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Question Number 64698 by Rio Michael last updated on 20/Jul/19

Commented by Rio Michael last updated on 20/Jul/19

$$thefigureaboveshowsawoodofmass50kgheldbyarope$$$$whichisinclinedat30°tothewoodandsupportedbyawall.$$$$calculatethetensionintherope$$$$(takeg=10{ms}^{-2})$$$$$$

Commented by mr W last updated on 20/Jul/19

$$pleaseuseasmallerfontsuchthat$$$$one{doesn}^{\prime }tneedtoscrollhorizontally$$$$toreadyourposts.$$

Commented by Rio Michael last updated on 20/Jul/19

$$okaysir$$

Commented by mr W last updated on 21/Jul/19

$$thanksforconsideringmyrequest!$$

Answered by mr W last updated on 20/Jul/19

Commented by mr W last updated on 20/Jul/19

$$T\sin 30°=\frac{Mg}{2}$$$$\Rightarrow T=\frac{Mg}{2\sin 30°}=\frac{50\times 10}{2\times 0.5}=500N$$

Commented by Rio Michael last updated on 20/Jul/19

$$thankyousir$$

Commented by Rio Michael last updated on 20/Jul/19

$$butsirwhydidyoutakeTsin30°andnotsin30°$$

Commented by Rio Michael last updated on 20/Jul/19

$$wheredidthetwocomefromsirplease.$$

Commented by Rio Michael last updated on 20/Jul/19

$$andsin30°=\frac{mg}{T}right?$$

Commented by mr W last updated on 20/Jul/19

$$theropecarrieshalfoftheweight$$$$ofthewood,tobeexact,thevertical$$$$componentofthetensionintherope$$$$carrieshalfthetheweightofthe$$$$wood.theverticalcomponentofthe$$$$tensionisT\sin 30°,thus$$$$T\sin 30°=\frac{Mg}{2}$$

Commented by Rio Michael last updated on 20/Jul/19

$$thankssomuchsir$$

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