Question

the resultant of 2 forces of magnitude 12 newton and 3 newton on a body is 13.74 newton find the angle between 2 forces. what is the direction of resultant forces with respect to 12 newton force​

Answer 1

Answer:-

From the vector formula

$r = \sqrt{ {a }^{2} } + {b}^{2} + 2ab \cos( \alpha )$

${13.74}^{2} = {12}^{2} + {3}^{2} + 72 \cos( \alpha )$

$188.7 = 144 + 9 + 72 \cos( \alpha )$

$188.7 - 153 = 72 \cos( \alpha )$

$35.7 = 72 \cos( \alpha )$

approximately 35.7 is equal to 36.

$\cos( \alpha ) = 36 \div 72$

$\cos( \alpha ) = \frac{1}{2}$

$\alpha = 60 \: degrees$

This the angle between vector 12N and 3N.

Let beta is the angle between resultant and 12N.

$\tan( \beta ) = \frac{b \sin( \alpha ) }{a + b \cos( \alpha ) }$

$\tan( \beta ) = \frac{ \frac{3 \sqrt{3} {} }{2} }{12 + \frac{3}{2} }$

as

$\sin(30) = \frac{ \sqrt{3} }{2}$

$\cos(60) = \frac{1}{2}$

simplifying gives

$\tan( \beta ) = \frac{ \sqrt{3} }{9}$

or

$\tan( \beta ) = \frac{1}{3 \sqrt{3} }$

this implies beta is equal to tan inverse 1/3root 3.

which is equal to,

$\beta = 29.6 \: degrees$

# refer the attachment for figure.