Friction

Does It Move? The friction verdict

Compare the push to the static limit: below it the block stays, above it the block slides and kinetic friction sets the acceleration.

mechanics Pushing 8 N then 12 N against a 10 N static limit makes the stay-or-slide verdict concrete, with a clean 3 m/s^2 once it slides.

Example

Compare the push to the static limit: below it the block stays, above it the block slides and kinetic friction sets the acceleration. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Push gently: it stays put

Push with 8 newtons. That is less than the 10 newton static limit, so static friction quietly matches it at 8 newtons and the block does not move.

8 N<10 N  ⇒  stays8\ \text{N} < 10\ \text{N} \;\Rightarrow\; \text{stays}
Push below the limit: no motionAn isolated block with a rightward push arrow and an equal leftward friction arrow.mpushf

Push harder: it breaks free

Push with 12 newtons instead. That beats the 10 newton limit, so the block breaks free and starts sliding.

12 N>10 N  ⇒  slides12\ \text{N} > 10\ \text{N} \;\Rightarrow\; \text{slides}

The sign of the excess decides the verdict

Compare the push to the static limit. Negative excess means the limit can still hold, zero is just at the edge, and positive excess means the block breaks free.

Pfsmax⁡P−fsmax⁡8 N10 N−2 N10 N10 N0 N12 N10 N2 N\begin{array}{c|c|c}P & f_s^{\max} & P - f_s^{\max} \\ \hline 8\ \text{N} & 10\ \text{N} & -2\ \text{N} \\ 10\ \text{N} & 10\ \text{N} & 0\ \text{N} \\ 12\ \text{N} & 10\ \text{N} & 2\ \text{N}\end{array}

Now kinetic friction sets the acceleration

While sliding, friction is the kinetic 6 newtons. The net force is the push minus friction, 12 minus 6, which is 6 newtons. Divide by the mass: 6 over 2 is 3 metres per second squared. The drag dropped from the 10 newton limit to 6 newtons the moment it broke free, which is why a heavy box lurches as it starts to slide.

a=12−62 kg=62 kg=3 m/s2a = \frac{12 - 6}{2\ \text{kg}} = \frac{6}{2\ \text{kg}} = \hl{3}\ \text{m}/\text{s}^{2}
Sliding: net force accelerates itAn isolated block with a longer rightward push arrow, a shorter leftward friction arrow, and an acceleration arrow.mpushfa

More sliding push gives more acceleration

Once it is sliding, kinetic friction is the fixed drag for this block and surface. More push leaves more net force, so acceleration grows.

PFneta12 N6 N3 m/s214 N8 N4 m/s216 N10 N5 m/s2\begin{array}{c|c|c}P & F_{\text{net}} & a \\ \hline 12\ \text{N} & 6\ \text{N} & 3\ \text{m}/\text{s}^{2} \\ 14\ \text{N} & 8\ \text{N} & 4\ \text{m}/\text{s}^{2} \\ 16\ \text{N} & 10\ \text{N} & 5\ \text{m}/\text{s}^{2}\end{array}