# The direction Of three forces 1N,2N,3N are acting at a point are parallel to the sides of an equilateral Triangle taken in order. The magnitude of their resultant??

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by syke

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by syke

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The three forces will be along three directions separated by 120 deg. from each other drawn at a point P. Let them be PA of 1 N. PB = 2 N and then PC = 3 N..

Let us draw PA horizontally along positive x axis.

PB is at 30 deg. anticlockwise from + y axis.

PC will be at 30 deg. clockwise from - y axis...

Resolving components along x axis and adding:

= 1 N + 2 N * Cos 120 + 3 N * Cos 240

= 1 - 2 * 1/2 - 3 /2 = - 3/2 Newtons

Resolving components along y axis and adding

= 0 N + 2 N * √3 /2 - 3 N * √3/2 = - √3/2 Newtons

So resultant is R = -1.5 i - √3 /2 j where i and j are unit vectors.

*magnitude = √[ 1.5² + 3/4 ] = √3 Newtons.*

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__The calculations are much simpler, if you take 2 Newtons force/vector along x axis and the other two at 120 deg. on either side to that.... __

Let us draw PA horizontally along positive x axis.

PB is at 30 deg. anticlockwise from + y axis.

PC will be at 30 deg. clockwise from - y axis...

Resolving components along x axis and adding:

= 1 N + 2 N * Cos 120 + 3 N * Cos 240

= 1 - 2 * 1/2 - 3 /2 = - 3/2 Newtons

Resolving components along y axis and adding

= 0 N + 2 N * √3 /2 - 3 N * √3/2 = - √3/2 Newtons

So resultant is R = -1.5 i - √3 /2 j where i and j are unit vectors.

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