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Answer :

mg, mg and 2 mg Solution :

Since `DC = a sqrt(3//2) and DE = a//2` it follows that `angleCDE = 90^(@)`. Resolve the force P = mg into two components in the directions CD and CE (Fig.). We obtain N = 2P, `F= P sqrt(3)`. Now resolve <br> <br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/ARG_AAP_PIP_PHY_C02_E01_008_S01.png" width="80%"> <br> the force F into components directed along the rods BC and AC (Fig.). Since in this case the parallelogram of forces is a rhombus with an apex angle of `60^(@)`, it follows that <br> `F_(1)=F_(2) = (F)/(2 cos 30^(@)) = (P sqrt(3))/(2 sqrt(3//2)) =P`