When to Include Air Resistance in HSC Force Diagrams

Drop a flat sheet of paper, then crumple the same sheet into a ball and drop it again. The mass hasn’t changed, so why does the flat sheet often fall much more slowly?

Predict what the force diagrams should show. Does the flat sheet experience less gravity?

No. Both have the same weight. The difference comes from how they interact with the surrounding air.

Here’s the rule for HSC Physics: include air resistance, also called drag, in a free-body diagram when it acts on your chosen object and matters to the motion you’re modelling. If the question explicitly tells you to neglect air resistance, leave it out. If the force is small enough to ignore for the required accuracy, you can also omit it, but that is an approximation.

The important detail is that drag depends on motion relative to the air, not simply whether an object is moving relative to the ground.

01Decide from the interactions, not just the motion

Before deciding whether to draw drag, identify the object you’re analysing.

Suppose a ball is released from someone’s hand and begins falling through still air. Choose the ball as your system.

The ball can interact with several things, but not all interactions act at every moment.

Here, \(m\) is mass in kilograms, \(g\) is gravitational acceleration, and \(D\) is the magnitude of the drag force in newtons.

Near Earth’s surface, we’ll use \(g=9.8\ \mathrm{m\,s^{-2}}\).

The force diagram includes forces acting on the ball. The ball also pushes on the surrounding air, but that reaction force acts on the air, not the ball.

For a fuller explanation of isolating an object and identifying its interactions, see Free-Body Diagrams: How to Draw Forces Correctly in HSC Physics. Here, we’ll concentrate on deciding when the air interaction matters.

Step 1: Work out the motion relative to the air

Imagine releasing a ball from rest in a room with no wind.

Immediately after release, its velocity relative to the air is zero. Ordinary motion-dependent drag is therefore zero at that instant.

Gravity then accelerates the ball downwards. As its downward speed increases, it begins moving through the air, which exerts an upward drag force.

For a simple drag-only model, the direction rule is:

\[
\vec v_{\mathrm{rel}}
=\vec v_{\mathrm{object}}-\vec v_{\mathrm{air}}
\]

Here, \(\vec v_{\mathrm{rel}}\) is the object’s velocity relative to the air. The drag force acts opposite this relative velocity.

That gives three possibilities:

• Moving down through still air: drag acts upwards.
• Moving up through still air: drag acts downwards.
• Stationary relative to the air: ordinary velocity-dependent drag is zero.

Air can also exert buoyancy or aerodynamic lift. These are different forces, so zero drag does not necessarily mean zero force from the air.

Step 2: Decide whether the force matters

An object can experience real drag even when a question tells you to ignore it.

That isn’t a contradiction. A free-body diagram represents a physical model, and models deliberately simplify reality.

Use this decision table.

The common mistake is assuming that moving through air automatically requires a drag arrow in every calculation.

For an idealised projectile question, you might be expected to neglect drag entirely. For a parachutist descending at constant speed, leaving it out would make the force diagram physically inconsistent.

Step 3: Identify the direction before using equations

For vertical motion, choose a positive direction and apply Newton’s second law.

If downward is positive for an object falling through still air,

\[
mg-D=ma
\]

where \(a\) is its downward acceleration in \(\mathrm{m\,s^{-2}}\).

Notice that drag subtracts from weight because the two forces point in opposite directions.

But don’t memorise \(mg-D\) as the equation for every falling object. In moving air, drag can point downwards even while the object is falling.

We’ll come back to that.

02An experiment that tests the no-drag model

How could you tell whether air resistance is important without assuming the answer?

Start with a prediction.

If air resistance is negligible and two objects are released from rest at the same height, both accelerate downwards at approximately \(g\), regardless of their masses.

They should reach the ground together, provided they begin with the same vertical velocity and fall through the same distance.

The first design: paper versus metal

Imagine a simple drop experiment.

A flat sheet of paper and a small metal washer are held at the same height above the floor. A camera is positioned side-on, facing a vertical measuring ruler behind the falling objects.

