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# Reaction forces

Take a step. You exert a stepping force on the ground. It is of course at an angle.1

• The horizontal part $F_{\text{step}\parallel}$ pushes along the surface. But your foot doesn’t slide since a static friction $f_s$ appears to balance it out.2 So, $f_s$ must be equal to the horizontal stepping component, but opposite.
• The vertical part $F_{\text{step}\perp}$ pushes directly down into the ground. But your foot doesn’t break through the ground since a normal force $n$ appears to balance it out. So, $n$ must be equal to the vertical stepping component, but opposite.

It now seems like static friction $f_s$3 and the normal force $n$ are components of the reaction force to your stepping force – the other half of the action/reaction force-pair – from Newton’s 3rd law:

\begin{align} \vec F_\text{step on surface} &=-\vec F_\text{surface reacting}\quad\Leftrightarrow \\ \begin{pmatrix} F_\parallel \\ F_\perp \end{pmatrix} &=- \begin{pmatrix}f_s \\ n \end{pmatrix} \end{align}

Note a crucial detail: The overall pattern here is that your stepping force does not push you forwards. The reaction to your stepping force does. We are able to take a step or drive a bike or car purely because the environment reacts to our forces.4