Drag — pressure (form) drag, skin friction drag: Question bank

Before we start, a few symbols and words you must not confuse — anchor them once, so every reveal below reads cleanly without hunting through other pages:

Definition Symbols and words used on this page

$D$ = the drag force: the total backward push (measured in newtons, N) the fluid exerts on the body, pointing opposite to the relative motion. It is the sum of form drag plus skin friction drag.
$ρ$ = fluid density (mass per unit volume, kg/m³). $v$ = the relative speed between body and fluid (m/s). $A$ = a chosen reference area (m², usually the frontal/projected area for bluff bodies).
$C_{D}$ = the drag coefficient: a dimensionless number that lumps all the messy shape/geometry effects into one experimentally-measured value. Together these give $D = \frac{1}{2} ρ v^{2} A C_{D}$ .
$μ$ (Greek "mu") = the dynamic viscosity: how strongly a fluid resists having its layers slide over one another (units Pa·s). It is the proportionality constant in Newton's law of viscosity, $τ = μ d u / d y$ . Bigger $μ$ → stickier fluid → more rubbing.
Normal force = push straight into the surface (perpendicular). This is pressure $p$ .
Tangential force = drag along the surface (parallel). This is shear stress $τ$ (Greek "tau", units Pa).
Wake = the churning, low-pressure pocket left behind a body when the flow can't follow its shape and separates.

To keep this bank self-contained, one picture pins down the whole distinction the questions test. Look at it before you start: the same arrows appear again and again in the reveals below.

Figure — Drag — pressure (form) drag, skin friction drag

Every question here is really asking: "Is this drag coming from the perpendicular push (form), the parallel rub (skin friction), or a bit of both — and did viscosity secretly cause it?"

True or false — justify

A perfectly inviscid (zero-viscosity) fluid in steady flow exerts zero drag on a sphere.

True — this is d'Alembert's paradox; with no viscosity the pressure is perfectly symmetric front-to-back so the pushes cancel, and there is no shear at all. You need viscosity to break that symmetry.

Skin friction drag can exist even when a body has no wake at all.

True — a flat plate held edge-on to the flow keeps the flow attached (essentially no wake) yet the fluid still rubs along both faces, so skin friction is the entire drag there.

Form drag can exist in a real fluid where the shear stress on the body is exactly zero everywhere.

False — in a real fluid form drag comes from separation, and separation only happens because of viscosity, which necessarily also produces shear. So "zero shear everywhere" quietly means inviscid, and an inviscid fluid gives zero form drag. You cannot have the separation without the shear.

A parachute's drag is mostly skin friction because so much fabric touches the air.

False — a parachute is a bluff body with an enormous wake, so it is ~100% form (pressure) drag; the rubbing on the fabric is negligible next to the front–back pressure difference.

Streamlining a body always reduces its total surface area.

False — a teardrop usually has more surface than a sphere; you accept extra skin friction to kill the wake, and the net total drag still drops.

Doubling the speed doubles the drag.

False — at high Reynolds number

C_{D}

is roughly constant, so

D = \frac{1}{2} ρ v^{2} A C_{D}

makes drag scale with

v^{2}

; doubling

v

quadruples

D

The drag coefficient

C_{D}

has units of newtons.

False —

C_{D}

is deliberately dimensionless; all the units live in

\frac{1}{2} ρ v^{2} A

(which already comes out in newtons), and

C_{D}

only carries the "shape mystery".

The no-slip condition means the fluid speed is zero everywhere near the wall.

False — the no-slip condition says the speed is zero exactly at the wall; just above it the speed climbs steeply, and that steep climb is precisely what creates the wall shear.

For a very streamlined body, skin friction can be the dominant part of the total drag.

True — when the wake is tiny, form drag is small, so the leftover skin friction (over a large wetted area) becomes the bigger share.

A turbulent boundary layer always means more total drag than a laminar one.

False — a turbulent boundary layer has more skin friction, but it also sticks to the surface longer, delaying separation and shrinking the wake. On a bluff body (like a golf ball) this can cut form drag so much that total drag drops — which is exactly why golf balls have dimples.

Spot the error

"Drag is just friction between the body and the fluid, like sliding a box on the floor."

The error is treating drag as one mechanism. There are two: the perpendicular pressure push (form) and the parallel viscous rub (skin friction). For most bluff bodies the pressure push dominates, not the rubbing.

"Since form drag comes from pressure, and pressure needs no viscosity, form drag exists even in an ideal fluid."

The hidden step is wrong: form drag needs an asymmetric pressure, and that asymmetry is produced by separation, which needs viscosity. In an ideal (inviscid) fluid the pressure stays symmetric, the front and back pushes cancel exactly, and the net drag is zero (d'Alembert's paradox). So the conclusion is backwards.

"A flat plate face-on and edge-on have the same drag because it's the same plate with the same area."