Both are released without an intentional push. The camera records their positions as they descend.

If the no-drag model describes both objects, their vertical motion should be nearly identical.

Suppose the video instead shows the washer falling noticeably faster.

That challenges the no-drag prediction, but there is a problem: the washer and paper differ in mass, shape, and surface area.

The first experiment does not isolate which difference is responsible for the observed motion.

The revised design: keep the mass the same

Repeat the investigation using the same sheet of paper in two different arrangements.

First release it flat. Then crumple it into a compact ball and release it from the same height.

The mass is unchanged, so both versions experience approximately the same gravitational force.

The main change is how the paper meets the airflow.

To make the comparison more reliable, use a room without noticeable draughts, keep the release height consistent, avoid pushing the paper, and repeat each trial. Flat paper can also flutter, so multiple recordings are more useful than one unusual fall.

The camera allows you to estimate speed from the change in position between successive frames.

Consider the following illustrative results. They are invented teaching data, not measurements from a real experiment.

With no air resistance, the predicted speed after \(0.30\ \mathrm{s}\) is

\[
v=gt=9.8(0.30)=2.94\ \mathrm{m/s}
\]

The crumpled paper comes much closer to that prediction. The flat paper does not.

This supports the explanation that the interaction with air matters much more for the flat sheet.

It doesn’t prove that projected area is the only cause. Crumpling also changes the shape, orientation, fluttering behaviour, and drag coefficient of the paper.

03Drag direction depends on the air, not the ground

Now consider a more surprising situation.

A lightweight parcel is descending vertically while a strong wind blows sideways.

You might predict that the air resistance points straight upwards, because the parcel is moving downwards.

But the parcel is also moving sideways relative to the air.

The correct direction comes from comparing the parcel’s velocity with the wind’s velocity.

Think of cycling at the same ground speed with a headwind and a tailwind. Your speed relative to the road is unchanged, but the airflow you experience is different. The comparison is useful; real aerodynamic forces can be more complicated because shape and orientation can also produce lift.

04Three distinctions that prevent wrong force diagrams

Zero speed does not always mean zero acceleration

Consider a ball thrown straight upwards through still air.

At its highest point, its instantaneous speed is zero.

Predict the air resistance and acceleration at that instant.

In the simple drag model, air resistance is zero because the ball is momentarily stationary relative to the air. But gravity is still acting, so the acceleration is \(9.8\ \mathrm{m\,s^{-2}}\) downwards.

Before the highest point, drag points downwards. After the highest point, drag points upwards.

Zero acceleration does not mean zero drag

A parachutist descending vertically at constant speed has zero acceleration.

Does that mean you should omit air resistance?

No. Assuming weight and drag are the only significant vertical forces, they must balance:

\[
D=mg
\]

This is terminal velocity. The parachutist is still moving through the air, and drag is essential to explaining why the speed is constant.

Without drag, the force diagram would predict downward acceleration, contradicting the stated motion.

Small compared with weight does not always mean negligible

Suppose a trolley moves horizontally on a level, very low-friction track.

Its weight and normal force nearly cancel vertically. Even a modest horizontal drag force could be a large part of its total horizontal force.

For example, if a horizontal pull is almost balanced by drag, omitting drag could change the predicted acceleration dramatically.

So the useful comparison isn’t always drag versus weight.

Ask how much omitting drag would change the quantity you’re trying to calculate. When other forces nearly cancel, a seemingly small force may control the result.

05Questions and solutions

Use \(g=9.8\ \mathrm{m\,s^{-2}}\) throughout. Treat all supplied force measurements and drag models as illustrative.

Question 1

A \(0.20\ \mathrm{kg}\) ball is descending vertically at \(4.0\ \mathrm{m/s}\) through still air. At that instant, air resistance has magnitude \(0.60\ \mathrm{N}\).

Assume gravity and drag are the only forces acting.

Draw a free-body diagram and calculate the ball’s acceleration, including its direction.

