Mechanical Engineering - Engineering Mechanics - Discussion

The resultant force is the same value of one of this either forces, so we can assume R=P, P1=P, P2=P.

Let using parallelogram therom,
R^2 = P1^2+P2^2+2P1P2COSθ
P^2 = P^2+P^2+2P^2COSθ
P^2 = 2P^2+2P^2COSθ
2P^2COSθ = -P^2
And θ=120°.

The two forces acting with the same magnitude and resultant too same magnitude. So the angle between each two forces is equal. So we can divide the circle into three equal segments. The angle of a circle is 360°. Then 360/3 = 120°each.

According to Lamis theorem : p/sinα+q/sinβ+r/sinγ.

To find the resultant

R= √(A2)+(B2)+2ABcosx

As given in the query, the magnitude of the force is the same and let it be f whose resultant be f too.

Substituting the variables in the above equation we get

f= √(f2)+(f2)+2(f2)cosx

Squaring on both sides we get

f2=2(f2)+2(f2)cosx
f2/f2 = 2 + 2 cos x
1 - 2 = 2 cosx
-1/2 = cos x
Therefore x=120°.

According to Lamis theorem, sine α = sine β = sine γ.

i.e. α + β + γ =360 so 360 /3 =120 =α = β= γ.