Question

what is the angle between 4N force and 6N so that their resultant force is 4N​

Answer 1

Answer:

$If the resultant of two vectors P⃗ P→ and Q⃗ Q→ is R⃗ ,R→, then,</p><p></p><p>|R⃗ |=|P⃗ |2+2|P⃗ ||Q⃗ |cosθ+|Q⃗ |2−−−−−−−−−−−−−−−−−−−−−√,|R→|=|P→|2+2|P→||Q→|cos⁡θ+|Q→|2,</p><p></p><p> where θθ is the angle between the two vectors P⃗ P→ and Q⃗ .Q→.</p><p></p><p>It is given that |P⃗ |=4|P→|=4 N, |Q⃗ |=6|Q→|=6 N and |R⃗ |=10|R→|=10 N.</p><p></p><p>⇒10=42+2(4)(6)cosθ+62−−−−−−−−−−−−−−−−−√⇒10=52+48cosθ−−−−−−−−−−√.⇒10=42+2(4)(6)cos⁡θ+62⇒10=52+48cos⁡θ.</p><p></p><p>⇒100=52+48cosθ⇒48cosθ=48.⇒100=52+48cos⁡θ⇒48cos⁡θ=48.</p><p></p><p>⇒cosθ=1⇒θ=0o.⇒cos⁡θ=1⇒θ=0o.</p><p></p><p>⇒⇒ The angle between the two vectors is 0o.</p><p></p><p></p><p>$

Therefore :-

What is given above is the general method.In this particular case, we can , by inspection, observe that the sum of the magnitudes of the two vectors is equal to the magnitude of the resultant vector.This can only happen when the two vectors are collinear and in the same direction.⇒ The angle between the two vectors is 0°

Answer 2

Explanation:

$resultant= \sqrt{ ({p}^{2} } + {q}^{2} + 2pq \cos( \alpha ) )$

where p and q are two different vectors and

α is angle between them,

ATQ,

$4 = \sqrt{ ({4}^{2} } + {6}^{2} + 2 \times 4 \times 6 \times \cos( \alpha ) )$

on squaring both side,

$16 = 16 + 36 + 48 \times \cos( \alpha )$

$0 = 36 + 48 \times \cos( \alpha )$

$- 36 = 48 \cos( \alpha )$

$\cos( \alpha ) = - 36 \div 48$

$\cos( \alpha ) = - 3 \div 4$

$\alpha = \cos ^{ - 1} ( - 3 \div 4)$

the value of angle between them is approx 139 degree