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AlephZero
#3
Feb5-13, 12:42 PM
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Quote Quote by CompuChip View Post
You mean that [itex]F_\mathrm{friction}(t) = \mu x' (t)[/itex]?

How would you solve it without the friction term? I guess you would make an Ansatz for the form of the solution such as [itex]x(t) = e^{\lambda t}[/itex] ?
Most engineers (at least in the UK and US) would call that "viscous damping", not "friction".

For the Coulomb model of friction, F in the OP's equation is constant, and its sign depends on the sign of the velocity.

You can easily solve the two separate cases where F is positive or negative. The solution is the same as if the mass and spring was vertical, and F was the weight of the mass.

For the complete solution, you start with one of the two solutions (depending on the initial conditiosn) until the velocity = 0, then you switch to the other solution, and so on. You can't easily get a single equation that gives the complete solution in one "formula" for x.

The graph of displacement against time will look like a sequence of half-oscillations of simple harmonic motion, with amplitudes that decrease in a linear progression (not exponentially). The mass will stop moving after a finite number of half-osciillations, at some position where the static friction force can balance the tension in the spring.