Question

If R is the resultant of two forces P and Q. If Q is reversed the resultant becomes S. Show that R2 + S2 = 2(P2 + Q2
)

Answer 1

Gɪᴠᴇɴ :-

• The resultant of two forces , $\sf \overrightarrow{P}$ and $\sf \overrightarrow{Q}$ is R .
• The dirⁿ of vector $\sf \overrightarrow{Q}$ is reversed and the resultant becomes S .

Tᴏ Pʀᴏᴠᴇ :-

• $\sf \overrightarrow{R^2}+\overrightarrow{S^2}=2(\overrightarrow{P^2}+\overrightarrow{Q^2})$

Pʀᴏᴏғ :-

Here its not given whether the whether the vectors ( forces ) are attached from head to tail or from tail to tail . Let's assume that they are attached from tail to tail .

$\tt \red{\longrightarrow}So\:here\:we\:can\:use\: parallelogram\:law\:of\: vector\: addition.$

$\boxed{\red{\bf \blue{\bigstar}\:\overrightarrow{R}=\sqrt{\overrightarrow{A^2}+\overrightarrow{B^2} +2\overrightarrow{A}\overrightarrow{B}\cos\theta}}}$

$\sf Using\: this \:, \:we\:have:,$

$\underline{\textsf{\textbf{\green{\orange{ \mapsto }\:Case\:\:1:-}}}}$

$\purple{\underbrace{\bf\pink{Figure:-}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(0,0){\vector(1,2){2}}\put(0,0){\vector(1,0){4}}\put(0,0){\vector(3,2){4}}\put(0,-0.34){ \sf O }\qbezier(0.6,0)(0.4,1)(0.3,0.65)\put(2, - 0.5){ \tt \overrightarrow{P} }\put(0.5,2){ \tt \overrightarrow{Q} }\put(2,1.7){ \tt \overrightarrow{R} }\end{picture}$

$\tt:\implies Resultant= \sqrt{\overrightarrow{A^2}+\overrightarrow{B^2} +2\overrightarrow{A}\overrightarrow{B}\cos\theta}$

$\tt:\implies Resultant =\sqrt{\overrightarrow{P^2}+\overrightarrow{Q^2} + 2\overrightarrow{A} +2\overrightarrow{P}\overrightarrow{Q}\cos\theta}$

$\underline{\boxed{\bf \red{\hookrightarrow}\:R^2= P^2 + Q^2+2PQ\cos\theta .}}$

$\rule{200}3$

$\underline{\textsf{\textbf{\green{\orange{ \mapsto }\:Case\:\:2:-}}}}$

In second case since Q is reversed it will become - Q .

$\purple{\underbrace{\bf\pink{Figure:-}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(0,0){\vector(-1,-2){2}}\put(0,0){\vector(1,0){4}}\put(0,0.34){ \sf O }\put(2, - 0.5){ \tt \overrightarrow{P} }\put( - 1.5, - 1.5){ \sf\overrightarrow{ - Q} }\put(0,0){\vector(3, - 2){3.5}}\put( 2.4, - 1.5){ \sf\overrightarrow{S} }\qbezier(0.5,0)(1,0.14)(-0.2,-0.4)\put( 0.2, - 0.7){ \tt \theta }\end{picture}$

$\tt:\implies Resultant = \sqrt{\overrightarrow{A^2}+\overrightarrow{B^2} +2\overrightarrow{A}\overrightarrow{B}\cos\theta}$

$\tt:\implies Resultant =\sqrt{\overrightarrow{P^2}+\overrightarrow{(-Q)^2} + 2\overrightarrow{A} +2\overrightarrow{P}\overrightarrow{-Q}\cos\theta}$

$\underline{\boxed{\bf \red{\hookrightarrow}\:S^2= P^2 + Q^2-2PQ\cos\theta .}}$

$\rule{200}3$

$\sf\pink{ \leadsto On\:adding\:both\: resultants:-}$

$\tt:\implies R^2+S^2=P^2+Q^2+\cancel{2PQ\cos\theta}+P^2+Q^2-\cancel{2PQ\cos\theta}$

$\tt:\implies R^2+S^2= P^2+Q^2+P^2+Q^2$

$\tt:\implies R^2+S^2=2P^2+2Q^2$

$\underset{\blue{\bf Hence\:Proved}}{\orange{\underbrace{\underline{\boxed{\red{\tt\longmapsto \overrightarrow{R^2}+\overrightarrow{S^2}=2\lgroup\overrightarrow{P^2}+\overrightarrow{Q^2}\rgroup }}}}}}$