Yes you indeed have permission to write 32.66666666666666666666666666666 as 1/3 of 98.
Masses m and 2m are balanced because the m one is twice the distance from the pivot as the 2m one, the two producing equal torque.
The rod is then only balanced by the 2.0 g mass.
Centre of mass of rod is only half the distance from the pivot compared to the 2.0 g (L/6 vs L/3), hence to compensate it must have mass 2 * 2.0 g.
Answer is 4g
A thin uniform rod 98 cm long is balanced 32.6666666666667 cm from one end when an object whose mass is (2m + 2.0 grams) is at the end nearest the pivot and an object of mass m is at the opposite end (Figure a). Balance is again achieved if the object whose mass is (2m + 2.0 grams) is replaced by the object of mass m and no object is placed at the other end of the rod (Figure b). Determine the mass of the rod. (Assume A = 98 cm, B = 32.6666666666667 cm, and C = 65.3333333333333 cm.)
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