Inclined planes

Resolving weight parallel and perpendicular to a slope, the normal force on an incline, the angle at which an object starts to slide, acceleration down a rough slope, and the mandatory inclined-plane practical with light gates to find the coefficient of friction.

Snow on a steep roof sits still for hours, then the whole sheet suddenly slides off. Nothing about the slope changed. A tiny rise in temperature made the snow slightly wetter, its grip on the roof dropped a little, and the balance between "pull down the slope" and "grip" tipped the other way. Inclined planes are all about that balance.

A slope is the standard HSC setting for combining forces, components and Newton's second law. The trick is to tilt your axes: one along the slope and one perpendicular to it. Then the motion is along one axis only.

Resolving weight on a slope

On a slope at angle $\theta$ to the horizontal, weight still points straight down. Split it into two components:

Parallel to the slope (pulling the object down it): $mg\sin\theta$
Perpendicular to the slope (pressing it into the surface): $mg\cos\theta$

The angle between the weight and the perpendicular-to-slope direction is the same $\theta$ as the slope itself. A quick check: when $\theta = 0$ (flat), the parallel part $mg\sin0 = 0$ and the perpendicular part is all of $mg$. When $\theta = 90°$ (a vertical wall), it is the other way around.

Forces on a block on a slope. The weight is resolved into mg sin θ down the slope and mg cos θ into the slope. Friction acts along the slope, opposing motion.

Perpendicular to the slope there is no acceleration, so the normal force balances the perpendicular component of weight:

$$N = mg\cos\theta$$

This is less than $mg$, so on a slope friction (which depends on $N$) is less than it would be on the flat.

A frictionless slope

With no friction, the only force along the slope is $mg\sin\theta$, so:

$$a = \frac{mg\sin\theta}{m} = g\sin\theta$$

The mass cancels. A loaded trolley and an empty one roll down a smooth ramp with the same acceleration, just as all objects fall with the same $g$.

Friction on a slope

Friction acts along the surface and opposes sliding (or the tendency to slide). There are two cases.

At rest. Static friction matches the parallel component exactly: $f_s = mg\sin\theta$. It can grow only up to a maximum, $f_{s,max} = \mu_s N = \mu_s mg\cos\theta$. The object is about to slip when these are equal:

$$mg\sin\theta = \mu_s mg\cos\theta \quad\Rightarrow\quad \tan\theta = \mu_s$$

So the steepest angle at which an object stays put (the angle of repose) depends only on the surfaces, not on the mass. Measuring that angle is a quick way to find $\mu_s$.

Sliding down. Kinetic friction $\mu_k N$ acts up the slope. Taking down the slope as positive:

$$ma = mg\sin\theta - \mu_k mg\cos\theta \quad\Rightarrow\quad a = g(\sin\theta - \mu_k\cos\theta)$$

Worked example: a box sliding down a ramp

A 15 kg box slides down a ramp inclined at 32°. The coefficient of kinetic friction is 0.25. Find the normal force, the friction force and the acceleration.

$N = mg\cos\theta = 15 \times 9.8 \times \cos32° = 125$ N.

normal force

$f_k = \mu_k N = 0.25 \times 124.7 = 31.2$ N, up the slope.

friction

$mg\sin\theta = 15 \times 9.8 \times \sin32° = 77.9$ N down the slope.

along the slope

$a = \dfrac{77.9 - 31.2}{15} = 3.1$ m s⁻² down the slope.

second law

Worked example: pushing a trolley up a ramp

A removalist pushes a 90 kg loaded trolley up a 12° ramp at constant speed, pushing parallel to the ramp. Rolling resistance is 60 N. What push is needed?

Equilibrium along the ramp: push = $mg\sin\theta$ + rolling resistance.

constant speed

$F = 90 \times 9.8 \times \sin12° + 60 = 183 + 60 = 243$ N, about $2.4 \times 10^2$ N up the ramp.

calculate

Lifting the trolley straight up would need 882 N. A ramp trades a smaller force for a longer distance.

compare

The inclined-plane practical

A typical set-up: a block slides down a smooth ramp through two light gates a measured distance apart, and a timer records the time.