Solution 1

The ball accelerates downwards at \(6.8\ \mathrm{m\,s^{-2}}\). Its free-body diagram contains weight downwards and air resistance upwards.

The weight is

\[
F_g=0.20(9.8)=1.96\ \mathrm{N}
\]

Taking downwards as positive,

\[
\begin{aligned}
F_{\mathrm{net}}&=1.96-0.60\\
&=1.36\ \mathrm{N}
\end{aligned}
\]

Newton’s second law gives

\[
a=\frac{1.36}{0.20}
=6.8\ \mathrm{m\,s^{-2}}
\]

The ball is accelerating downwards because its weight exceeds drag.

It is tempting to say that air resistance makes the ball slow down. But here it is still speeding up. Drag simply makes its downward acceleration smaller than \(g\).

Question 2

A \(0.10\ \mathrm{kg}\) ball is thrown vertically upwards in still air.

A motion tracker records the following:

• While the ball is moving upwards at \(8.0\ \mathrm{m/s}\), its acceleration is \(14.0\ \mathrm{m\,s^{-2}}\) downwards.
• Later, while the ball is moving downwards at \(8.0\ \mathrm{m/s}\), its acceleration is \(5.6\ \mathrm{m\,s^{-2}}\) downwards.

Assume only gravity and drag act.

Calculate the drag magnitude at each instant. Explain whether the results are consistent with drag depending on speed, and determine the acceleration at the highest point.

Solution 2

The drag magnitude is \(0.42\ \mathrm{N}\) at both recorded instants. At the highest point, drag is zero and the ball accelerates downwards at \(9.8\ \mathrm{m\,s^{-2}}\).

First calculate the weight:

\[
mg=0.10(9.8)=0.98\ \mathrm{N}
\]

During the upward journey, both gravity and drag point downwards.

Using downward as positive,

\[
mg+D=ma
\]

Therefore,

\[
\begin{aligned}
D_{\mathrm{up}}&=0.10(14.0)-0.98\\
&=0.42\ \mathrm{N}
\end{aligned}
\]

During the downward journey, drag points upwards.

\[
mg-D=ma
\]

So

\[
\begin{aligned}
D_{\mathrm{down}}&=0.98-0.10(5.6)\\
&=0.42\ \mathrm{N}
\end{aligned}
\]

The equal values are consistent with a drag model in which force magnitude depends on relative speed. Equal speeds in opposite directions give equal drag magnitudes but opposite drag directions.

They do not, by themselves, prove a particular mathematical drag law.

At the highest point, the ball has zero instantaneous speed relative to the still air. Ordinary drag therefore vanishes, leaving gravity as the only force.

Its acceleration is still \(g\) downwards.

The tempting error is to assume that the acceleration must be zero because the ball briefly stops. Zero velocity does not require zero net force.

Question 3

A \(0.15\ \mathrm{kg}\) instrument is released from rest and falls vertically through still air.

A simplified laboratory model gives the drag magnitude as

\[
D=kv^2
\]

where \(v\) is the instrument’s speed in \(\mathrm{m/s}\), \(D\) is in newtons, and \(k=0.0030\ \mathrm{N\,s^2\,m^{-2}}\).

A modelling decision has been made: air resistance may be omitted only if its magnitude remains below \(5\%\) of the instrument’s weight throughout the drop.

Two possible drop heights are considered:

• Test A: \(1.0\ \mathrm{m}\)
• Test B: \(10.0\ \mathrm{m}\)

For Test B, a camera records a speed of \(8.0\ \mathrm{m/s}\) at one point during the fall.

Without solving a differential equation, determine which test can be guaranteed to meet the \(5\%\) condition. Use the camera reading to decide whether the no-drag model meets that condition for Test B.

Explain why using the maximum speed predicted by a no-drag calculation can be useful even though drag is actually present.

Solution 3

Test A is guaranteed to satisfy the specified \(5\%\) force criterion. Test B does not: the recorded speed shows that drag exceeds \(5\%\) of the instrument’s weight at that instant.

