Physics, asked by 276jainpalak, 9 months ago

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

Answers

Answered by Ayush4101
4

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

Answered by RitaNarine
0

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.

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