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Three equal masses of m kg each are fixe...

Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC.
(a) What is the force acting on a mass 2m placed at the centroid G of the triangle?
(b) What is the force if the mass at the vertex A is doubled ?
Take AG = BG = CG = 1 m

Text Solution

Verified by Experts

(a) The angle between GC and the positive x-axis is `30^(@)` and so is the angle between GB and the negative x-axis. The individual forces in vector notation are
`F_(GA) = (Gm(2m))/(1) hatj`
`F_(GB) = (Gm(2m))/(1) ( - hati cos 30^(@) - hatj sin 30^(@))`
`F_(GC) = (Gm(2m))/(1) ( + hati cos 30^(@) - hatj sin 30^(@))`
From the principle of superposition and the law of vector addition, the resultant gravitational force `F_(R)` on (2m) is
`F_(R) = F_(GA) + F_(GB) + F_(GC)`
`F_(R) = 2Gm^(2) hatj + 2Gm^(2) ( - hati cos 30^(@) - hatj sin 30^(@))`
`+ 2Gm^(2) (hati cos 30^(@) - hatj sin 30^(@)) = 0 `
Alternatively. one expects on the basis of symmetry that the resultant force ought ot the zero.
(b) Now if the mass at vertex A is doubled then
`F_(GA) = (G2m.2m)/(1) hatj = 4Gm^(2) hatj`
`F_(GB) = F_(GB) and F_(GC) = F_(GC)`
`F_(R)= F_(GA) + F_(GB) + F_(GC)`
`F_(R) = 2 GM^(2) hatj`
For the gravitational force between an extended object (like the earth) and a point mass, Eq. (8.5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in the same direction. We have to add up these forces vectorially for all the point masses in the extended object to get the total force. This is easily done using calculus. For two special cases, a simple law results when you do that :
(1) The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell. Qualitatively this can be understood as follows: Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components prependicular to this line cancel out when summing over all regions of the shell leaving only a resultant force along the line joining the point to the centre. The magnitude of this force works out to be as stated above.
(2) The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero. Qualitatively, we can again understand this result. Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely
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