# Which of the following sets of forces acting simultaneously on a particle keep it in equilibrium?...

## Question:

Which of the following sets of forces acting simultaneously on a particle keep it in equilibrium?

a) 3N, 5N, 1ON

2) 4N, 7N, 12N

3) 2N, 6N, 3N

4) 5N, 8N, 1N

## Balanced Forces:

When we have three forces acting simultaneously on a particle, the particle is in dynamic or static equilibrium if the three forces are in balance, the geometry of the forces is such that they form a closed triangle or polygon in general. Only the sets that are capable of this can establish equilibrium at certain angles.

The forces are in equilibrium if they can make a triangle. The geometric condition of a triangle is that the longest side must be less than the sum of the two shorter sides. Let's test that with the sets.

a) 3 N, 5 N, 10 N

{eq}\begin{align*} 3\ \text{N}+ 5\ \text{N} = 8\ \text{N} < 10\ \text{N} \end{align*} {/eq}

Not possible since the longest side is bigger.

b) 4 N, 7 N, 12 N

{eq}\begin{align*} 4\ \text{N}+ 7\ \text{N} = 11\ \text{N} < 12\ \text{N} \end{align*} {/eq}

Not possible since the longest side is bigger.

c) 2 N, 6 N, 3 N

{eq}\begin{align*} 2\ \text{N}+ 3\ \text{N} = 5\ \text{N} < 6\ \text{N} \end{align*} {/eq}

Not possible since the longest side is bigger.

d) 5 N, 8 N, 1 N

{eq}\begin{align*} 5\ \text{N}+ 1\ \text{N} = 6\ \text{N} < 8\ \text{N} \end{align*} {/eq}

Not possible since the longest side is bigger.

None of the forces can keep a particle in equilibrium