A uniform wire with mass M and length L is bent into a semicircle. Find the magnitude and direction of the gravitational force this wire exerts on a point with mass m placed at the center of curvature of the semicircle.

• The mass of the uniform wire is M.
• The length of the uniform wire is L.
• The mass of the point is m.

The law illustrates that the force exerted by a mass is directly proportional to the product of the masses and inversely proportional to the square of the distances of the masses.

The product of the masses multiplied by the gravitational constant and divided by the square of the distances gives the force exerted on the point by the wire.

From Newton’s law of gravitation, the force exerted by the sphere is expressed as:

$F=G\frac{m_1{m}_2}{r^2}$

Here, F is the force exerted by the sphere, G is the gravitational constant, $m_1$and$m_2$are the mass of the object and the sphere are r is the distance between the objects.

The free-body diagram of the uniform wire has been described below-

where L is the length of the mass and R is the radius of the semi-circle.

As we have bent the wire into the infinitesimal pieces, we will have the differential

mass:

$dM=\frac{M}{L}Rd\theta$

Here,$\theta$is the angle of the semi-circle.

Due to the symmetry of the net force in the positive Y direction, the total force is being expressed as:

Integrating the equation on both sides after substituting the values will result in,

$$\begin{gathered} \int dF=\int G\frac{mdM}{R^2}\sin \theta \\ F={\int }_0^{n}G\frac{mMR}{R^{2}L}\sin \theta d\theta \\ F=G\frac{mM\mathrm{\pi }}{L^2}{\int }_0^n\sin \theta d\theta \\ F=G\frac{2\mathrm{\pi mM}}{L^2} \end{gathered}$$

Hence, the gravitational force exerted by the wire is$F=G\frac{2\mathrm{\pi mM}}{L^2}$and is in the positive direction.