Practical: motion on an inclined plane

Aim: to predict and measure the acceleration of a block sliding down an incline, and use the results to find the coefficient of kinetic friction.

Variables: independent: angle $\theta$. Dependent: acceleration $a$. Controlled: block, surface, distance between gates, release point, mass.

Method:

Set a smooth plank on a height adjuster. Measure the angle with a protractor app, or more precisely as $\sin\theta = h/L$ from the height and length of the ramp.
Raise the ramp slowly until the block just starts to slide by itself. Record the angle and repeat five times: $\mu_s = \tan\theta$.
Set the ramp steeper than this. Release the block from rest at the first light gate and time it over 1.00 m to the second gate. Repeat five times and average.
Calculate $a = \dfrac{2s}{t^2}$ (from $s = ut + \tfrac12 at^2$ with $u = 0$).
Repeat for at least five angles.

Analysis: For each angle, $\mu_k = \tan\theta - \dfrac{a}{g\cos\theta}$. Alternatively, plot $a$ against $\sin\theta$ for a trolley on wheels: the model $a = g\sin\theta$ predicts a straight line through the origin with gradient $g$.

Risks: the block can slide off the end and fall: put a padded stop at the bottom. Keep fingers clear when adjusting the ramp height.

Validity and reliability: Releasing from rest exactly at the first gate matters, because $a = 2s/t^2$ assumes $u = 0$. A data logger with the gates gives two speeds and lets you use $v^2 = u^2 + 2as$ instead. Any dust or moisture on the ramp changes the friction, so clean it and use the same face of the block every time. Repeat trials and average to reduce random timing errors.

Worked example: processing incline data

At 35.0°, a wooden block released from rest travels 1.00 m in a mean time of 0.791 s. Find the acceleration and the coefficient of kinetic friction.

$a = \dfrac{2s}{t^2} = \dfrac{2(1.00)}{0.791^2} = 3.20$ m s⁻².

acceleration

$a = g(\sin\theta - \mu_k\cos\theta)$, so $\mu_k = \tan\theta - \dfrac{a}{g\cos\theta}$.

rearrange

$\mu_k = 0.700 - \dfrac{3.20}{9.8 \times 0.819} = 0.700 - 0.399 = 0.30$.

substitute

InteractiveInclined plane lab

Try this

With Wood on wood at 20°, the block stays put. Static friction is 6.70 N, exactly matching $mg\sin\theta$. Raise the angle slowly: it starts sliding just above 26.6°, where $\tan\theta = 0.50$.
Set 35° and press Release a few times. The model predicts 3.21 m/s², so each trial takes about 0.79 s, and the measured $\mu_k$ in the table comes out at about 0.30.
Change the mass from 2 kg to 5 kg at 35°. The forces all grow, but the acceleration stays at 3.21 m/s².
Try Waxed ski on snow: it slides above 5.7°. Rubber on concrete holds until 42.0°.

Test yourself

A 4.0 kg block rests on a slope at 30°. The normal force on it is
A box just begins to slide when a ramp is raised to 22°. The coefficient of static friction is
Two trolleys, of mass 1 kg and 3 kg, roll down the same frictionless ramp. Compared with the 1 kg trolley, the 3 kg trolley has
A student times a block over 0.80 m from rest and gets 0.65 s. The acceleration is

Label the forces on the block on the slope.

Order the steps for finding the coefficient of kinetic friction on a ramp.

On a slope, the normal force equals $mg$. On a frictionless slope, the acceleration is $g$ times of the angle. An object starts to slide when $\tan\theta$ becomes equal to the coefficient of friction. Changing the mass of the object does change its acceleration down the slope.

6 marks

A student slides a 0.50 kg block down a ramp at 30.0° and measures an acceleration of 2.4 m s⁻². (a) Calculate the coefficient of kinetic friction. (3 marks) (b) The student then repeats the experiment at 15.0°, but the block does not move. Explain this observation and predict the smallest angle at which the block would start to slide if $\mu_s = 0.40$. (3 marks)

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