Balancing the Plank
A 3.0-meter uniform plank is supported by a single pivot 1.2 meters from its left end. A 4.0-kilogram mass sits at the left end. A 6.0-kilogram mass must be placed somewhere on the plank so that the plank balances horizontally. One student begins by treating the plank as massless. The instructor points out that the plank actually has a mass of 2.0 kilograms spread evenly along its length, and asks whether the first answer is close enough to use. The 6.0-kilogram mass must rest on the plank itself, not beyond either end, and nothing may be clamped to the underside.
1. Find the balance position under the massless-plank assumption and confirm it is physically possible.
Read the worked answer
Take torques about the pivot. The 4.0 kg mass sits at the left end, 1.2 m to the left:
torque = 4.0 x 1.2 = 4.8 kg-m, counterclockwise.
Let the 6.0 kg mass sit a distance d to the right of the pivot:
6.0 x d = 4.8, so d = 0.80 m.
Position from the left end: 1.2 + 0.80 = 2.0 m.
Check the constraint: the plank runs from 0 to 3.0 m, and 2.0 m is on the plank with a meter to spare. Feasible.
Check the sense of the answer: the heavier mass must sit closer to the pivot than the lighter one, and 0.80 m < 1.20 m, which is what we expect.
2. Redo the problem with the plank's 2.0 kg mass included.
Read the worked answer
A uniform plank acts as if all its mass sits at its center: 1.5 m from the left end, which is 1.5 - 1.2 = 0.30 m to the right of the pivot.
Plank torque = 2.0 x 0.30 = 0.60 kg-m, clockwise - the same direction as the 6.0 kg mass.
Balance equation:
4.8 (counterclockwise) = 6.0 x d + 0.60 (clockwise)
6.0 x d = 4.2
d = 0.70 m right of the pivot, i.e. 1.9 m from the left end.
The plank helps tip the right side down, so the 6.0 kg mass must move toward the pivot to compensate. The correction is 0.10 m.
3. The student used the 2.0 m position anyway. What happens, and when does the 0.1 m error matter?
Read the worked answer
At 2.0 m, compute the net torque about the pivot:
- Counterclockwise: 4.0 x 1.2 = 4.8 kg-m
- Clockwise from the 6.0 kg mass: 6.0 x 0.8 = 4.8 kg-m
- Clockwise from the plank: 2.0 x 0.3 = 0.6 kg-m
- Net: 0.6 kg-m clockwise
The right end dips. The plank is not marginally off; it is off by the full weight of the plank acting at its center, which is exactly the term the student ignored. Ignoring a body's mass is only safe when that mass is small compared with the loads, and here it is one-fifth of the total.
When the 0.1 m matters:
- It always matters for a demo that must visibly balance. A 0.6 kg-m residual torque on a free pivot will rotate the plank.
- It matters more as the plank mass grows relative to the loads. Scaling the plank to 4.0 kg would triple the correction, not nudge it.
- It matters least when the plank is short and symmetric about the pivot, because then its own torque is near zero.
The judgment point: a 5 percent error in position is not automatically harmless. Convert any neglected term into a torque and ask whether it is small compared with the load torques - 0.6 against 4.8 is 12.5 percent, which is not.