Recent content by Steve4Physics

This somewhat repeats what has already been said, but there is an important underlying point that's maybe worth making explicitly... Suppose you have any function ##y=f(x)## and some (say positive) constant ##a##. The graphs of ##y=f(x)## and ##y = f(x-a)## are the same shape - but the graph...

I get the same equation as you - but don’t forget that you haven’t completely answered the question. You are given values for the masses, so you will need to calculate a value for ##a_1##’s magnitude and state its direction. We commonly take upwards as positive and to-the-right as positive...

Yes. Agree 100%. In case there's any confusion, I certainly wasn't (in Post #10) promoting the OP's (Post #1) approach. But in Post #8, they wrote: No one had specifically replied to this. My Post #10 was an attempt to do this in order to help the OP better understand what was going on.

@michelp, this is a bit late in the day, but it might explain the misunderstanding... In the case of a simple Atwood machine (with ##m_2 > m_1## say), it is not uncommon to find equations written in the form: ## T – m_1 g = m_1 a## and ##m_2 g – T = m_2 a## For example...

A few thoughts... 20MPa is not exceptional. E.g. 20MPa is the tension in a 1mm diameter vertical string with 1.6kg hanging from it. Think of that as a ‘pressure’ of -20MPa. The article refers to ‘extremely powerful ultrasound’ which sounds dramatic but may be misleading. The author probably...

I would like to add something as I too remember being confused by this many decades ago. ##\hat {\rho}## is not a unique unit vector and it would be clearer (IMO) to write it as a function, i.e. to write it as ##\hat {\rho}{(\phi)}##. Referring back to Post #1 of this thread, this is what the...

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...