2023 A/L Physics Past Paper · Question 44 (English Medium)

Answer: (3) $\dfrac{1 - \sin\theta}{\cos\theta}$

What the question asks

We want: the coefficient of dynamic friction $\mu$ between the slope and the mass $M$.

What each letter means:

$M$ ($\text{kg}$): the mass of each of the two equal masses.

$\theta$: the angle of the slope to the horizontal.

$T$ ($\text{N}$): the pull in the string.

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

$F_d$ ($\text{N}$): the dynamic friction on the mass.

$\mu$: the coefficient of dynamic friction.

Relevant theory

The dynamic friction acting on an object is also proportional to the normal contact force between surfaces. If the dynamic friction is $F_d$ and normal contact force is R,

Newton's first law states that a resultant force can change the state of motion of an object. Newton's second law states about this resultant force.

The first quote says that the dynamic friction on a body is proportional to the normal contact force, $F_d = \mu_d R$. The second quote gives Newton's second law, $F = ma$. For a body at a steady velocity, $a = 0$, so the forces along the line of motion are balanced.

For example, push a box up a ramp at a steady speed with a rope. The rope pull balances the part of the weight along the ramp plus the friction. The rougher the ramp, the harder you pull.

Required equations

$T = Mg$

$R = Mg\cos\theta$

$F_d = \mu R = \mu Mg\cos\theta$

$T = Mg\sin\theta + \mu Mg\cos\theta$

$\mu = \dfrac{1 - \sin\theta}{\cos\theta}$

Reasoning and calculation

The hanging mass:

It moves at a steady speed, so its acceleration is zero and the forces on it balance. The string pulls it up with $T$ and its weight $Mg$ pulls it down.

So $T = Mg$.

$T = Mg$

$T - Mg = 0$

The mass on the slope:

The string pulls it up the slope with $T$, so it moves up the slope. The dynamic friction opposes the motion, so it acts down the slope. The weight part along the slope, $Mg\sin\theta$, also acts down the slope.

Across the slope there is no motion, so $R = Mg\cos\theta$ and $F_d = \mu Mg\cos\theta$. Along the slope the speed is steady, so $T = Mg\sin\theta + \mu Mg\cos\theta$.

$Mg\cos\theta$

$R = Mg\cos\theta$

$\mu Mg\cos\theta$

$F_d = \mu R$

$Mg = Mg\sin\theta + \mu Mg\cos\theta$

$T = Mg\sin\theta + \mu Mg\cos\theta$

$\mu = \dfrac{1 - \sin\theta}{\cos\theta}$

$\mu \cos\theta = 1 - \sin\theta$

$0.4 = 0.8\mu$

$Mg = 0.6\,Mg + \mu \times 0.8\,Mg$

$0.5$

$\mu = \dfrac{0.4}{0.8}$

$(1 - s) / c$

Reading the options

Now compare the form with the five options.

Option (3) is $\dfrac{1 - \sin\theta}{\cos\theta}$. It matches.

Option (4) has the opposite sign and option (5) has a plus. Option (2) has no $\cos\theta$, and option (1) is $\tan\theta$, which has no string in it. None of these match.

Checks

Check the units. The coefficient of friction is a pure number. $\sin\theta$ and $\cos\theta$ are pure numbers too, so $\dfrac{1 - \sin\theta}{\cos\theta}$ is a pure number, as it must be.

Does it make sense?

If $\theta = 0$, the plane is flat. The string pull $Mg$ balances the friction $\mu Mg$ alone, so $\mu = 1$. The formula gives $\dfrac{1 - 0}{1} = 1$.

If $\theta = 90^{\circ}$, the block just hangs beside the other mass, and the weights balance with no friction. The formula gives $0$ over $0$, which fits: friction is not needed.

Conclusion

$\mu = \dfrac{1 - \sin\theta}{\cos\theta}$

The coefficient of dynamic friction is $\mu = \dfrac{1 - \sin\theta}{\cos\theta}$.

The answer is (3).

The quotes give the dynamic friction rule and Newton's second law. That the block moves up the slope, so the friction acts down it, and that the string pull is $Mg$, are our reasoning from them and from the figure.

Teacher's note

At a steady speed the forces balance. Write one balance for each mass: the hanging mass, then the mass along the slope. Friction acts against the motion, and $R = Mg\cos\theta$.

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