Contact forces & friction | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Split contact into normal and friction components

The full contact force is the single push from one surface on the other. In calculations it is represented by two perpendicular components: normal reaction and friction.

Normal reaction $R$ is perpendicular to the surface.
Friction $F$ is parallel to the surface.
On a horizontal surface, $R=mg$ only when no other force has a vertical component and there is no vertical acceleration.
On a slope, $R=mg\cos\theta$ only when no other force has a perpendicular component.

Key idea

“Reaction” is not automatically equal to weight. Find

R

by resolving perpendicular to the contact surface.

Worked example

Rebuild the full contact force

A 6 kg block is at rest on rough horizontal ground while an 8 N horizontal force acts on it. There are no other vertical forces.

R=60\text{ N},\qquad F=8\text{ N}

C^2=60^2+8^2

C=60.5\text{ N}

\alpha=\tan^{-1}\!\left(\dfrac{F}{R}\right)=7.59^\circ

The full contact force has magnitude 60.5 N and is 7.59° from the normal, towards the friction direction.

Make friction oppose the tendency to slide

First imagine the surface with no friction. Decide which way the body would move or try to move; friction points the other way.

Swipe left or right to see the whole diagram →

Before limiting equilibrium, friction supplies only the amount needed. The value μR is the maximum, not the automatic value.

Worked example

Friction supplies only what is needed

A block remains at rest on a horizontal floor while a 7 N force pulls it to the right. If the maximum available friction is 10 N, the actual friction is 7 N to the left—not 10 N.

Answer

F=7\text{ N to the left}

Common mistake

Friction does not always oppose the applied force. It opposes the actual or impending relative motion. In “least and greatest force” questions, the possible friction direction must be considered twice.

Know what the smooth model removes

A smooth contact sets $F=0$ . It does not remove the normal reaction, and it does not mean the surface is horizontal.

Smooth slope: keep $R$ perpendicular to the slope and weight vertically downward; omit friction.
Rough slope: include friction only after deciding the direction of motion or tendency.
Model limitation: a real surface is never perfectly smooth, but the approximation is useful when friction is negligible compared with the other forces.

Your turnIndependent check [3 marks]

A block is at rest on a rough horizontal surface. A horizontal force $P$ acts to the right. The maximum available friction is 14 N. State the friction force when (a) $P=9\text{ N}$ and (b) $P=18\text{ N}$ , and say whether equilibrium is possible in each case.

Show worked answer

For $P=9\text{ N}$ , friction can match the applied force, so $F=9\text{ N}$ left and equilibrium is possible.

For $P=18\text{ N}$ , the greatest possible friction is 14 N. It cannot balance 18 N, so equilibrium is impossible and the block moves right.

(a) $F=9\text{ N}$ left; equilibrium is possible.

(b) $F_{\max}=14\text{ N}$ left; equilibrium is impossible.

Split contact into normal and friction components

The full contact force is the single push from one surface on the other. In calculations it is represented by two perpendicular components: normal reaction and friction.

Normal reaction $R$ is perpendicular to the surface.
Friction $F$ is parallel to the surface.
On a horizontal surface, $R=mg$ only when no other force has a vertical component and there is no vertical acceleration.
On a slope, $R=mg\cos\theta$ only when no other force has a perpendicular component.

Key idea

“Reaction” is not automatically equal to weight. Find

R

by resolving perpendicular to the contact surface.

Worked example

Rebuild the full contact force

A 6 kg block is at rest on rough horizontal ground while an 8 N horizontal force acts on it. There are no other vertical forces.

R=60\text{ N},\qquad F=8\text{ N}

C^2=60^2+8^2

C=60.5\text{ N}

\alpha=\tan^{-1}\!\left(\dfrac{F}{R}\right)=7.59^\circ

The full contact force has magnitude 60.5 N and is 7.59° from the normal, towards the friction direction.

Make friction oppose the tendency to slide

First imagine the surface with no friction. Decide which way the body would move or try to move; friction points the other way.

Swipe left or right to see the whole diagram →

Before limiting equilibrium, friction supplies only the amount needed. The value μR is the maximum, not the automatic value.

Worked example

Friction supplies only what is needed

A block remains at rest on a horizontal floor while a 7 N force pulls it to the right. If the maximum available friction is 10 N, the actual friction is 7 N to the left—not 10 N.

Answer

F=7\text{ N to the left}

Common mistake

Know what the smooth model removes

A smooth contact sets $F=0$ . It does not remove the normal reaction, and it does not mean the surface is horizontal.

Smooth slope: keep $R$ perpendicular to the slope and weight vertically downward; omit friction.
Rough slope: include friction only after deciding the direction of motion or tendency.
Model limitation: a real surface is never perfectly smooth, but the approximation is useful when friction is negligible compared with the other forces.

Your turnIndependent check [3 marks]

Show worked answer

For $P=9\text{ N}$ , friction can match the applied force, so $F=9\text{ N}$ left and equilibrium is possible.

For $P=18\text{ N}$ , the greatest possible friction is 14 N. It cannot balance 18 N, so equilibrium is impossible and the block moves right.

(a) $F=9\text{ N}$ left; equilibrium is possible.

(b) $F_{\max}=14\text{ N}$ left; equilibrium is impossible.