Question

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

(i)
`\sf \sqrt{3 {p}^(2) + {q}^(2) }`

(ii)
`\sf \sqrt{2( {p}^(2) + {q}^(2) )}`

(iii)
`\sf \sqrt{ {p}^(2) + {q}^(2) }`





(hint : use law of cosin will be used)

please help especially in (iii) part I have been trying to solve it for hours ^^"

ty in advance!
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Answer 1

See below ~

Step-by-step 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 =
`(-(P^(2)+ Q^(2)) )/(2(P^(2)-Q^(2)) )`

`\theta = cos^(-1) (-(P^(2)+ Q^(2)) )/(2(P^(2)-Q^(2)) )`

Answer 2

Use resultant formula

`\boxed{\sf R=√(A^2+B^2+2ABcos\theta)}`

So

#1

A be p+q and B be p-q

`\\ \rm\Rrightarrow R=√(3p^2+q^2)`

`\\ \rm\Rrightarrow √((p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha)=√(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=(1)/(2)`

`\\ \rm\Rrightarrow \alpha=(\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=(\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 =(q^2-p^2)/(2(p^2-q^2))`

`\\ \rm\Rrightarrow \gamma=cos^(-1)\left((q^2-p^2)/(2(p^2+q^2))\right)`