2022 A/L Physics Past Paper · Question 20 (English Medium)

Answer: (1) $\dfrac{\sqrt{3}}{4}$

Given information

What we know:

The mass of the block is $m = 1\ \text{kg}$. We take $g = 10\ \text{m s}^{-2}$, so its weight is $mg = 10\ \text{N}$.

The pull is $P = 4\ \text{N}$, at $\theta = 30^\circ$ above the horizontal.

The block is just about to slip: this is limiting equilibrium, and nothing accelerates.

What the question asks

We want: the coefficient of limiting friction $\mu$.

What each letter means:

$P$ ($\text{N}$): the pulling force.

$\theta$ (degrees): the angle of the pull above the horizontal.

$R$ ($\text{N}$): the normal reaction of the surface on the block.

$F$ ($\text{N}$): the limiting friction.

$\mu$ (no unit): the coefficient of limiting friction.

Relevant theory

The frictional force acting when the surfaces just begin to slip past one another is called the limiting frictional force.

limiting friction between surfaces $\propto$ normal contact force between surfaces

$F = \mu R$

For example, think of pulling a suitcase along the floor by its handle.

If you pull a little upward, the suitcase presses less on the floor, and it slides more easily. The sideways part of your pull is what drags it. The upward part only lifts a little of its weight.

So we split the pull into two parts, one along the floor and one straight up.

$$P_x = P\cos\theta \qquad P_y = P\sin\theta$$

In words: the part along the floor is $P\cos\theta$ and the part upward is $P\sin\theta$.

The block does not move up or down. So the upward forces balance the downward force.

$$R + P\sin\theta = mg$$

In words: the reaction plus the upward part of the pull equals the weight.

At the point of slipping the friction has its limiting value and balances the part along the floor.

$$F = P\cos\theta \qquad F = \mu R$$

In words: the limiting friction equals the horizontal part of the pull, and it is $\mu$ times the reaction.

Required equations

$R = mg - P\sin\theta$

$F = P\cos\theta$

$F = \mu R$

$\mu = \dfrac{P\cos\theta}{mg - P\sin\theta}$

Applying it to this question

All forces are in newtons, so $\mu$, a ratio of two forces, has no unit.

$2$

$P\sin 30^\circ = 4 \times \dfrac{1}{2}$

$8$

$R = 10 - 2$

$3.4641016$

$P\cos 30^\circ = 4 \times \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$

$F = 2\sqrt{3}$

$0.4330127$

$\mu = \dfrac{F}{R} = \dfrac{2\sqrt{3}}{8} = \dfrac{\sqrt{3}}{4}$

Reasoning and calculation

The reaction is smaller than the weight, $8\ \text{N}$, because the pull lifts the block a little.

The limiting friction balances $2\sqrt{3}\ \text{N}$. So $\mu = \dfrac{\sqrt{3}}{4}$. That is option (1).

Checks

$f / n$

Does it make sense?

The value $\dfrac{\sqrt{3}}{4} \approx 0.43$ is a normal coefficient for two dry surfaces.

The upward part of the pull makes the reaction smaller, $8\ \text{N}$ and not $10\ \text{N}$. Dividing by a smaller reaction makes $\mu$ larger than $\dfrac{2\sqrt{3}}{10}$, as we found.

Conclusion

$\mu = \dfrac{\sqrt{3}}{4}$

The reaction is $R = 8\ \text{N}$ and the limiting friction is $F = 2\sqrt{3}\ \text{N}$. So $\mu = \dfrac{F}{R} = \dfrac{\sqrt{3}}{4}$.

The answer is (1).

When the pull is slanted upward, the reaction is not $mg$. Subtract the upward part of the pull: $R = mg - P\sin\theta$.

Then put the horizontal part of the pull equal to $\mu R$.

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