Recent content by dsol

Maples is right. You should be using the equations for PLANE stress/strain and not AXIAL stress/strain. For axial strain ONLY, you would solve \epsilonxx=\sigmaxx/E. For plane strain, use \epsilonxx=\sigmaxx/E - \upsilon/E*\sigmayy and comparably \epsilonyy=\sigmayy/E -...

My first thoughts are: Dampers of this design work by the loss that occurs when a viscous fluid is forced through a small orifice (most of the losses occur because the jet that is induced through the holes creates turbulence and drag). A first shot would be to find the velocity of those jets...

As an aside, most engineers do not need to know what energy "is". I assume you are an early undergraduate student in engineering, and I am basing the following statements on that assumption: Take it from someone who spent way too much time trying to understand the minutia of physics in his...

In your case, sumF=mg-kx=0 at x=d, mg=kd --> k=mg/d To find the maximum distance the spring stretches, you need to use an energy perspective (@Chi Meson) sum E=mgx-(k*x^2)/2-(m*v^2)/2=0 The first term is the potential energy of the mass. The second term is the potential energy of the...

The first method F = mg = kx is what you would use to find the spring constant k. The second equation is used for the dynamic problem "How fast is the mass moving at distance x?". To solve, equate the potential energy "lost" by the downward motion to the potential energy of the spring AND the...