The key is to use the no-drag model as an upper bound on the speed.

If the instrument falls from rest through a vertical distance \(h\), conservation of energy without drag gives

\[
v^2=2gh
\]

Real drag removes mechanical energy during a fall through still air, so the actual speed cannot exceed the speed predicted when drag is omitted, under the stated assumptions.

For a full drop of height \(H\),

\[
v^2\leq 2gH
\]

Since \(D=kv^2\), the drag-to-weight ratio satisfies

\[
\frac{D}{mg}\leq\frac{2kH}{m}
\]

For Test A,

\[
\begin{aligned}
\frac{D}{mg}&\leq
\frac{2(0.0030)(1.0)}{0.15}\\
&=0.040=4.0\%
\end{aligned}
\]

Even the upper bound is below \(5\%\). Test A therefore satisfies the stated force criterion throughout its fall.

For Test B, the same calculation gives

\[
\frac{D}{mg}\leq0.40=40\%
\]

This bound is too large to establish that drag is negligible.

But a large upper bound does not prove that the actual drag reaches that value. We need additional information.

The camera measurement supplies it.

At \(v=8.0\ \mathrm{m/s}\),

\[
\begin{aligned}
D&=0.0030(8.0)^2\\
&=0.192\ \mathrm{N}
\end{aligned}
\]

The instrument’s weight is

\[
mg=0.15(9.8)=1.47\ \mathrm{N}
\]

Therefore,

\[
\frac{D}{mg}=\frac{0.192}{1.47}
\approx13.1\%
\]

Drag exceeds the allowed \(5\%\) at the recorded instant, so Test B fails the criterion.

The interesting point is that we used the no-drag prediction without pretending it was exact. Since drag opposes the fall and removes mechanical energy, the no-drag speed provides a conservative upper limit.

Also, the \(5\%\) tolerance was specified for this problem. It is not a universal HSC rule for neglecting drag, particularly when other forces nearly cancel.

Question 4

A \(5.0\ \mathrm{kg}\) trolley fitted with a broad sail travels east along a level, very low-friction track at \(2.0\ \mathrm{m/s}\).

A cord pulls it east with a force of \(3.0\ \mathrm{N}\). The sail experiences horizontal air resistance of \(2.5\ \mathrm{N}\) westwards.

Neglect wheel resistance and vertical aerodynamic forces.

(a) Identify all forces acting on the trolley and calculate its acceleration.

(b) A student argues that drag can be ignored because it is only about \(5\%\) of the trolley’s weight. Calculate the acceleration this model predicts and assess the argument.

(c) The cord suddenly goes slack. At the first instant afterwards, the trolley is still moving east at \(2.0\ \mathrm{m/s}\), and drag remains \(2.5\ \mathrm{N}\) westwards. Determine its acceleration and explain how the force diagram changes.

Solution 4

The trolley initially accelerates east at \(0.10\ \mathrm{m\,s^{-2}}\). Ignoring drag would predict \(0.60\ \mathrm{m\,s^{-2}}\) east, six times the actual acceleration. After the cord goes slack, the trolley accelerates west at \(0.50\ \mathrm{m\,s^{-2}}\).

For part (a), the forces on the trolley are:

• Weight, \(49\ \mathrm{N}\), downwards.
• Normal force, \(49\ \mathrm{N}\), upwards.
• Cord tension, \(3.0\ \mathrm{N}\), eastwards.
• Air resistance, \(2.5\ \mathrm{N}\), westwards.

The vertical forces balance because there is no vertical acceleration.

Taking east as positive,

\[
\begin{aligned}
F_{\mathrm{net},x}&=3.0-2.5\\
&=0.50\ \mathrm{N}
\end{aligned}
\]

Therefore,

\[
a_x=\frac{0.50}{5.0}
=0.10\ \mathrm{m\,s^{-2}}
\]

For part (b), omitting drag gives

\[
a_x=\frac{3.0}{5.0}
=0.60\ \mathrm{m\,s^{-2}}
\]

This is six times the correct acceleration.

