Question

Three forces 2p 3p and 4p act along three sides of equilateral

Answer 1

Answer:

resolving all the forces horizontally

=-1.5p

resolving all the forces vertically

=-0.866p

resultant force =1.732p

Answer 2

Given:

Three forces 2p 3p and 4p act along three sides of equilateral.

To Find:

The magnitude of the resultant force

Solution:

All the angles of an equilateral triangle are equal

∠A = ∠B = ∠C = 60°

Sum of forces in the horizontal direction

F1 = 0 + 2P (-cos60) + 3P  ($\frac{-1}{2}$)

F1 = P +2P(-1/2) +3P (-1/2)

F1 = -1.5P

Sum of the forces in the vertical direction

F2 = 0 + 2P(sin60) + 3P(-sin60)

F2 = 2P(0.86) +3P(-0.86)

F2 = -0.866P

Now,

Magnitude of the resultant force = $\sqrt{F1+F2}$

                                                     R = $\sqrt{(-1.5)^{2}+(-0.866)^{2} }$

                                                     R = $\sqrt{3}$

                                                    R = 1.732P

Therefore, the magnitude of the resultant force = 1.732P.