Single Choice

Three concurrent forces of the same magnitude are in equillibrium. What is the angle between the force? Also name the triangle formed by the force as sides

A$$\;60^{\circ}$$ equilateral triangle
B$$\;120^{\circ}$$ equilateral triangle
Correct Answer
C$$\;120^{\circ},30^{\circ},30^{\circ}$$ an isosceles triangle
D$$\;120^{\circ}$$ an obtuse angled triangle

Solution

The concurrent forces always passing through a common point.
According to Lami's theorem , $$\dfrac{C}{\sin a}=\dfrac{B}{\sin b}=\dfrac{A}{\sin c}$$
As the magnitude of forces are same so: $$A=B=C$$
Thus, $$a=b=c$$
Also, $$a+b+c=360^o$$ or $$a=b=c=360/3=120^o$$
As all forces same magnitude and same angle between them so it will make equilateral triangle.


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