Contact forces and smooth surfaces

8 minAS Level4.1.44.1.5

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can split a contact force into its normal reaction and friction, explain what "smooth" means in a model, and resolve a weight along and perpendicular to a slope.

One push, two parts

Put a book on a table. The table pushes back on the book. That push is the contact force: one force from one touch.

It's easiest to work with in two perpendicular parts (components):

the normal reaction $R$ , at right angles to the surface ("normal" means perpendicular). It stops the book sinking into the table;
the frictional force $F$ , along the surface. If something tries to slide the book (like the pull $P$ in the diagram), friction acts along the surface against that. A book left alone on a level table feels no friction at all.

$R$ and $F$ aren't two separate contacts. They are two components of the same push, just as a slanted arrow can be split into "across" and "up" parts. In the left panel of the diagram the dashed arrow is the whole contact force, and $R$ and $F$ are its parts.

All forces here are measured in newtons (N), and $g$ is the acceleration due to gravity, which 9709 takes as $10\text{ m s}^{-2}$ , so an object of mass $m$ kg has weight $mg$ newtons.

Definition

Smooth contact

A smooth contact is a modelling assumption: there is no friction, so the contact force is entirely the normal reaction $R$ , at right angles to the surface.

Smooth means slippery, not absent

"Smooth" doesn't mean the surface has stopped pushing. Ice still holds you up; it just can't grip you sideways. A smooth surface still gives a normal reaction, possibly a big one. Only the along-the-surface part is switched off.

Real surfaces always have some friction. Calling one smooth is a simplification that makes the maths easier and is close enough when friction is small, like a puck on ice or a well-oiled track.

Its limitation: a smooth model can't explain anything that relies on grip. A block resting on a slope with nothing holding it would, in a smooth model, slide straight down; in real life friction can hold it still.

Splitting the weight on a slope

On a slope, $R$ is at right angles to the slope, not straight up. So it pays to use two new directions: along the slope and perpendicular to it. $R$ already lies in one of them. The weight doesn't, so split it.

Why is the angle between $mg$ and the perpendicular equal to $\alpha$ ? Tilt a horizontal floor up by $\alpha$ to make the slope. Its perpendicular, which was straight up and down, tilts by the same $\alpha$ . So the line "into the slope" makes angle $\alpha$ with the vertical, and the weight is vertical.

Now use a right-angled triangle with $mg$ as the hypotenuse:

the side next to $\alpha$ (into the slope) is $mg\cos\alpha$ ;
the side opposite $\alpha$ (down the slope) is $mg\sin\alpha$ .

Sanity check: on flat ground, $\alpha = 0$ gives $mg\sin 0 = 0$ down the slope (nothing pulls you sideways) and $mg\cos 0 = mg$ into the ground ✓

Worked example: A block held on a smooth slope

A block of mass $4\text{ kg}$ rests on a smooth plane inclined at $25^\circ$ to the horizontal. It is held in equilibrium by a light string parallel to the plane. Take $g = 10\text{ m s}^{-2}$ . Find the tension $T$ in the string and the normal reaction $R$ .

Have a go first?

Worked example: R is not always mg, even on the flat

A crate of mass $5\text{ kg}$ is on smooth horizontal ground. A rope pulls it with a force of $20\text{ N}$ at $30^\circ$ above the horizontal, and the crate stays on the ground. Take $g = 10\text{ m s}^{-2}$ . Find $R$ .

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

DefinitionQuestion 2 · Recall

CalculateQuestion 3 · Apply

Answer the 5 questions above to continue.