The student’s argument fails because weight is not the useful comparison for horizontal motion here.

Weight and normal force cancel vertically. Horizontally, the pull is almost cancelled by drag. A force that is small compared with weight can still dominate the horizontal force balance.

For part (c), the cord can no longer exert tension once it goes slack.

The horizontal force diagram now contains only air resistance.

\[
\begin{aligned}
F_{\mathrm{net},x}&=-2.5\ \mathrm{N}\\
a_x&=\frac{-2.5}{5.0}\\
&=-0.50\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

The negative sign means westwards.

Immediately after the cord goes slack, the trolley still moves east. Its westward acceleration means it starts slowing down.

The trap is treating the existing eastward velocity as evidence of an eastward force. The cord’s force has disappeared, but the trolley’s velocity cannot change instantaneously merely because the cord goes slack.

Question 5

Two packages are connected vertically by a light, inextensible string.

Package A is above Package B.

Their masses are:

\[
m_A=0.40\ \mathrm{kg},\qquad
m_B=0.60\ \mathrm{kg}
\]

Initially, both packages descend together at \(10.0\ \mathrm{m/s}\) through still air, with the string taut.

The upward drag forces are:

• Package A: \(2.0\ \mathrm{N}\)
• Package B: \(1.0\ \mathrm{N}\)

An automatic canopy can be attached to either package. When it opens, it adds an extra \(11.0\ \mathrm{N}\) of upward drag to whichever package carries it, without changing the drag on the other package at that instant.

Compare two separate arrangements:

• Arrangement I: the canopy opens on upper package A.
• Arrangement II: the canopy opens on lower package B.

Assume the quoted drag forces apply immediately after opening, the canopy produces no opening impulse, and neither package affects the airflow around the other. Ignore buoyancy and lift. The string can pull but cannot push, and becomes slack if the packages move closer together.

(a) Calculate the common acceleration and string tension immediately before the canopy opens.

(b) Calculate the acceleration of the combined system immediately after opening in each arrangement.

(c) Determine whether the string remains taut in each arrangement. Where it remains taut, calculate the tension. Where it becomes slack, calculate each package’s acceleration.

(d) Explain why the two arrangements can have the same centre-of-mass acceleration but different individual accelerations.

Solution 5

Before opening, both packages accelerate downwards at \(6.8\ \mathrm{m\,s^{-2}}\), with a string tension of \(0.80\ \mathrm{N}\).

After opening, both arrangements have a centre-of-mass acceleration of \(4.2\ \mathrm{m\,s^{-2}}\) upwards. But the string remains taut only when the canopy opens on upper package A.

When the canopy opens on lower package B, the string slackens, and the packages accelerate differently.

The key is that the string can exert tension but cannot push the packages apart.

Part (a): Before opening

Choose upwards as positive.

The total mass is

\[
m_A+m_B=1.0\ \mathrm{kg}
\]

Total drag is \(3.0\ \mathrm{N}\) upwards, while total weight is \(9.8\ \mathrm{N}\) downwards.

Treating both packages as one system removes the internal tension forces:

\[
\begin{aligned}
a&=\frac{3.0-9.8}{1.0}\\
&=-6.8\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

So both accelerate downwards at \(6.8\ \mathrm{m\,s^{-2}}\).

To find tension, analyse lower package B.

The string pulls B upwards. Its weight is \(0.60(9.8)=5.88\ \mathrm{N}\) downwards.

\[
\begin{aligned}
T+1.0-5.88&=0.60(-6.8)\\
T&=0.80\ \mathrm{N}
\end{aligned}
\]

The positive tension confirms that the initial taut-string arrangement is physically possible.

Part (b): Combined acceleration after opening

In either arrangement, total upward drag becomes

\[
D_{\mathrm{total}}=2.0+1.0+11.0
=14.0\ \mathrm{N}
\]

Total weight remains \(9.8\ \mathrm{N}\).

