_p={\int }_{3/2}^{t}Adt=\frac{40}{11}{\int }_{3/2}^t(t-\frac{3}{2})dt$$$$=\frac{20}{11}{(t-\frac{3}{2})}^2$$$$s_p=\frac{20}{11}{\int }_{3/2}^{2.6}{(t-\frac{3}{2})}^{2}dt$$$$=\frac{20}{11}\times \frac{1}{3}\times {\left(\frac{11}{10}\right)}^3=\frac{121}{150}m\approx 0.8m$$$$\left(s_{p}bythetimeblockBislifted\right).$$$$stillthesameanswer..!$$

Commented by mrW1 last updated on 05/Nov/17

$$Icheckedagain.Theanswer\frac{121}{150}is$$$$correct.$$$$Canyoualsofindthedisplacementof$$$$pulleyatt=5sec?$$

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