Question

At what angle two forces P + Q and (P - Q) act so that their resultant is :

Answer 1

Use resultant formula

$\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}$

So

#1

A be p+q and B be p-q

$\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2$

$\\ \rm\Rrightarrow 2cos\alpha=1$

$\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}$

$\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}$

#2

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)$

$\\ \rm\Rrightarrow 2cos\beta=0$

$\\ \rm\Rrightarrow cos\beta=0$

$\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}$

#3

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2$

$\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)$

$\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}$

$\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)$

Answer 2

Answer:

See below ~

Explanation:

Here, let's apply the resultant vector formula :

$R = \sqrt{A^{2} + B^{2} + 2ABcos\theta}$

Part (i) :

√3P² + Q² = √(P + Q)² + (P - Q)² + 2(P + Q)(P - Q)cosθ

3P² + Q² = P² + Q² + 2PQ + P² + Q² - 2PQ + 2(P² - Q²)cosθ

2(P² - Q²)cosθ = P² - Q²

cosθ = 1/2

θ = π/3 or 60°

Part (ii) :

√2(P² + Q²) = √(P + Q)² + (P - Q)² + 2(P + Q)(P - Q)cosθ

2(P² + Q²) = P² + Q² + 2PQ + P² + Q² - 2PQ + 2(P² - Q²)cosθ

2(P² - Q²)cosθ = 0

cosθ = 0

θ = π/2 or 90°

Part (iii) :

√P² + Q² = √(P + Q)² + (P - Q)² + 2(P + Q)(P - Q)cosθ

P² + Q² = P² + Q² + 2PQ + P² + Q² - 2PQ + 2(P² - Q²)cosθ

2(P² - Q²)cosθ = -P² - Q²

cosθ = $\frac{-(P^{2}+ Q^{2}) }{2(P^{2}-Q^{2}) }$

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