Question

Forces equal to 3P, 7P and 5P act along the sides AB, BC and CA respectively of an equilateral

triangle ABC. Find the magnitude direction and line of action of the resultant.​

Answer 1

Given : Forces equal to 3P, 7P and 5P act along the sides AB, BC and CA respectively of an equilateral triangle ABC.

To Find :  the magnitude direction and line of action of the resultant.​

Solution:

Let say AB is along X axis

Then BC is at 120°

CA is at 240°

Along x axis  = 3P  + 7Pcos120°  + 5PCos240°

= 3P + 7P(-0.5) + 5P(-0.5)

= 3P - 3.5P - 2.5P

= - 3P

Along Y axis  = 0   + 7Psin120°  + 5Psin240°

= 0 + 7P(√3 / 2) + 5P(-√3 / 2)

=   2P(√3 / 2)

= P√3

Along x axis  = - 3P

Along Y axis  = P√3

Magnitude = √(-3P)²  + P(√3)²   = 2√3P

Angle = 180° + tan⁻¹ (P√3 /- 3P)  = 150°  Hence  angle bisector of B

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