Recent content by Steve4Physics

Note that this is only justified because section DG is in equlibrium with only 2 forces acting on it. So: a) the 2 forces must have equal magnitudes and opposite directions; b) the 2 forces must have a common line-of-action (dashed line, below).so that these forces have zero net moment. Edit...

I agree. And well done for using LaTeX! A couple of points: 1. You have assumed, without justification, that some reactions are at 45 degrees to the horizontal. It happens to be correct in this case but do you see why? (Using the separate xy components is safer.) 2. The question asks for the...

The Post #1 question seems to have only 20 minutes (exam’) working time advised, so a quick method is needed. There is a fairly quick approach which uses (almost) no equations – essentially just diagrams. I hope I’m not giving the OP too much assistance but they seem to have made reasonable...

Happy 2026 @haruspex! If B is pinned and A is on a horizontal track, I suspect this would happen: It satisfies the (intuitive) requirement that the system's geometrical centre moves right. Yes. I believe so (i.e. that ##A_x## and ##B_x## are non-zero and can be determined). In fact I've just...

Happy New Year @Lnewqban! Yes, I see what you mean. So, for example, with A remaining pinned and with B free to move horizontally, the frame could be deformed into something like this

Isn’t the system statically indeterminate? [EDIT - No it isn't! See later posts.] Considering only external forces, it is trivial to show that ##A_y = - F, B_y = F## and ## A_x + B_x = -F##: ##A_x## and ##B_x## cannot be uniquely determined because they have the same line of action...

@jojosg, there are a number of issues, so it's complicated to explain. But here is an attempt... I've written some of your equations from Post 14, so that LaTeX renders them correctly (for me anyway). This required putting a pair of hash symbols (## \# \# ##) at the beginning and end of each...

I'm out for most of today but will take a look when I return.

You are on the right track but it looks like you have ignored some forces when expressing ##\Sigma F_x=0, ~\Sigma F_y=0## and ##\Sigma M=0##. (But see EDIT at bottom). You found that ##O_x=-F## (i.e. F acting to the left) and ##O_y = F##. With the bottom-right section removed, the new object...

The 'Method of Sections' is a way around this. See Post #9 which is intended to start the OP solving the problem using this method.

The way you've marked the 'cut' in your Post #3 diagram suggests that you may have misunderstood the method of sections. A suitable cut is th red line: Consider the section above the red line as a single object. It has known external forces at O, E and C. There are 3 additional unknown...

That would only be true if the tension in member OH were zero. But in the x-direction, the force-balance at point O is: ##-F~ +~ T_{OA}~ +~ T_{OH} \cos 45^ \circ = 0##

FWIW there is a typo'. It should be ##\chi_n=e^{-i \frac { E_n}{\hbar} t}##.

Why did the girl-eigenvector and the boy-eigenvector split up? They found that they had very different eigenvalues. And they both needed more Hilbert space.

No. The shape of a stress-strain graph doesn't depnd on the wire's diameter. So the breaking stress and strain are 'constants' for the material (in the same way that a material's density doesn't depend on its size/shape). E.g. wire-1 has cross-sectional area A and wire-2 has cross-sectional...