Recent content by Motorbiker

I added them because there are three forces acting on the system, F1, F2 and F3 right?

Yes they are.

Please can I get someone help with this? I don't really understand the solution fully, where does the shear force come from?

I'm guessing this equation applies to both sides? Also how do we know that there is shear force acting on the beam?

Thanks for this. Could you please explain how you got this equation? I want to understand this better.

I’ve attached another picture of my working, please let me know if you can read this one.

I’ve attached my finished solution to this problem, can you just check that it looks okay? I just want to make sure I have a good understanding.

Okay thanks a lot for that, I understand it better and it looks like my labelling for the x's is correct for this case.

I think zero x1,x2 and x3 give zero potential energy.

I tend to use their names because that's way it's done in my notes however I could just use the values of the spring constants if that makes the problem simpler or clearer. When the system is in equilibrium there are 3 x's for 3 rigid bodies? The x is simply a frame of reference for the body...

I think I showed the equations in my working?

In my working, I have drawn a free body diagram of the LHS of the beam, I obtained some equations but after that not sure what I need to do. Usuaully in vibration problems you are given BC's then you plug them into the general solution to obtain the frequency equation. Please can you help me...

I'm not sure how to define the x1,x2,x3, usually the x is placed above the respective rigid body-so x1 on top of m1 and so on.And the arrow would be pointing to the right matching the arrow on the force. So would this method be suitable in this problem?I understand that all k's except for k3...

Yes I have incorporated your comment. What am I doing wrong?

I have attached my work on this problem so far. Let me know if it looks correct. @anuttarasammyak