Question

Three particles each of mass m are placed at the corners of equilateral triangle of side l. The force due to this system of particle on another particle of mass m placed at the midpoint of any side is

Answer 1

eqaul to thta side of mod point

Answer 2

Dear student,

Given :-

• masses of particles = m
• mass of the particle kept at the midpoint = m
• length of sides = l
• angle between two sides = 60°

To find :-

• the force due to this system of particle on another particle of mass m placed at the midpoint of any side

Solution :-

The two masses m acting on the mass m' kept at the midpoint on the bottom line of the triangle cancel out each other for they act in equal and opposite direction with equal magnitudes. We consider only the forces between mass m' and the mass at the uppermost vertex of the equilateral triangle.

We join the the point m' and m with a line. For the value of that line, we have;

          sin 60° = perpendicular/ hypo.

          √3/ 2 = P/ L

           P = √3 L/2

So, we have the length of the line denoted by p as √3 L/2. Now, we find the force acting between the mass m' and m. We know that;

     F = G m₁m₂/r²

Putting the given values in the above formula, we obtain;

    F = G mm/ (√3 L/2)²

   F = 4 Gm²/ 3 L

Hence, the force due to the system of particle on another particle of mass m placed at the midpoint of any side is  4 Gm²/ 3 L.