Question

The resultant of two forces 3p and 2p is R. If the first force is doubled then the resulting is also doubled. find the angle between two forces.

Answer 1

angle is 120°. Tell me if it's correct or not?

Answer 2

120° is the “angle between the two forces”.

Given:

a=3p; b=2p

To find:

We have to find the angle between two forces

Solution:

We have

Resultant, $R=\sqrt{\left(a^{2}+b^{2}+2 a b \cos x\right)}$

By the” triangular law of vector addition”,

We get

$r=\sqrt{\left\{(3 p)^{2}+(2 p)^{2}+2 \times 3 p \times 2 p \times \cos x\right\}}$

$r=\sqrt{\left(9 p^{2}+4 p^{2}+12 p^{2} \cos x\right)}$

by simplification we can get,

$r=\sqrt{\left(13 p^{2}+12 p^{2} \cos x\right)}$

According to question we get

$2 r=\sqrt{\left\{(6 p)^{2}+(2 p)^{2}+2 \times 6 p \times 2 p \times \cos x\right\}}$

$2 r=\sqrt{\left(36 p^{2}+4 p^{2}+24 p^{2} \cos x\right)}$

$2 r=\sqrt{\left(40 p^{2}+24 p^{2} \cos x\right)}$

$2\left\{\sqrt{\left(13 p^{2}+12 p^{2} \cos x\right)}\right\}$

$=\sqrt{\left(40 p^{2}+24 p^{2} \cos x\right)}$

$=4\left(13 p^{2}+12 p^{2} \cos x\right)$

$=\left(40 p^{2}+24 p^{2} \cos x\right)$

$52 p^{2}+48 p^{2} \cos x=40 p^{2}+24 p^{2} \cos x$

By simple simplification we have the above expression as

$12 p^{2}=-24 p^{2} \cos x$

$\cos x=-\frac{1}{2}$

$x=\cos ^{-1}\left(-\frac{1}{2}\right)$

x = 120°

Therefore the angle between the forces is 120°.