Question

At what angle do the two forces (P+Q) and (P-Q) act so that resultant is root 3P​

Answer 1

Answer:

The angle must be of,

$cos^{-1} (\frac{1}{2} (1- \frac{Q^{2} }{(P+Q)(P-Q)}))$

Explanation:

Let the angle between them be "a",

Now using the parallelogram addition law,

We'll end up here writing the net force (resultant) as,

$\sqrt{(P+Q)^{2} + (P-Q)^{2} + 2(P+Q)(P-Q)cos(a) }$

Which according to question is equal to,

$\sqrt{3}P$

So the equation we'll have is,

$\sqrt{(P+Q)^{2} + (P-Q)^{2} + 2(P+Q)(P-Q)cos(a) }= \sqrt{3P}$

Squaring both sides gives,

${(P+Q)^{2} + (P-Q)^{2} + 2(P+Q)(P-Q)cos(a) }={3P^{2} }$

Solving it further,

$\sqrt{(P+Q)^{2} + (P-Q)^{2} + 2(P+Q)(P-Q)cos(a) }= \sqrt{3P}\\\\P^{2} +2PQ+Q^{2} +P^{2} -2PQ+Q^{2} +2cosa(P-Q)(P+Q)= 3P^{2} \\\\ 2Q^{2} + 2(P+Q)(P-Q)cos a = P^{2} \\\\Q^{2} +2(P+Q)(P-Q)cos a = P^{2}-Q^{2}\\\\Q^{2}+2(P+Q)(P-Q)cos a= (P+Q)(P-Q)$

Dividing both sides by,

(P+Q)(P-Q)

$\frac{Q^{2} }{(P+Q)(P-Q)} +2cos a =1$

exchanging sides gives,

$2cos a =1- \frac{Q^{2} }{(P+Q)(P-Q)} \\\\cos a= \frac{1}{2} (1- \frac{Q^{2} }{(P+Q)(P-Q)})\\\\a= cos^{-1} (\frac{1}{2} (1- \frac{Q^{2} }{(P+Q)(P-Q)}))$

Hence the angle.

Answer 2

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We know that,

Resultant = √(A²+B²+2ABcosФ)   where, A and B are the 2 vectors

                                                                       Ф is tha angle between A and B

In the given case,

A =P+Q

B = P-Q

Resultant = 3P

Ф = ?

Hence,

3P = √ [ (P+Q)² + (P-Q)² + 2(P+Q)(P-Q)cosФ ]  

Squaring both sides

(3P)² = (P+Q)² + (P-Q)² + 2(P+Q)(P-Q)cosФ

Using the algebraic identities,

9P² = P² + Q² + 2PQ + P² + Q² - 2PQ + 2(P²-Q²)cosФ

2PQ and -2PQ gets cancelled to 0,

9P² = 2P² + 2Q² + 2(P²-Q²)cosФ

2(P²-Q²)cosФ = 7P² - 2Q²

cosФ = (7P² - 2Q²) / 2(P² - Q²)

Ф = $cos^{-1}[\frac{7P^{2}-2Q^{2} }{2(P^{2} - Q^{2} )} ]$

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