Torque problem...help!!!!?

Question

a uniform meter stick supported at the 30-cm. mark is an equilibrium when an 1-kg rock is suspended at the 0-cm. compare the mass of the meter stick to the rock.

Answer 1

ml + .3Ml/2 = .7M(1 - l)/2; where m = 1 kg, l = .3 m, and M = ? the mass of the stick.  ml is the torque on the l = .3 m end from the mass.  .3Ml/2 is the torque on the l = .3 m side from the center of mass of 30% of the stick's mass.  .7M(1 - l)/2 is the opposing torque from the center of mass of 70% of the stick mass acting (1 - l)/2 = .35 m from the fulcrum.

Solve for M.  .7M(1 - l)/2 - .3Ml/2 = (1/2)M[.7 - .7l - .3l] = (1/2)M(.7 - l) = ml; so that M = 2ml/(.7 - l); m = 1 kg, l = .3 m...you can do the math.

The physics is that the torque on the LHS of the stick is offset by the torque on the RHS of the stick.  That's why the stick is balanced.  The trick is to recognize that the mass on the LHS of the stick also contributes to the torque and the stick's mass on the RHS is the only thing balancing both the mass m and the 30% of the stick mass on the LHS.  The final trick is to recognize that the centers of mass on either side of the stick are half-way from the fulcrum to the end points on both sides.  That's where the 1/2 factor comes into play.

Answer 2

balance of torqs is m1*a1 +m2*a2 = m3*a3, where m1=1kg rock, a1=30cm is arm of the rock, m2=p*a1 is mass of the shorter end of the stick, p=M/100 is linear density of the stick, M is mass of the stick, a2=a1/2 is center of mass of shorter end of the stick, m3=p*(100 –a1) is mass of longer end of the stick, a3=(100 –a1)/2 is center of mass of longer end of the stick;

thus 1*30 +p*30*15 = p*70*35, hence p=30/2000, hence M=1.5 kg;