2018 A/L Physics Past Paper · Question 28 (English Medium)

Answer: (5) $\dfrac{m}{4}$

Given information

The plank is $5l$ long. Its mass is $5m$.

It is uniform. So its weight acts at the middle, $2.5l$ from the left end.

The supports are at $l$ and $3l$ from the left end.

The painter has mass $m$. The bucket has mass $M$.

What the question asks

We want: the biggest bucket mass $M$ that lets the painter stand anywhere without the plank tipping over.

Relevant theory

The algebraic sum of the moments of the three forces about any point in their plane should be zero.

The moment of a force is the turning effect produced by the force.

Just before the plank tips, it rests on one support only.

The other support pushes with zero force.

Then the turning effects (moments) about the support it rests on are equal.

Required equations

At that moment three forces act:

The weight of the plank, $5mg$, at the middle.

The painter and bucket, $(m + M)g$, at the end.

The push $R$ of the support it rests on.

The other support pushes with zero force.

$(m + M)g \times d_{end} \le 5mg \times d_{centre}$

Applying it to this question

$2l$

$d_{end} = 5l - 3l$

$0.5l$

$d_{centre} = 3l - 2.5l$

$d_{end,L} = l$

$1.5l$

$d_{centre,L} = 2.5l - l$

Reasoning and calculation

$(m + M) \times 2l \le 5m \times 0.5l$

$2m + 2M \le 2.5m$

$\tfrac{m}{4}$

$M \le \dfrac{2.5m - 2m}{2}$

$(m + M) \times l \le 5m \times 1.5l$

$6.5m$

$M \le 7.5m - m$

Both ends must be safe.

The right end is the harder one, because $2l$ of plank sticks out there.

So $M \le \dfrac{m}{4}$.

Checks

$(5 * m * 0.5 * l - m * 2 * l) / (2 * l)$

Check the turning directions.

About the right support, the plank's weight turns it one way. It keeps the plank steady.

The painter and bucket turn it the other way. They try to tip it.

They turn in opposite directions, so comparing them is right.

Does it make sense?

With no bucket: $2ml < 2.5ml$. So the painter alone can walk to the end.

Only $0.5ml$ is left over. So only a small bucket fits: $M \le m/4$.

The answer is reasonable.

Conclusion

$M_{max} = \frac{m}{4}$

At the right end:

$$(m + M) \times 2l \le 5m \times 0.5l$$

So $M_{max} = \dfrac{m}{4}$.

The answer is option (5).

Checking only one end. Or forgetting the painter's own mass.

Check both ends.

Use the end that sticks out more (here the right end, $2l$).

Painter and bucket together weigh $(m + M)g$.

Explore

The painter starts at the left end of the plank with the bucket.

As they walk to the right, the reaction of the left support gets smaller.

At the right end that reaction becomes zero: the plank is about to tip.

There the moment of the plank’s weight about the right support equals the moment of the painter and bucket; a bucket heavier than $m/4$ tips the plank.