Question 2.7.1: Consider a spring–mass system with sliding friction describe......
Consider a spring–mass system with sliding friction described by equation (2.97) with stiffness k = 1.5 × 10^4 N/m, driving harmonically a 10-kg mass by a force of 90 N at 25 Hz. Calculate the approximate amplitude of steady-state motion assuming that both the mass and the surface are steel (unlubricated).
m \ddot{x}+\mu m g \operatorname{sgn}(\ddot{x})+k x=F_0 \sin \omega t (2.97)
First, look up the coefficient of friction in Table 1.5, which is μ = 0.3. Then from inequality (2.112),
4 \mu\, m g<\pi F_0 (2.112)
\begin{aligned} 4 \mu\ m g & =4(0.3)(10\ kg )\left(9.8\ m / s ^2\right)=117.6\ N \\ & <(90\ N )(3.1415)=282.74=\pi F_0 \end{aligned}
so that the approximation developed previously for the steady-state-response amplitude is valid. Converting 25 Hz to 157 rad/s and using equation (2.109) then yields
X=\frac{F_0}{k} \frac{\sqrt{1-\left(4 \mu\, m g / \pi F_0\right)^2}}{\left|\left(1-r^2\right)\right|} (2.109)
X=\frac{90\ N }{1.5 \times 10^4\ N / m } \frac{\sqrt{1-(117.6\\ N / 282.74\ N )^2}}{\left|1-\left(2.467 \times 10^4 / 1.5 \times 10^3\right)\right|}=3.53 \times 10^{-4}\ m
Thus the amplitude of oscillation will be less than 1 mm.
| Table 1.5 Approximate Coefficients of Friction For Various Objects Sliding Together | ||
| Material | Kinetic | Static |
| Metal on metal (lubricated) | 0.07 | 0.09 |
| Wood on wood | 0.2 | 0.25 |
| Steel on steel (unlubricated) | 0.3 | 0.75 |
| Rubber on steel | 1.0 | 1.20 |