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.
Answers
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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