Recent content by tanujkush

k (and kt) are in N/mm so their values in SI units would be 30N/mm or 30x10^3 N/m

Ok so assuming downward direction as positive and also assuming that the spring damper push the sprung mass upwards(i.e. the spring gets compressed), here are the ODEs: F = ma f(t)-k(y - y1)-c(\frac{dy}{dt} - \frac{dy1}{dt}) = m_{s}\frac{d^{2}y}{dt^{2}} and c(\frac{dy}{dt} - \frac{dy1}{dt})...

Even so, the way you have used f(t) in the expressions, it appears to be the road displacement rather than a road reaction force. If that is the case then your formulation seems correct. If however, f(t) is a force on the sprung mass (like you showed in your first illustration), then f(t) should...

Hey, at first look, I found these equations to be fishy: is the f(t) a force being applied on the sprung mass m_s? If so, then shouldn't the expression f(t) appear in the equation of motion of m_s as opposed to m_u as you have formulated?

:cool:

Everything tries to reach a state where its energy (total energy, maybe thermal, electrical, potential, whatever) is minimum. Contrary to what you quote, particles have to *lose* energy as opposed to *get* energy to move from high to low. And as we all know, losing is easier than gaining. How...

Aren't you stating the law of action and reaction?

That kinda sums it up. Whether there is equal pressure on all sides of an object or not, fluids exert pressure when the sum total of its molecules exert a non zero force on a surface which may be exterior to it, or may be an imaginary surface that cuts through the fluid bulk. But be warned, the...

I of course, was talking about the second part of your quote.

See the units of the formula you have got, its got mass times velocity squared by length. That is (M*(L/T)^2)/L = M*L/T^2 - That is by definition a force not pressure. Pressure is force per unit area - Force/Area = (ML/T^2)/(L^2)=M/(LT^2) Pressure is never the *same* thing as a load, although...

like cmdro said, v^2 - u^2 = 2as v = 0 (the cars stopped) u = 26.38m/s a = -4m/s^2 s is the unknown: this gives s = 86.98m now the driver takes 1s to put the brakes on, so distance traveled in one sec = u*t=26.38m. So total s = 86.98+26.38 = 113.36m For the second part...

Oh sorry, I meant the Dirac delta function, not the Kronecker, my bad.

You could take the first approach provided the derivative of both g(x) and h(x) exists. This is reminiscent of the Taylor expansion series. Only, your equation does not have the higher order terms in it.

Well, how I would go about it is this: Take a infinitesimally small portion of the chain say at some x, of length dx and do a free body diagram of this. You should end up with a differential equation which you can then 1. integrate to find out the tension at each point in the chain and thus...

Yeah I missed out the equation for conservation of energy! Sorry! The angles though would be known (if one assumes a non dissipative collision, which I'm sure you've been told to assume)