Orientation changes which source dominates. Face-on gives a big wake (form drag,

C_{D} \approx 1.2

); edge-on gives almost pure skin friction (

C_{D} \approx 0.005

) — hundreds of times less.

"Wall shear stress depends on the whole velocity profile, so you need the entire boundary layer to compute it."

Only the slope of velocity at the wall matters:

τ_{w} = μ (d u / d y) ∣_{y = 0}

. Skin friction "sees" just that gradient at

y = 0

, nothing deeper.

"The reference area

A

D = \frac{1}{2} ρ v^{2} A C_{D}

is always the surface area the fluid touches."

For bluff bodies

A

is the frontal (projected) area, not the wetted area. Wetted area is what matters for skin friction, but the drag formula's

A

is a chosen reference, usually frontal.

"Bernoulli says pressure drops where speed rises, so the fast wake behind a car should have high pressure and push the car forward."

The wake is slow and swirly, not fast, and Bernoulli doesn't apply across the separated, turbulent region. The wake is low-pressure, so it fails to push the back forward — that's exactly the form drag.

Why questions

Why does a teardrop shape reduce total drag even though it has more surface to rub against?

The gentle tail lets the pressure recover slowly so the flow stays attached, shrinking the wake and slashing form drag. You lose a little to extra skin friction but win far more by killing the wake.

Why is

\frac{1}{2} ρ v^{2}

(dynamic pressure) the natural combination, rather than

ρ v

ρ v^{3}

Build it step by step. In one second the body sweeps out a tube of fluid of length

v

and cross-section

A

, so mass

ρ (v A)

passes it each second (that's the mass flux). Each bit of that mass arrives carrying speed

v

, so the momentum it brings per second is (mass per second)

\times v = ρ v A \cdot v = ρ v^{2} A

— a force. That fixes the

v^{2}

and the

ρ A

. The extra

\frac{1}{2}

is a convention that makes the group equal the dynamic pressure

\frac{1}{2} ρ v^{2}

, so

C_{D}

becomes a clean dimensionless corrector.

Why does a faster or smaller body tend to have more skin friction stress at the wall?

A thinner boundary layer squeezes the same speed change into a smaller gap, so

d u / d y

is steeper, and

τ_{w} = μ d u / d y

grows.

Why do we bother packaging everything into

C_{D}

instead of a formula from pure physics?

Form drag depends on separation, which is too messy to solve exactly for real shapes. We predict the clean part (

ρ, v, A

) and stuff all the geometric mystery into an experimentally-measured $C_D$.

Why is viscosity called the "hidden source of both drag types"?

Skin friction is directly viscous shear; form drag exists only because viscosity causes separation. Remove viscosity and both vanish (d'Alembert). So one property secretly drives both.

Why does whether the boundary layer is laminar or turbulent change the split between form and skin friction?

A laminar layer is smooth and thin (low skin friction) but separates early, giving a wide wake and large form drag. A turbulent layer mixes fast fluid down to the wall (higher skin friction) but stays attached longer, shrinking the wake and cutting form drag. So the flow regime shifts which of the two sources dominates.

Why does terminal velocity depend on the same

C_{D}

that governs drag here?

At terminal velocity gravity balances drag

\frac{1}{2} ρ v^{2} A C_{D}

, so the same coefficient that sets how hard the fluid pushes back also sets the speed at which the push equals the weight.

Edge cases

At the exact instant the fluid speed relative to the body is zero, what is the drag?

Zero — with

v = 0

the dynamic pressure

\frac{1}{2} ρ v^{2}

vanishes, so both form and skin friction drag vanish. No relative motion, no rub, no front–back pressure difference.

What happens to skin friction drag in the limit of zero viscosity (

μ \to 0

It goes to zero, since

τ_{w} = μ d u / d y

has

μ

as a direct factor — no viscosity means no shear rub at all.

For a body perfectly aligned with the flow (like a razor-thin plate edge-on), what fraction of drag is form drag?

Almost none — with no frontal projected area to build a wake, the flow stays attached, so the drag is essentially all skin friction.

For an ideal flat plate held exactly face-on (perpendicular) in a real fluid, which drag dominates and why?

Form drag, overwhelmingly — the flow separates right at the sharp edges leaving a huge low-pressure wake, so the front–back pressure difference dwarfs the tiny edge shear.

In the limit of a perfectly symmetric pressure distribution (no separation), what is the net form drag?

Zero — the forward push on the front is exactly cancelled by an equal push on the back, so the pressure integral around the body gives no net rearward force.

At the boundary between laminar and turbulent flow (the "drag crisis"), what happens to

C_{D}

of a sphere as speed rises through the critical Reynolds number?

It suddenly drops, not rises — the boundary layer trips to turbulent, clings to the surface longer, narrows the wake, and form drag falls sharply. This counter-intuitive dip is the drag crisis.