(Q&A) Torque question...

Q:-
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.

A:-
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;