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find-the-resultant-force-of-a-system-of-three-forces-O-P-9N-O-R-10N-and-O-Q-10N-acting-at-point-O-where-angle-POR-is-135-angle-POQ-is-135-and-QOR-is-90-




Question Number 62585 by Jmasanja last updated on 23/Jun/19
find the resultant force of a system  of three forces O^− P^→  =9N,O^− R^→  =10N and O^− Q^→  10N   acting at point O where angle POR is  135°,angle POQ is 135°  and QOR is   90°
$${find}\:{the}\:{resultant}\:{force}\:{of}\:{a}\:{system} \\ $$$${of}\:{three}\:{forces}\:\overset{−} {{O}}\overset{\rightarrow} {{P}}\:=\mathrm{9}{N},\overset{−} {{O}}\overset{\rightarrow} {{R}}\:=\mathrm{10}{N}\:{and}\:\overset{−} {{O}}\overset{\rightarrow} {{Q}}\:\mathrm{10}{N}\: \\ $$$${acting}\:{at}\:{point}\:{O}\:{where}\:{angle}\:{POR}\:{is} \\ $$$$\mathrm{135}°,{angle}\:{POQ}\:{is}\:\mathrm{135}°\:\:{and}\:{QOR}\:{is}\: \\ $$$$\mathrm{90}° \\ $$
Answered by MJS last updated on 23/Jun/19
OP^(⇀) = ((9),(0) )  OR^(⇀) = (((10cos 135°)),((10sin 135°)) )= (((−5(√2))),((5(√2))) )  OQ^(⇀) = (((−5(√2))),((−5(√2))) )  their sum is  (((9−10(√2))),(0) )  their magnitude is 10(√2)−9
$$\overset{\rightharpoonup} {{OP}}=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{0}}\end{pmatrix}\:\:\overset{\rightharpoonup} {{OR}}=\begin{pmatrix}{\mathrm{10cos}\:\mathrm{135}°}\\{\mathrm{10sin}\:\mathrm{135}°}\end{pmatrix}=\begin{pmatrix}{−\mathrm{5}\sqrt{\mathrm{2}}}\\{\mathrm{5}\sqrt{\mathrm{2}}}\end{pmatrix}\:\:\overset{\rightharpoonup} {{OQ}}=\begin{pmatrix}{−\mathrm{5}\sqrt{\mathrm{2}}}\\{−\mathrm{5}\sqrt{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{9}−\mathrm{10}\sqrt{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:\mathrm{their}\:\mathrm{magnitude}\:\mathrm{is}\:\mathrm{10}\sqrt{\mathrm{2}}−\mathrm{9} \\ $$

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