Menu Close

A-light-rope-is-passed-over-a-pulley-such-that-at-its-one-end-a-block-is-attached-and-on-the-other-end-a-boy-is-climbing-up-with-acceleration-g-2-relative-to-rope-Mass-of-the-block-is-30-kg-and-th

Tinku Tara 0 Comments
Pin




Question Number 18940 by Tinkutara last updated on 01/Aug/17
A light rope is passed over a pulley such that at its one end a block is attached, and on the other end a boy is climbing up with acceleration (g/2) relative to rope. Mass of the block is 30 kg and that of the boy is 40 kg. Find the tension and acceleration of the rope.
$$\mathrm{A}\:\mathrm{light}\:\mathrm{rope}\:\mathrm{is}\:\mathrm{passed}\:\mathrm{over}\:\mathrm{a}\:\mathrm{pulley}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{at}\:\mathrm{its}\:\mathrm{one}\:\mathrm{end}\:\mathrm{a}\:\mathrm{block}\:\mathrm{is}\:\mathrm{attached}, \\ $$$$\mathrm{and}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other}\:\mathrm{end}\:\mathrm{a}\:\mathrm{boy}\:\mathrm{is}\:\mathrm{climbing} \\ $$$$\mathrm{up}\:\mathrm{with}\:\mathrm{acceleration}\:\frac{{g}}{\mathrm{2}}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{rope}. \\ $$$$\mathrm{Mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{block}\:\mathrm{is}\:\mathrm{30}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{that}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{boy}\:\mathrm{is}\:\mathrm{40}\:\mathrm{kg}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{and} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rope}. \\ $$
Commented by ajfour last updated on 01/Aug/17
Commented by ajfour last updated on 01/Aug/17
T−mg=ma T−Mg=M((g/2)−a) ⇒ (M−m)g=a(m+M)−((Mg)/2) a=(((((3M)/2)−m)g)/(M+m)) = ((((3/2)×40−30)g)/(40+30)) a=((3g)/7) . T=m(g+a)=30(g+((3g)/7))=((300g)/7) .
$$\mathrm{T}−\mathrm{mg}=\mathrm{ma} \\ $$$$\mathrm{T}−\mathrm{Mg}=\mathrm{M}\left(\frac{\mathrm{g}}{\mathrm{2}}−\mathrm{a}\right) \\ $$$$\Rightarrow\:\:\:\left(\mathrm{M}−\mathrm{m}\right)\mathrm{g}=\mathrm{a}\left(\mathrm{m}+\mathrm{M}\right)−\frac{\mathrm{Mg}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{a}=\frac{\left(\frac{\mathrm{3M}}{\mathrm{2}}−\mathrm{m}\right)\mathrm{g}}{\mathrm{M}+\mathrm{m}}\:=\:\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{40}−\mathrm{30}\right)\mathrm{g}}{\mathrm{40}+\mathrm{30}} \\ $$$$\:\:\:\:\:\:\mathrm{a}=\frac{\mathrm{3g}}{\mathrm{7}}\:.\: \\ $$$$\mathrm{T}=\mathrm{m}\left(\mathrm{g}+\mathrm{a}\right)=\mathrm{30}\left(\mathrm{g}+\frac{\mathrm{3g}}{\mathrm{7}}\right)=\frac{\mathrm{300g}}{\mathrm{7}}\:. \\ $$
Commented by Tinkutara last updated on 01/Aug/17
Wonderful Sir!
$$\mathrm{Wonderful}\:\mathrm{Sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *