Lets consider the radius of the circle = r where r=L/pie.Consider a infinitesimally small part of the semicircular wire d (thita) located at a angle of thita form one end. The mass of this Mrd(thita)/(pie)r= Md(thita)/(pie).Then the force of attraction on the mass m due to this part of the wire is (GmM(d thita)/pie)/(r^2). Integrating the sin component of this force with respect to thita.We get (Gm M/pie)/r^2)cos(thita) from 0 to pie.= (2Gm M/pie)/r^2) putting l/(pie) in place of r. We get (2Gm M)pie/(l^2))
Can you please sent me a diagram of the sum that you have used to solve it. please!!!
Here is the required image.
Three identical balls are placed on a frictionless
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