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.)

Question

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.)

Answer 1

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

Answer 2

To determine the mass of the rod, we need to consider the principle of rotational equilibrium. In rotational equilibrium, the sum of the clockwise torques must be equal to the sum of the counterclockwise torques.

Let's denote the mass of the rod as M (in grams). We will use the given values A, B, and C, which represent the distances in centimeters. Assume the center of the rod is at point O.

In Figure a, the torques can be calculated as follows:
Clockwise torques: (2m + 2.0 grams) × B
Counterclockwise torques: m × (A - B)

Since the rod is balanced, the sum of these torques is equal to zero:
(2m + 2.0 grams) × B = m × (A - B) ---- Equation 1

In Figure b, the torques can be calculated as follows:
Clockwise torques: m × C
Counterclockwise torques: M × (A - C)

Again, since the rod is balanced, the sum of these torques is equal to zero:
m × C = M × (A - C) ---- Equation 2

Now we have two equations (Equation 1 and Equation 2) with two unknowns (M and m).

Let's substitute the given values into the equations:
Equation 1: (2m + 2.0 grams) × 32.6666666666667 cm = m × (98 cm - 32.6666666666667 cm)
Equation 2: m × 65.3333333333333 cm = M × (98 cm - 65.3333333333333 cm)

We can simplify the equations:
(65.3333333333333 - 21.3333333333333) m = 2 (32.6666666666667 - 0.6666666666667) M

Now we can solve for M:
44 m = 2 × 32 M
M = (44/2) m

Therefore, the mass of the rod is equal to half the mass of the object at the other end of the rod.
M = 22 m.