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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

the figure above shows a wood of mass 50kg held by a rope which is inclined at 30° to the wood and supported by a wall. calculate the tension in the rope (take g = 10ms^(−2) )

$${the}\:{figure}\:{above}\:{shows}\:{a}\:{wood}\:{of}\:{mass}\:\mathrm{50}{kg}\:{held}\:{by}\:{a}\:{rope}\: \\ $$$${which}\:{is}\:{inclined}\:{at}\:\mathrm{30}°\:{to}\:{the}\:{wood}\:{and}\:{supported}\:{by}\:{a}\:{wall}. \\ $$$${calculate}\:{the}\:{tension}\:{in}\:{the}\:{rope} \\ $$$$\:\:\:\left({take}\:{g}\:=\:\mathrm{10}{ms}^{−\mathrm{2}} \right) \\ $$$$ \\ $$

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

please use a smaller font such that one doesn′t need to scroll horizontally to read your posts.

$${please}\:{use}\:{a}\:{smaller}\:{font}\:{such}\:{that} \\ $$$${one}\:{doesn}'{t}\:{need}\:{to}\:{scroll}\:{horizontally} \\ $$$${to}\:{read}\:{your}\:{posts}. \\ $$

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

okay sir

$${okay}\:{sir} \\ $$

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

thanks for considering my request!

$${thanks}\:{for}\:{considering}\:{my}\:{request}! \\ $$

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

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

T sin 30°=((Mg)/2) ⇒T=((Mg)/(2 sin 30°))=((50×10)/(2×0.5))=500 N

$${T}\:\mathrm{sin}\:\mathrm{30}°=\frac{{Mg}}{\mathrm{2}} \\ $$$$\Rightarrow{T}=\frac{{Mg}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{30}°}=\frac{\mathrm{50}×\mathrm{10}}{\mathrm{2}×\mathrm{0}.\mathrm{5}}=\mathrm{500}\:{N} \\ $$

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

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

but sir why did you take Tsin30° and not sin30°

$${but}\:{sir}\:{why}\:{did}\:{you}\:{take}\:\:{Tsin}\mathrm{30}°\:{and}\:{not}\:{sin}\mathrm{30}°\: \\ $$

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

where did the two come from sir please.

$${where}\:{did}\:{the}\:{two}\:{come}\:{from}\:{sir}\:{please}. \\ $$

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

and sin30° = ((mg)/T) right?

$${and}\:\:{sin}\mathrm{30}°\:=\:\frac{{mg}}{{T}}\:\:{right}? \\ $$

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

the rope carries half of the weight of the wood, to be exact, the vertical component of the tension in the rope carries half the the weight of the wood. the vertical component of the tension is T sin 30°, thus T sin 30°=((Mg)/2)

$${the}\:{rope}\:{carries}\:{half}\:{of}\:{the}\:{weight} \\ $$$${of}\:{the}\:{wood},\:{to}\:{be}\:{exact},\:{the}\:{vertical} \\ $$$${component}\:{of}\:{the}\:{tension}\:{in}\:{the}\:{rope} \\ $$$${carries}\:{half}\:{the}\:{the}\:{weight}\:{of}\:{the} \\ $$$${wood}.\:{the}\:{vertical}\:{component}\:{of}\:{the} \\ $$$${tension}\:{is}\:{T}\:\mathrm{sin}\:\mathrm{30}°,\:{thus} \\ $$$${T}\:\mathrm{sin}\:\mathrm{30}°=\frac{{Mg}}{\mathrm{2}} \\ $$

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

thanks so much sir

$${thanks}\:{so}\:{much}\:{sir} \\ $$

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