Apply Newton's second law to objects on horizontal surfaces and inclined planes, including problems with static and kinetic friction ($f_s \le \mu_s N$, $f_k = \mu_k N$)

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 $a = g(\sin\theta - \mu_k \cos\theta)$.