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I have a question regarding the answer from this topic Is the force of gravity always directed towards the center of mass?

One of the answers states the following problem.

"Three spherically symmetric bodies (or point masses if you can tolerate this) are at the three vertices of a 45°, 90°, 45° triangle, ABC. The masses of the bodies are: $m_A=m$, $m_B=M$, $m_C=2M$. Regard the bodies at B and C as a single body, BC; join them, if you like, by a light rod.

The centre of mass of body BC is at point P, 2/3 of the way between B and C.

But the pull due to BC experienced by $m$, at A, is not directed towards P, as one can easily show by vector addition of the forces due to B and C. [In this case the forces are of equal magnitude, so the resultant bisects angle BAC and clearly doesn't pass through P!] The reason for the discrepancy is the inverse square law of gravitation."

I do not quite get why the forces on A due to B and C are equal in magnitude. Distances AC and AB are the same (since the triangle is isosceles one), but since the mass of C is 2 times mass of B, the force due to C should be twice force due to B. So, after vector addition, mass $m$ is indeed attracted to the center of mass of B and C.

Where am I wrong?

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The triangle is right angled at B. So if AB = 1, AC = √2 , whereas mass at B is M and C is 2M, hence just apply inverse square law both forces should be equal.

As both the forces are equal, resultant force should be an angle bisector of vertex A. By basic theorem of Euclidean geometry, the angle bisector should cut the side BC in a ratio of AB : AC = 1 : √2 but the center of mass is clearly at a point whose distance is in ratio 1:2 .
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