Question

Three forces equal to 3P, 5P and 7p act simultaneously along three sides of an equilateral triangle
of side 'a'. Find the magnitude, direction and position of the resultant force.​

Answer 1

Answer:

Explanation:

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

I hope it helps..! :)

Answer 2

Answer:

Magnitude is $2\sqrt{3} P$ and angle is 150°.

Explanation:

It is given that  Forces equal to 3P, 7P and 5P act along the sides AB, BC and CA respectively of an equilateral triangle ABC.

We need to determine the magnitude direction and line of action of the resultant.​

Assume AB is along X axis.

BC is at 120°.

So, CA will be at 240°

Along C axis, the equation will = 3P  + 7Pcos120°  + 5PCos240°.

= 3P - 3.5P - 2.5P= - 3P

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

=   2P(√3 / 2)

= P√3

Hence, magnitude will be $=\sqrt{(-3P)} ^{2} +\sqrt{(P\sqrt{3} )} ^{2} =2\sqrt{3} P$

Angle is 180° + tan⁻¹ (P√3 /- 3P)  = 150°

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