Friction and inclined planes (VCE Physics Unit 2)

What this dot point is asking

VCAA wants you to apply Newton's second law to objects on horizontal and inclined surfaces, including the friction terms, and to set up the equations using the standard axis choice (along and perpendicular to the slope).

Types of friction

Static friction. Opposes motion that would otherwise begin. $f_s \le \mu_s N$. The inequality reflects that static friction takes whatever value is needed to keep the object stationary, up to the threshold.

Kinetic friction. Acts on a moving object opposing motion. $f_k = \mu_k N$. Constant magnitude (independent of speed, to a good approximation).

Typically $\mu_s > \mu_k$.

On a horizontal surface

Normal force: $N = mg$.

If a horizontal applied force $F$ is overcoming friction:

Just about to slip: $F = \mu_s mg$.

Sliding: $F - \mu_k mg = ma$, so $a = (F - \mu_k mg)/m$.

On an inclined plane

Choose $x$-axis along the slope and $y$-axis perpendicular. Weight resolves into:

• Parallel to slope (down): $mg \sin\theta$.
• Perpendicular to slope: $mg \cos\theta$.

Normal force balances the perpendicular component: $N = mg \cos\theta$.

Frictionless ramp. Net force along slope = $mg \sin\theta$. Acceleration $a = g \sin\theta$.

With kinetic friction. Net force along slope = $mg \sin\theta - \mu_k mg \cos\theta$. Acceleration $a = g(\sin\theta - \mu_k \cos\theta)$.

If $\sin\theta < \mu_s \cos\theta$, the object will not start to slide (static friction can hold it).

Worked example (horizontal pull)

A $5.0$ kg block on a horizontal surface is pulled by a horizontal force of $30$ N. $\mu_k = 0.30$. $g = 9.8$ m s$^{-2}$.

$N = mg = 49$ N. $f_k = \mu_k N = 0.30 \cdot 49 = 14.7$ N.

Net force: $30 - 14.7 = 15.3$ N. $a = 15.3/5.0 = 3.06$ m s$^{-2}$.

Common traps

Using $N = mg$ on an incline

Wrong. On an incline, $N = mg \cos\theta$ (smaller than weight).

Treating $f_s = \mu_s N$

Static friction equals this only at the threshold. Below the threshold it equals whatever force is opposing motion.

Mixing up which direction friction acts

Friction always opposes relative motion (or the tendency of motion). Identify the direction of motion first.

Forgetting friction's direction on an incline

A block sliding down has friction pointing up the slope.

In one sentence

On an inclined plane, weight resolves into parallel ($mg \sin\theta$) and perpendicular ($mg \cos\theta$) components, the normal force balances the perpendicular component, and the net force along the slope is $mg \sin\theta - \mu_k mg \cos\theta$ when kinetic friction is present, giving acceleration $a = g(\sin\theta - \mu_k \cos\theta)$.

Examples in context

Example 1. Truck on the West Gate Freeway descent. A $20\,000$ kg truck descends the West Gate Bridge approach at a $5^\circ$ gradient. The component of weight along the slope is $mg\sin 5^\circ = 20\,000 \times 9.8 \times 0.0872 = 1.71 \times 10^4$ N. If the coefficient of kinetic friction on the tyres-tarmac contact is $\mu_k = 0.6$, friction can supply up to $\mu_k m g \cos 5^\circ = 0.6 \times 20\,000 \times 9.8 \times 0.996 = 1.17 \times 10^5$ N. Engine braking and engine-exhaust retarders are sized to handle the difference, while emergency runaway-truck ramps with steep upgrades and deep gravel beds dissipate the kinetic energy if friction alone is insufficient.

Example 2. Snowy 2.0 tunnel-boring machine on the Yarrangobilly incline. Snowy 2.0's tunnel-boring machine, mass approximately $2200$ tonnes, moves on rails laid at a $9^\circ$ incline through Yarrangobilly. The slope-parallel weight is $2.2 \times 10^6 \times 9.8 \times \sin 9^\circ = 3.37 \times 10^6$ N. Static friction between steel wheel and steel rail has $\mu_s \approx 0.3$, providing a maximum holding force of $0.3 \times 2.2 \times 10^6 \times 9.8 \times \cos 9^\circ = 6.39 \times 10^6$ N. Friction alone can hold the machine, but hydraulic locking pins are added because dynamic loads (vibration from the cutter head) can briefly exceed the static limit.

Try this

Q1. State the difference between static and kinetic friction. [2 marks]

• Cue. Static friction prevents motion up to $\mu_s N$; kinetic friction acts during motion at $\mu_k N$ (typically $\mu_k < \mu_s$).

Q2. A $10$ kg box rests on a $30^\circ$ ramp with $\mu_s = 0.4$. Determine (a) whether the box slides, and (b) if not, the friction force required to hold it. [4 marks]

• Cue. Slope force $mg \sin 30^\circ = 49$ N; max static friction $\mu_s mg \cos 30^\circ = 0.4 \times 10 \times 9.8 \times 0.866 = 33.9$ N. Slope force exceeds max static, so the box slides.

Q3. Refer to the West Gate truck descent. (a) Calculate the slope-parallel weight component. (b) Determine the maximum kinetic-friction force on the tyres. (c) Explain why engine retarders are still required on long descents. [2+2+2 marks]

• Cue. (a) $1.71 \times 10^4$ N. (b) $1.17 \times 10^5$ N. (c) To prevent brake-fade heating; friction alone heats discs to dangerous levels.