9.6 Dynamic Friction and Vibrations: Natural Frequency and Damping

9.6 Dynamic Friction and Vibrations: Natural Frequency and Damping

These are the last two sub-topics in the NCEES Dynamics list — "Dynamic friction" and "Vibrations (e.g., natural frequency)" — and vibrations is the one candidates most often reach the exam without having studied. It is a small, highly formulaic topic: three or four equations cover almost everything asked.

Dynamic (Kinetic) Friction

Once sliding begins, the friction force is:

$F_k = \mu_k N$

Three properties, all tested as conceptual items:

1. Independent of contact area. Doubling the footprint does not change the friction force, because the real contact area at asperities is set by the load, not the apparent area.
2. Independent of sliding speed (in the classical Coulomb model).
3. Always opposes relative sliding velocity — its direction is set by the motion, not by the applied force.

Static vs. Kinetic

$\mu_k < \mu_s \qquad\text{typically } \mu_k \approx 0.7\text{–}0.8\,\mu_s$

This inequality has a directly observable consequence: a block on the verge of sliding needs a force $\mu_s N$ to break free, but once moving it needs only $\mu_k N$ to keep going, so the surplus accelerates it. This is the "stick-slip" jerk of a dragged object and the reason antilock braking exists — a rolling tire uses the larger $\mu_s$, while a locked skidding tire is stuck with the smaller $\mu_k$ and stops in a longer distance.

Worked Example: Skidding Stop Distance

A vehicle at 25 m/s locks its brakes on a road with $\mu_k = 0.62$. Find the stopping distance.

$a = -\mu_k g = -0.62(9.81) = -6.08\ \text{m/s}^2$

$v^2 = v_0^2 + 2as \;\Rightarrow\; 0 = (25)^2 + 2(-6.08)s$

$s = \frac{625}{12.16} = \boxed{51.4\ \text{m}}$

Mass does not appear — friction force scales with weight and so does inertia, so they cancel. Note also that $s \propto v^2$: doubling speed quadruples stopping distance, to 206 m.

Springs in Series and Parallel

Getting the equivalent stiffness right precedes every natural-frequency calculation, and the rule is the opposite of resistors:

$\text{Parallel (side by side, share the load): } k_{eq} = k_1 + k_2$ $\text{Series (end to end, share the deflection): } \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$

Physical check: springs in series are softer than either one alone (each stretches), while springs in parallel are stiffer. Two 100 N/m springs give 200 N/m in parallel and 50 N/m in series.

Undamped Free Vibration

The equation of motion $m\ddot{x} + kx = 0$ yields simple harmonic motion at the natural frequency:

$\omega_n = \sqrt{\frac{k}{m}}\ \ [\text{rad/s}], \qquad f_n = \frac{\omega_n}{2\pi}\ \ [\text{Hz}], \qquad \tau = \frac{1}{f_n} = 2\pi\sqrt{\frac{m}{k}}$

$x(t) = A\sin(\omega_n t + \phi)$

The static-deflection shortcut is worth memorizing: since $k = mg/\delta$ for a spring or beam sagging $\delta$ under its own load, $\omega_n = \sqrt{g/\delta}$. Measure the sag and you have the natural frequency without ever knowing $k$ or $m$ separately. A floor deflecting 5 mm under load has $\omega_n = \sqrt{9.81/0.005} = 44.3$ rad/s = 7.1 Hz — uncomfortably close to the 2–4 Hz range of walking harmonics, which is why floor vibration is a serviceability check.

Damped and Forced Vibration

$m\ddot{x} + c\dot{x} + kx = F_0\sin\omega t$

$\zeta = \frac{c}{2\sqrt{km}} = \frac{c}{2m\omega_n}, \qquad \omega_d = \omega_n\sqrt{1-\zeta^2}$

Because $\omega_d = \omega_n\sqrt{1-\zeta^2}$, light damping barely shifts the frequency: at $\zeta = 0.05$ (typical for a steel structure), $\omega_d = 0.9987,\omega_n$. For lightly damped systems you may use $\omega_n$ for $\omega_d$, which is why the undamped formula is so widely applicable.

Resonance

When the forcing frequency approaches $\omega_n$, the amplitude is amplified by:

$MF = \frac{1}{\sqrt{\left[1-(\omega/\omega_n)^2\right]^2 + \left[2\zeta(\omega/\omega_n)\right]^2}}$

At $\omega = \omega_n$ this reduces to $MF = 1/(2\zeta)$. With $\zeta = 0.02$, the amplification is 25× — which is why resonance destroys machines and why the standard remedies are to shift $\omega_n$ (change mass or stiffness), avoid the forcing frequency, or add damping.

Worked Example: Natural Frequency and Damping

A 40 kg machine sits on two identical parallel springs, each $k = 5{,}000$ N/m, with a damper $c = 180$ N·s/m. Find $\omega_n$, $f_n$, $\zeta$, and $\omega_d$.

Equivalent stiffness (parallel): $k_{eq} = 5{,}000 + 5{,}000 = 10{,}000\ \text{N/m}$

Natural frequency: $\omega_n = \sqrt{\frac{10{,}000}{40}} = \sqrt{250} = 15.81\ \text{rad/s}, \qquad f_n = \frac{15.81}{2\pi} = \boxed{2.52\ \text{Hz}}$

Damping ratio: $\zeta = \frac{c}{2\sqrt{km}} = \frac{180}{2\sqrt{(10{,}000)(40)}} = \frac{180}{2\sqrt{400{,}000}} = \frac{180}{2(632.5)} = \boxed{0.142}$

Underdamped, as most machinery mounts are.

Damped frequency: $\omega_d = 15.81\sqrt{1-(0.142)^2} = 15.81\sqrt{0.9798} = 15.65\ \text{rad/s}$

Only 1% below $\omega_n$. Resonant amplification if forced at $\omega_n$: $MF = 1/(2 \times 0.142) = 3.5\times$ — modest, because the damper is doing its job. Remove the damper and the amplification is unbounded.

Trap: adding the two springs' deflections instead of their stiffnesses treats parallel springs as series, giving $k_{eq} = 2{,}500$ N/m and $f_n = 1.26$ Hz — exactly half the correct value.