Topic 2.7 Kinetic and Static Friction: distinguish static from kinetic friction, and calculate friction forces using the coefficient of friction and the normal force.

What this topic is asking

The College Board (Topic 2.7) wants you to distinguish static friction from kinetic friction, calculate each using the coefficient of friction and the normal force, and incorporate friction into Newton's second law. The subtle point the exam loves is that static friction is a variable force, adjusting itself up to a maximum, while kinetic friction has a fixed value once sliding begins.

Static versus kinetic friction

The everyday experience captures the difference: it takes a hard initial push to get a heavy box moving (overcoming static friction), but once it slides it is easier to keep going (only kinetic friction acts). This is because the maximum static friction is usually larger than the kinetic friction.

The friction equations

The crucial difference in how the two are used: $f_k = \mu_k N$ is an equality you can plug into directly once an object slides, but $\mu_s N$ gives only the threshold. The actual static friction is found from the equilibrium condition (it balances the applied force) and never exceeds $\mu_s N$.

Friction depends on the normal force

Both friction forces are proportional to the normal force $N$, not to the contact area or the object's weight directly. On a level surface with no vertical applied force, $N = mg$, so friction grows with weight. But if you press down on the object, $N$ (and friction) increases; if you pull up at an angle, $N$ (and friction) decreases. Always find the normal force from the vertical force balance first, then compute friction. On an inclined plane, the normal force is $N = mg\cos\theta$, smaller than the full weight, which is why objects slide more easily on steeper slopes.

Why static friction being variable matters

The single most common friction mistake is to treat static friction as always equal to $\mu_s N$. In fact, static friction is a response force: it provides exactly enough force to keep an object from sliding, no more. If you push a heavy crate gently, static friction matches your push and the net force stays zero; push harder and static friction grows to match; only when your push exceeds $\mu_s N$ does the surface "let go" and the crate begins to slide, at which point friction drops to the smaller kinetic value $\mu_k N$. This is why an object can be on the verge of moving (static friction at its maximum) and then accelerate suddenly once it breaks free. To decide whether motion occurs, compare the applied force with $\mu_s N$: if it is smaller, the object stays put and static friction equals the applied force; if it is larger, the object accelerates under a net force of (applied force minus $\mu_k N$). Getting this logic right, threshold first, then kinetic friction, is essential for friction problems on the exam.

Try this

Q1. A $10$ kg box on a level floor has $\mu_k = 0.25$. Calculate the kinetic friction force while it slides ($g = 9.8$ m/s squared). [2 points]

• Cue. $f_k = \mu_k mg = (0.25)(10)(9.8) = 24.5$ N.

Q2. A box needs $40$ N to start moving but the applied force is only $30$ N. State the static friction force. [1 point]

• Cue. $30$ N; static friction matches the applied force (below the $40$ N maximum), so the box stays still.