A Plank, Two Bricks and a Learner Who Wants to Sit on the End
A carpentry club builds a bench for the school garden from a uniform wooden plank 3.0 m long and of mass 12 kg. It is laid flat across two concrete bricks, one 0.50 m from the left end and one 2.50 m from the left end. The plank is not fixed to the bricks; it simply rests on them, overhanging 0.50 m at each end. A learner of mass 40 kg wants to sit on the plank while painting a sign, and the club is not sure the arrangement is safe. The teacher says the plank will only tip if the learner sits too far outside the bricks, and asks the club to work out the limits on paper before anyone climbs on. Take g = 10 N/kg. Treat the plank as rigid and the learner as a point load, but remember that a brick can crush or slide, and that a person does not sit down gently.
1. The learner sits 0.75 m from the left end. Find the force carried by each brick and comment on whether the arrangement is comfortable.
Read the worked answer
Set up: plank weight W = 12 x 10 = 120 N acting at its centre, 1.50 m from the left end. Learner weight = 40 x 10 = 400 N at 0.75 m. Total load = 520 N.
Take moments about the left brick (0.50 m):
- Learner: 400 N at (0.75 - 0.50) = 0.25 m to the right -> 400 x 0.25 = 100 N m clockwise
- Plank: 120 N at (1.50 - 0.50) = 1.00 m to the right -> 120 x 1.00 = 120 N m clockwise
- Right brick: reaction R at (2.50 - 0.50) = 2.00 m to the right -> 2.00R anticlockwise
So 2.00R = 220, giving R = 110 N on the right brick.
Vertical balance: left brick = 520 - 110 = 410 N.
Check: 410 + 110 = 520 N, correct.
Comment: the left brick carries 410 N (about 41 kg) while the right brick carries only 110 N. The load is badly shared. The arrangement stands, but nearly all the load is on one brick sitting on soil, and the right end is close to lifting - any shift of the learner to the left will unload it further. A sensible club would ask the learner to sit between the bricks, near 1.5 m, where each brick carries close to 260 N.
2. Work out how far to the left of the left brick the learner could sit before the plank tips, and explain what the answer means in practice.
Read the worked answer
Tipping condition: the plank pivots about the left brick when the right brick carries zero force. At that instant only the plank's own weight and the learner act.
Take moments about the left brick:
- Plank weight 120 N acts 1.00 m to the right of the brick -> 120 N m, which holds the right end down (restoring moment).
- Learner 400 N at distance x to the left of the brick -> 400x N m, which lifts the right end.
Tipping begins when 400x = 120, so x = 120/400 = 0.30 m.
In practice: the left brick is 0.50 m from the end, so 0.30 m beyond it leaves the learner's centre only 0.20 m from the end of the plank. The limit is the far edge of a small 0.20 m strip.
Judgement: 0.30 m is a mathematical limit, not a working limit. It assumes the learner sits perfectly still, sits down slowly, and that the plank is rigid and horizontal. Sitting down suddenly can briefly double the effective load, which would cut the safe distance roughly in half - about 0.15 m. The recommendation is therefore not to use the overhang at all: keep the learner between the two bricks, or add a third brick and clamp the plank so that the calculation is no longer the only thing preventing an accident.
3. The teacher says the calculation is useful but should not be treated as a guarantee. Give three reasons why the real plank may behave differently from the model.
Read the worked answer
1. The learner is not a point load and does not load the plank gently. A 40 kg learner plonking down can produce a momentary force well above 400 N. The model uses a static load, so every reaction force calculated is a lower bound.
2. The plank bends. A real wooden plank sags between the bricks, so the ends tilt up and the contact with a brick becomes an edge rather than a flat surface. Bending also means the plank's own weight is not exactly at 1.50 m once it curves, and a long overhang can snap rather than tip.
3. The supports are not ideal. A brick on soft ground can sink on one side or slide sideways; the brick itself can crack under a concentrated load. The model assumes frictionless, immovable, uncrushable supports - none of which is true.
Also worth noting: the plank may not be uniform (a knot or a join alters where its weight acts), and the learner's mass was taken from a guess rather than a balance.
Judgement: the calculation is valuable because it shows the arrangement is badly balanced and near the tipping point, but it should be used to change the design, not to license sitting on the overhang. Safe practice adds a margin and a physical fixing.