Question

Two forces of 4N and 6N are inclined at an angle of 60° with each other . Find the magnitude and direction of resultant vecrors.

Answer 1

Answer:

, P and Q be the two forces having magnitudes 15 Newtons and 20 Newtons respectively. They are acting at an angle α = 60°.

Let, R be the resultant vector. Vector R acts at an angle θ with vector P.

So, |R^2| = 15^2 + 20^2 + 2 * 15 * 20 * cos 60°

Or, |R^2| = 225 + 400 + 300 = 925

Or, |R| = 30.4138 Newtons

So, tan θ = (20 * sin 60°) / {15 + 20 * cos 60°}

Or, tan θ = (10√3) / (15 + 10) = (10√3) / 25

Or, tan θ = (2√3) / 5 = 0.69282

Or, θ = 34.715°

So, magnitude of the resultant force = 30.4138 Newtons and the resultant force acts at an angle 34.715° with the 15 Newton magnitude force.

Explanation:

i hope this helps u

Answer 2

Given,

$F_{1}$ = 4N

$F_{2}$ = 6N

Angle of inclination (n) = 60°

To Find,

Magnitude and Direction of Resultant

Solution,

We know the formula for the magnitude of resultant of two vectors is

= $\sqrt{F_{1}^{2}+F_{2}^{2}+2*F_{1}*F_{2}*cos(n) }$

Substituting the given values we get,

= $\sqrt{4^{2}+6^{2}+2*4*6 *cos(60) }$

= $\sqrt{76}$

= 8.7177 N

For Direction,

tan (n) = $F_{2}$/$F_{1}$

tan (n) = 6/4

tan (n) = 1.5

    n    =  $tan^{-1}$(1.5)

    n    =  $56.30^{0}$

The magnitude and direction of the resultant force are 8.718 N and 56.30 degree respectively.