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Question Number 17530 by Tinkutara last updated on 07/Jul/17

$$\mathrm{Two}\mathrm{masses}5\mathrm{kg}\mathrm{and}M\mathrm{are}\mathrm{hanging}$$$$\mathrm{with}\mathrm{the}\mathrm{help}\mathrm{of}\mathrm{light}\mathrm{rope}\mathrm{and}\mathrm{pulley}$$$$\mathrm{as}\mathrm{shown}\mathrm{below}.\mathrm{If}\mathrm{the}\mathrm{system}\mathrm{is}\mathrm{in}$$$$\mathrm{equilibrium}\mathrm{then}M=$$

Commented by Tinkutara last updated on 07/Jul/17

Commented by mrW1 last updated on 07/Jul/17

$$\mathrm{Tcos}53=5\times \cos 37$$$$\mathrm{T}=5\times \frac{\cos 37}{\cos 53}=5\times \frac{\cos 37}{\sin 37}$$$$\mathrm{M}=5\times \sin 37+\mathrm{T}\times \sin 53$$$$=5\times \sin 37+5\times \frac{{\cos }^{2}37}{\sin 37}$$$$=\frac{5}{\sin 37}=8.31\mathrm{kg}$$

Commented by 1234Hello last updated on 07/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!\mathrm{I}\mathrm{forgot}\mathrm{the}\mathrm{options}.\mathrm{But}$$$$\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{why}M=5\sin 37°$$$$+T\sin 53°?$$

Commented by mrW1 last updated on 07/Jul/17

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{what}\mathrm{you}\mathrm{mean}\mathrm{with}$$$$\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{in}\mathrm{the}\mathrm{options}.\mathrm{You}\mathrm{can}\mathrm{solve}$$$$\mathrm{it}\mathrm{in}\mathrm{different}\mathrm{ways}.\mathrm{Since}37+53=90,$$$$\mathrm{the}\mathrm{three}\mathrm{forces}\mathrm{build}\mathrm{a}\mathrm{right}\mathrm{angeled}$$$$\mathrm{triangle},\mathrm{so}\mathrm{you}\mathrm{can}\mathrm{directly}\mathrm{get}$$$$\mathrm{M}=5/\sin 37°.\mathrm{If}\mathrm{you}\mathrm{have}\mathrm{no}\mathrm{calculator},$$$$\mathrm{you}\mathrm{can}\mathrm{use}\sin 37\approx \frac{\sin 30+\sin 45}{2}\approx \frac{0.5+0.7}{2}=0.6$$$$\Rightarrow \mathrm{M}\approx 5/0.6\approx 8.3.$$

Commented by ajfour last updated on 07/Jul/17

$$5\mathrm{mgcos}37=\mathrm{Tcos}53$$$$\Rightarrow 4\mathrm{mg}=3\mathrm{T}$$$$\mathrm{mgsin}37+\mathrm{Tsin}53=\mathrm{Mg}$$$$3\mathrm{mg}+4\mathrm{T}=5\mathrm{Mg}$$$$\Rightarrow 3\mathrm{mg}+\frac{16\mathrm{mg}}{3}=5\mathrm{Mg}$$$$\mathrm{M}=\frac{5\mathrm{m}}{3}=\frac{5\times 5\mathrm{kg}}{3}=8\frac{1}{3}\mathrm{kg}.$$

Commented by mrW1 last updated on 07/Jul/17

$$\mathrm{equilibrium}\mathrm{in}\mathrm{x}-\mathrm{direction}:$$$$\mathrm{Tcos}53=5\times \cos 37$$$$\mathrm{equilibrium}\mathrm{in}\mathrm{y}-\mathrm{direction}:$$$$\mathrm{M}=5\times \sin 37+\mathrm{T}\times \sin 53$$

Commented by Tinkutara last updated on 08/Jul/17

$$\mathrm{Thanks}\mathrm{ajfour}\mathrm{and}\mathrm{mrW}1\mathrm{Sir}!$$

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