Help Gravitational Force Question.

[eco742]

A mass M is ring shaped with radius r. A small mass m is placed at a distance x along the ring's axis. Show that the gravitational force on the mass m due to the ring is directed inward along the axis and has magnitude

F= GMmx/(x^2+r^2)^3/2

Hints:
-Think of the ring as made up of many small point masses dM
-Sum over the forces due to each dM
-Use symmetry

I am just confused on where to begin! Any help would be greatly apperciated!

 

[housemartin]

draw a ring (or a circle) on paper, take one little part of it, the force of that part, which mass is dM is F = GmdM/(L^2), and L is distance from dM to m. You can find this distance using Pythagoras theorem (L^2 = x^2 + r^2). Also notice, that the little element dM has his opposite element on the other side of the circle, which exerts same force (in magnitude) as dM. Draw those two forces in other paper ;] then you maybe notice that component of those forces along rings radius cancels, and stays only the component along rings axis if you sum them.
hope this help a little, I am not very good at writing thoughts ;]

 

[jack action]

To make the problem easier, imagine that instead of a ring you have 4 different masses of mass M/4, at a distance r from the axis. Find the resultant gravity force for each mass M with the mass m at a distance x (don't forget a force is a vector). Then separate all those forces into 2 components: one along the axis, the other perpendicular to it. You will find that all components perpendicular to the axis will cancel each other and the net resultant force will be the sum of all the ones parallel to the axis.

 

[eco742]

Great, thanks that was helpful!

 

[rdl]

Hello,

Thank you for reaching out for help with this gravitational force question. I will do my best to guide you through the steps to solve it.

Firstly, it is important to understand the concept of gravitational force. According to Newton's law of universal gravitation, the force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This can be represented by the equation F = GmM/r^2, where G is the gravitational constant.

Now, let's break down the problem. We have a ring with mass M and radius r, and a small mass m placed at a distance x along the ring's axis. Our goal is to find the gravitational force on the mass m due to the ring. To do this, we can think of the ring as made up of many small point masses, each with a mass dM.

Next, we need to consider the forces acting on the mass m due to each of these small point masses. Since the mass m is placed at a distance x along the ring's axis, we can assume that the forces due to the point masses on either side of m will cancel out (due to symmetry). This means that the only forces we need to consider are the ones acting on m from the point masses directly above and below it on the ring.

We can now use the equation F = GmM/r^2 to calculate the force due to each of these point masses. However, we need to take into account the distance between the point mass and the mass m, which is given by the Pythagorean theorem as (x^2 + r^2)^1/2. This distance will be the same for all the point masses on the ring.

Since we are considering all the point masses on the ring, we need to sum up the forces due to each of them. This can be represented by an integral, as follows:

F = ∫ (Gm dM)/(x^2 + r^2)^1/2

where the integral is taken over the entire ring.

Now, we need to substitute the value of dM in terms of M and r. Since the ring is uniformly distributed, we can write dM = Mdl, where dl is the length of a small element of the ring. We can also represent dl in terms of the angle θ, as dl = r dθ.

Substituting these values, we