Recent content by rannasquaer

Given the triangle above where $$V < v'_{1}$$, prove that the \[ v_{1}=V \cos(\psi)+v'_{1} \cos(\theta - \psi) \] It is said that $$v_{1}$$ is equal to the sum of the orthogonal projections on $$v_{1}$$ of $$V$$ and of $$v'_{1}$$ and that is precisely the expression that show. But I couldn't...

Great, I understood how to continue to do the math. Thank you!😄

I think yes, if I rewrite like \[ \mathscr{L}^{-1} \frac{a(s+\lambda)}{(s+ \lambda)^2- \omega^2} + \mathscr{L}^{-1} \frac{b + \lambda a}{(s+ \lambda)^2- \omega^2} \] but I have \[ (s+\lambda)^2-\omega^2 \] and not \[ (s+\lambda)^2+\omega^2 \] The table of Laplace transforms lists that \[...

How to solve the transforms below \[ \mathscr{L}^{-1} \frac{a(s+2 \lambda)+b}{(s+ \lambda)^2- \omega^2} \]

Thank you so much, I was having trouble understanding what to use as f(x), I thought I should use q(x), and everything was going wrong. Thank you!

A beam of length L with fixed ends, has a concentrated force P applied in the center exactly in L / 2. In the differential equation: \[ \frac{d^4y(x)}{dx^4}=\frac{1}{\text{EI}}q(x) \] In which \[ q(x)= P \delta(x-\frac{L}{2}) \] P represents an infinitely concentrated charge distribution...