Question Number 64698 by Rio Michael last updated on 20/Jul/19
Commented by Rio Michael last updated on 20/Jul/19
$${the}\:{figure}\:{above}\:{shows}\:{a}\:{wood}\:{of}\:{mass}\:\mathrm{50}{kg}\:{held}\:{by}\:{a}\:{rope}\: \\ $$$${which}\:{is}\:{inclined}\:{at}\:\mathrm{30}°\:{to}\:{the}\:{wood}\:{and}\:{supported}\:{by}\:{a}\:{wall}. \\ $$$${calculate}\:{the}\:{tension}\:{in}\:{the}\:{rope} \\ $$$$\:\:\:\left({take}\:{g}\:=\:\mathrm{10}{ms}^{−\mathrm{2}} \right) \\ $$$$ \\ $$
Commented by mr W last updated on 20/Jul/19
$${please}\:{use}\:{a}\:{smaller}\:{font}\:{such}\:{that} \\ $$$${one}\:{doesn}'{t}\:{need}\:{to}\:{scroll}\:{horizontally} \\ $$$${to}\:{read}\:{your}\:{posts}. \\ $$
Commented by Rio Michael last updated on 20/Jul/19
$${okay}\:{sir} \\ $$
Commented by mr W last updated on 21/Jul/19
$${thanks}\:{for}\:{considering}\:{my}\:{request}! \\ $$
Answered by mr W last updated on 20/Jul/19
Commented by mr W last updated on 20/Jul/19
$${T}\:\mathrm{sin}\:\mathrm{30}°=\frac{{Mg}}{\mathrm{2}} \\ $$$$\Rightarrow{T}=\frac{{Mg}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{30}°}=\frac{\mathrm{50}×\mathrm{10}}{\mathrm{2}×\mathrm{0}.\mathrm{5}}=\mathrm{500}\:{N} \\ $$
Commented by Rio Michael last updated on 20/Jul/19
$${thank}\:{you}\:{sir} \\ $$
Commented by Rio Michael last updated on 20/Jul/19
$${but}\:{sir}\:{why}\:{did}\:{you}\:{take}\:\:{Tsin}\mathrm{30}°\:{and}\:{not}\:{sin}\mathrm{30}°\: \\ $$
Commented by Rio Michael last updated on 20/Jul/19
$${where}\:{did}\:{the}\:{two}\:{come}\:{from}\:{sir}\:{please}. \\ $$
Commented by Rio Michael last updated on 20/Jul/19
$${and}\:\:{sin}\mathrm{30}°\:=\:\frac{{mg}}{{T}}\:\:{right}? \\ $$
Commented by mr W last updated on 20/Jul/19
$${the}\:{rope}\:{carries}\:{half}\:{of}\:{the}\:{weight} \\ $$$${of}\:{the}\:{wood},\:{to}\:{be}\:{exact},\:{the}\:{vertical} \\ $$$${component}\:{of}\:{the}\:{tension}\:{in}\:{the}\:{rope} \\ $$$${carries}\:{half}\:{the}\:{the}\:{weight}\:{of}\:{the} \\ $$$${wood}.\:{the}\:{vertical}\:{component}\:{of}\:{the} \\ $$$${tension}\:{is}\:{T}\:\mathrm{sin}\:\mathrm{30}°,\:{thus} \\ $$$${T}\:\mathrm{sin}\:\mathrm{30}°=\frac{{Mg}}{\mathrm{2}} \\ $$
Commented by Rio Michael last updated on 20/Jul/19
$${thanks}\:{so}\:{much}\:{sir} \\ $$