Thus,

\[
\begin{aligned}
a_{\mathrm{CM}}&=\frac{14.0-9.8}{1.0}\\
&=4.2\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Here \(a_{\mathrm{CM}}\) means the acceleration of the centre of mass.

It is upwards in both arrangements.

But have we proved that both packages individually accelerate upwards at \(4.2\ \mathrm{m\,s^{-2}}\)?

Not yet. That would require the string to remain taut.

Part (c): Test the string tension

In Arrangement I, the canopy opens on upper package A.

The drag forces become:

\[
D_A=13.0\ \mathrm{N},\qquad
D_B=1.0\ \mathrm{N}
\]

Assume initially that the string remains taut. Both packages would then accelerate upwards at \(4.2\ \mathrm{m\,s^{-2}}\).

For B,

\[
\begin{aligned}
T+1.0-5.88&=0.60(4.2)\\
T&=7.4\ \mathrm{N}
\end{aligned}
\]

This tension is positive, so the taut-string assumption is valid.

In Arrangement I, both packages accelerate upwards at \(4.2\ \mathrm{m\,s^{-2}}\), with tension \(7.4\ \mathrm{N}\).

Now consider Arrangement II.

The canopy opens on lower package B, so

\[
D_A=2.0\ \mathrm{N},\qquad
D_B=12.0\ \mathrm{N}
\]

Again, temporarily assume the string stays taut.

For B,

\[
\begin{aligned}
T+12.0-5.88&=0.60(4.2)\\
T&=-3.6\ \mathrm{N}
\end{aligned}
\]

A negative tension would mean that the string needs to push the lower package downwards to make both packages share the same acceleration.

But an ordinary string cannot push.

Therefore, the taut-string assumption is impossible in Arrangement II.

The string begins to slacken, making its tension zero. Immediately afterwards, calculate the two accelerations separately.

For upper package A,

\[
\begin{aligned}
a_A&=\frac{2.0-3.92}{0.40}\\
&=-4.8\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Package A accelerates downwards at \(4.8\ \mathrm{m\,s^{-2}}\).

For lower package B,

\[
\begin{aligned}
a_B&=\frac{12.0-5.88}{0.60}\\
&=10.2\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

Package B accelerates upwards at \(10.2\ \mathrm{m\,s^{-2}}\).

Both packages were initially descending at the same speed. Immediately after opening, A speeds up downwards while B slows its downward motion.

Their separation begins decreasing, allowing the string to go slack.

Part (d): Why the combined acceleration is unchanged

For Arrangement II, calculate the centre-of-mass acceleration from the individual accelerations:

\[
a_{\mathrm{CM}}
=\frac{m_Aa_A+m_Ba_B}{m_A+m_B}
\]

Substituting, with upward positive,

\[
\begin{aligned}
a_{\mathrm{CM}}
&=\frac{0.40(-4.8)+0.60(10.2)}{1.0}\\
&=4.2\ \mathrm{m\,s^{-2}}
\end{aligned}
\]

This agrees with the combined-system calculation.

Both arrangements have the same external vertical forces, so their centres of mass accelerate in exactly the same way.

But the distribution of those forces determines what happens to the string and the individual packages.

In Arrangement I, strong drag on the upper package stretches the string and transmits an upward pull to the lower package.

In Arrangement II, strong drag on the lower package makes it decelerate downwards much more rapidly than the upper package. The string cannot force them to maintain their separation, so it slackens.

The combined-system calculation was correct for the centre of mass in both cases. The mistake would be assuming that the centre-of-mass acceleration must also be the acceleration of each individual package.

This is the larger lesson for force diagrams: choosing the correct forces is not quite enough. You must also check which objects are constrained to move together.

And once drag becomes important, the next useful question is whether it eventually balances the other forces. That leads to terminal velocity, where an object can continue moving through air with zero acceleration precisely because the air resistance has not been omitted.