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'three parallel forces P,Q & R act at the vertices A,B & C of a triangle and as proportional to a,b,c show that their resultant forces through in

centre of the triangle?'
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Three forces P, Q and R are acting on A, B and  C respectively.
P, Q and R are parallel.
P/a = Q/b = R/c = k

Net force through the center of the triangle = P+Q+R = P(1+Pb/a+Pc/a)
                 = (a+b+c)P/a  = (a+b+c) k

Net force is proportional to the perimeter of the triangle. It is in the same direction as the parallel forces P, Q and R.

As there are multiple forces acting on the triangle at different points, there will be a net Torque on the triangle and So  the triangle ABC will rotate.

Let the angle made by P, Q and R with the side is Ф. Then the net torque about the axis perpendicular to the plane of ABC and passing through G (centroid) is :

    = d_P Χ P + d_R Χ Q + d_R Χ R   ,   where       d_P, d_Q and d_R are perpendicular lengths from G onto P, Q and R respectively.
    net torque = [ P (a + b - 2c) Sin Φ - R (b + c - 2 a) Sin (B-Φ) ] / 3 

If the triangle is free to move and rotate, then it will translate with the net force and the instantaneous torque is given by above formula.

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