Laplace transforms: Under-damped Mass-Spring System on an Incline

Brandon.plubell 05:44, 26 October 2009 (UTC)

Under-Damped Mass-Spring System on an Incline

Part 1 - Use Laplace Transformations

Problem Statement

Find the equation of motion for the mass in the system subjected to the forces shown in the Free Body Diagram (FBD). The inclined surface is coated in SAE 30 oil.

Initial Conditions and Values

• A is the area of the box in contact with the surface
• g is the gravitational acceleration field constant
• b_t is the thickness of the fluid covering the inclined surface
• μ is the viscosity constant of the fluid;
• m is the mass of the box
• k is the spring constant

$A={\frac {1}{2}}m^{2}$

$g=9.81{\frac {m}{s^{2}}}$

$b_{t}=1mm{\frac {}{}}$

$\mu =0.06{\frac {N\cdot s}{m^{2}}}$

$m=45kg{\frac {}{}}$

$k=200{\frac {N}{m}}$

Let the initial conditions be:

$x(0)=-0.5m{\frac {}{}}$

${\dot {x}}(0)=0{\frac {m}{s}}$

Equations of Equilibrium

The sum of the moments in the x direction yields the equatiom

$+\swarrow \sum F_{x}=m{\ddot {x}}$

$\Rightarrow \ m{\ddot {x}}=F_{s}+F_{f}+mg\sin \theta$

Where

$F_{s}=-k\,x$

$F_{f}=-{\frac {\mu \,A}{b_{t}}}\,{\dot {x}}$

To make the algebra easier, let

$\lambda ={\frac {\mu \,A}{b_{t}}}$

Then, from the sum of forces equation:

$m\,{\ddot {x}}+\lambda \,{\dot {x}}+k\,x=mg\sin \theta$

$\Rightarrow \ {\ddot {x}}+{\frac {\lambda }{m}}\,{\dot {x}}+{\frac {k}{m}}\,x=g\sin \theta$