= Static friction
{title2=$|f_s|\le\mu_sN$}
<Static friction> is the tangential contact <force> that maintains zero <contact slip>. “Static” describes the contact state: both surfaces can move together in the chosen <inertial frame>. In the ideal dry-contact model, the actual <static friction> adjusts to the <equations of motion>, subject to a bound, rather than always taking its maximum value.<University Physics Volume 1 by William Moebs and others/6.2 Friction>[Static and Kinetic Friction; Magnitude of Static Friction, equation 6.1 and responsive-force paragraph]{ref}
$$
|f_s|\le\mu_sN,\qquad N\ge0,\quad\mu_s\ge0.
$$
Here $N$ is the magnitude of the <normal force> and $\mu_s$ the <coefficient of static friction>. Equality means limiting <static friction>; it is a threshold, not an instruction to replace every unknown contact <force> by $\mu_sN$.
For a stipulated example, place a nonrotating $2\,\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take <gravitational acceleration> $g=9.8\,\mathrm{m/s^2}$ and <coefficient of static friction> $\mu_s=0.40$, with no other vertical <force>. The vertical <force balance> gives $N=mg=19.6\,\mathrm N$, so the largest allowed <static friction> magnitude is $7.84\,\mathrm N$. Choose right as positive and apply a horizontal <force> $F$. Remaining at rest requires <Newton's second law> to give
$$
F+f_s=0,\qquad f_s=-F,\qquad |F|\le7.84\,\mathrm N.
$$
At $F=2\,\mathrm N$, the <static friction> is $-2\,\mathrm N$; at $F=6\,\mathrm N$, it is $-6\,\mathrm N$. Both states have zero <acceleration>. At $F=10\,\mathrm N$, balancing would require $f_s=-10\,\mathrm N$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent <acceleration> requires a <kinetic friction> law; setting $f_s=-7.84\,\mathrm N$ after <contact slip> begins would still use the wrong contact state. At exactly $F=7.84\,\mathrm N$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.
\Image[/static-friction-adjustable-response.png]
{title=<Static friction> adjusts to the applied <force> within its admissible bound}
{description=Original signed horizontal model for the stipulated $2\,\mathrm{kg}$ block: the shaded band is the allowed <static friction>, the solid line satisfies <force balance>, and the dashed line shows the demand outside that band. The crossed points at $F=\pm10\,\mathrm N$ are impossible static demands, not sliding <forces>.}
The toy model used in <kinetic friction> makes the contact-state distinction explicit: $m=1\,\mathrm{kg}$, $N=10\,\mathrm N$, $\mu_s=0.5$ and <coefficient of kinetic friction> $\mu_k=0.3$. On stationary ground, an initially resting box pushed right with $4\,\mathrm N$ has $f_s=-4\,\mathrm N$. At $6\,\mathrm N$ it cannot remain at rest; once it slides right, $f_k=-3\,\mathrm N$ and $a=3\,\mathrm{m/s^2}$. At $5\,\mathrm N$, rest is allowed with $f_s=-5\,\mathrm N$, whereas an already right-sliding box has $f_k=-3\,\mathrm N$ and $a=2\,\mathrm{m/s^2}$. The applied <force> alone does not specify the <friction> branch: <contact slip> and motion history matter.
For a flat contact with two tangential directions, the bound applies to a <vector norm>, not separately to each component:
$$
\|\mathbf f_s\|\le\mu_sN,\qquad
f_{s,x}^{\,2}+f_{s,y}^{\,2}\le(\mu_sN)^2.
$$
For the stipulated $7.84\,\mathrm N$ bound, balancing horizontal components $4\,\mathrm N$ and $4\,\mathrm N$ requires a <force> magnitude $\sqrt{32}\,\mathrm N\simeq5.657\,\mathrm N$, which is allowed. Components $6\,\mathrm N$ and $6\,\mathrm N$ require $\sqrt{72}\,\mathrm N\simeq8.485\,\mathrm N$, which is forbidden even though each component separately is below $7.84\,\mathrm N$. This circular bound assumes the same <coefficient of static friction> in every tangential direction. Near impending <contact slip>, the limiting <static friction> opposes the relative contact motion that would otherwise begin. Once actual <contact slip> exists, <kinetic friction> instead opposes its tangential <relative velocity>, which need not have the direction of the body's <center of mass> motion.
The <normal force> must also come from the <equations of motion>. On horizontal ground, an upward pull of magnitude $P$ at <angle> $\alpha$ above horizontal gives, for a block remaining at rest with no other <forces>,
$$
N=mg-P\sin\alpha,\qquad f_s=-P\cos\alpha,\qquad
|P\cos\alpha|\le\mu_s(mg-P\sin\alpha).
$$
The <sine> component unloads the contact and the <cosine> component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative <normal force>. For the stipulated $2\,\mathrm{kg}$ block and $\alpha=30^\circ$, the positive threshold is $P=\mu_smg/(\cos\alpha+\mu_s\sin\alpha)\simeq7.3544\,\mathrm N$, with $N\simeq15.9228\,\mathrm N$ and $|f_s|\simeq6.3691\,\mathrm N$. A downward pull increases $N$ instead. An <inclined plane> gives another reason why $N$ is not always $mg$.
To solve a new problem, impose zero <contact slip>, find the required <static friction> and <normal force> from the <equations of motion>, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing <rolling without slipping> on stationary ground, a forward <static friction> can accelerate the vehicle. Its role is to prevent <contact slip>, rather than to oppose all motion of the whole body.
Nor does the word “static” by itself make <friction> do zero <work>. The contact <power> delivered to a body is $P_s=\mathbf f_s\cdot\mathbf v_c$, a <dot product> with its material contact-point <velocity> in the chosen <inertial frame>. At an ideal contact undergoing <rolling without slipping> on stationary ground, $\mathbf v_c=\mathbf0$, so this <power> is zero. For a <rigid body>, adding the translational and rotational contributions gives $\mathbf f_s\cdot\mathbf V_{\mathrm{CM}}+\boldsymbol\tau_{\mathrm{CM}}\cdot\boldsymbol\omega=\mathbf f_s\cdot\mathbf v_c$, where $\boldsymbol\tau_{\mathrm{CM}}$ is the contact <torque> about the <center of mass> and $\boldsymbol\omega$ its <angular velocity>. Counting only the first term omits the rotational contribution.
At a moving contact with zero <contact slip>, the two material contact points share a <velocity> $\mathbf v_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact <power> $f_sU$ when the belt's rightward <velocity> $U$ and rightward <static friction> $f_s$ are positive. The belt receives $-f_sU$ by <Newton's third law>. These contact contributions sum to zero in the ideal no-slip model, while the block's <kinetic energy> can increase. Zero <contact slip> and zero <work> on each body are different conditions.
Back to article page
