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Question Number 18940 by Tinkutara last updated on 01/Aug/17

$$\mathrm{A}\mathrm{light}\mathrm{rope}\mathrm{is}\mathrm{passed}\mathrm{over}\mathrm{a}\mathrm{pulley}\mathrm{such}$$$$\mathrm{that}\mathrm{at}\mathrm{its}\mathrm{one}\mathrm{end}\mathrm{a}\mathrm{block}\mathrm{is}\mathrm{attached},$$$$\mathrm{and}\mathrm{on}\mathrm{the}\mathrm{other}\mathrm{end}\mathrm{a}\mathrm{boy}\mathrm{is}\mathrm{climbing}$$$$\mathrm{up}\mathrm{with}\mathrm{acceleration}\frac{g}{2}\mathrm{relative}\mathrm{to}\mathrm{rope}.$$$$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{is}30\mathrm{kg}\mathrm{and}\mathrm{that}\mathrm{of}$$$$\mathrm{the}\mathrm{boy}\mathrm{is}40\mathrm{kg}.\mathrm{Find}\mathrm{the}\mathrm{tension}\mathrm{and}$$$$\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{rope}.$$

Commented by ajfour last updated on 01/Aug/17

Commented by ajfour last updated on 01/Aug/17

$$\mathrm{T}-\mathrm{mg}=\mathrm{ma}$$$$\mathrm{T}-\mathrm{Mg}=\mathrm{M}(\frac{\mathrm{g}}{2}-\mathrm{a})$$$$\Rightarrow (\mathrm{M}-\mathrm{m})\mathrm{g}=\mathrm{a}(\mathrm{m}+\mathrm{M})-\frac{\mathrm{Mg}}{2}$$$$\mathrm{a}=\frac{(\frac{3\mathrm{M}}{2}-\mathrm{m})\mathrm{g}}{\mathrm{M}+\mathrm{m}}=\frac{(\frac{3}{2}\times 40-30)\mathrm{g}}{40+30}$$$$\mathrm{a}=\frac{3\mathrm{g}}{7}.$$$$\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})=30(\mathrm{g}+\frac{3\mathrm{g}}{7})=\frac{300\mathrm{g}}{7}.$$

Commented by Tinkutara last updated on 01/Aug/17

$$\mathrm{Wonderful}\mathrm{Sir}!$$

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