Question

At what forces P + Q and p - Q act so that the resultant √2p^2+Q^2

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Answer 1

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Q1) At what forces P+Q and P-Q

act so that the resultant :

$\sqrt{2 {p}^{2} + {q}^{2} }$

$\huge{\blue{\pink{Answer}}}$

$\: \pink \bigstar \boxed {\bf \red{ \: by \: using \: parallelogram \: of \: law \: of \: addition \: of \: two \: vectors}}$

$\implies\bf{R^2=\sqrt{P^2+Q^2+2PQcos\theta}}$

$\: \implies \bf{r = \sqrt{ {(p}^{2} + {q}^{2} + 2pqcos \theta) }}$

$\implies \bf{ \sqrt{ {2p}^{2} + q} = \sqrt{ {(p + q)}^{2} + {(p - q)}^{2} + 2(p + q)(p - q)cos \theta}}$

$\implies \bf{2 {p}^{2} + q = {(p + q)}^{2} + {(p - q) }^{2} + 2(p - q)(p + q)cos \theta}$

$\: \implies \bf{2 {p}^{2} + q = {p}^{2} + {q}^{2} + 2pq + {p}^{2} + {q}^{2} - 2pq + 2(p(p - q) + q(p - q)cos \theta}$

$\implies \bf{ {2p}^{2} + {q}^{2} = {p}^{2} + {q}^{2} + {p}^{2} + {q}^{2} + 2( {p}^{2} - pq + pq - {q}^{2} )cos \theta}$

$\implies \bf{2 {p}^{2} + {q}^{2} - {2p}^{2} - {2q}^{2} + 2 {p}^{2} + 2{q}^{2} cos \theta}$

$\implies \bf{ - {q}^{2} +( {2p}^{2} + {2q}^{2}) cos \theta}$

$\: \implies \boxed{\bf{cos \theta = \frac{ {q}^{2} }{2( {p}^{2} + {q}^{2}) } }}$

$\: \bigstar \bf{at \: \theta = 0}$

$\: \: \implies \bf{ \cos(0 \degree) = \frac{ {q}^{2} }{2 {p}^{2} + {2q}^{2} } }$

$\: \: \implies \bf{1 = \frac{ {q}^{2} }{ {2p}^{2} + {2q}^{2} } }$

$\: \implies \bf{2 {p}^{2} + {2q}^{2} = {q}^{2} }$

$\implies \bf{2 {p}^{2} + {2q}^{2} - {q}^{2} = 0}$

$\: \implies \bf{2 {p}^{2} + {q}^{2} = 0}$

$\: \implies \bf{2 {p}^{2} = - {q}^{2} }$

$\: \implies \bf{2p = q}$

$\: \implies\boxed{ \bf{p = \frac{q}{2} }}$