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Three equal masses m are placed at the three corners of an equilateral triangle of side a. The force exerted by this system on another particle of mass m placed at the mid-point of a side. |
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Answer» Solution :If m is placed at the mid point of a side .AB. Then `vec(F_(OA)) = (4Gm^(2))/(a^(2))` in OA DIRECTION and `vec(F_(OB)) = (4Gm^(2))/(a^(2))` in O1B direction SINCE both are equal in OPPOSITE, cancel each other `vec(F_(OC)) = (Gm^(2))/(((sqrt(3))/(2)a)^(2)) = (4Gm^(2))/(3a^(2))` in OC direction `therefore` Net gravitational FORCE on `m = (4Gm^(2))/(3a^(2))` |
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