Question

the resultant of two forces 15 Newton and F is 20N inclined at 60 degree to the 15 Newton force find the magnitude and direction of force [ do this question by parallelogram law

Answer 1

Answer: The magnitude of resultant force is 30.41 N and is at an angle of 27.45 deg to the 20 N force.

Explanation:

Given that,

First force $f_{1}= 15 N$

Second force $f_{2}= 20 N$

We know,

The magnitude of the resultant force is

$R = \sqrt{f_{1}^2+f_{2}^2+2f_{1}f_{2}\ cos\theta}$

$R = \sqrt{15^{2}+20^{2}\times2\times15\times20\times cos\ 60^{0}}$

$R = \sqrt{15^{2}+20^{2}\times2\times15\times10}$

$R = \sqrt{925}$

$R = 30.41 N$

The direction of the resultant force is

$tan\theta = \dfrac{f_{1}sin\theta}{f_{1}+f_{2}cos\theta}$

$tan\theta = \dfrac{15\times sin\ 60^{0}}{15+20\ cos 60^{0}}$

$tan\theta = \dfrac{15\times\dfrac{\sqrt{3}}{2}}{15+20\times\dfrac{1}{2}}$

$tan\theta = 0.5196$

$\theta = tan^{-1}0.5196$

$\theta = 27.45^{0}$

Hence, The magnitude of resultant force is 30.41 N and is at an angle of 27.45 deg to the 20 N force.

Answer 2

Answer: resolution of vector

Explanation: