M3 Statics killing me slowly

Hi guys,

I have literally tried so hard to understand this part of M3 but my brain isn't cooperating. Does anyone know why we have to take a 'strip' out for the calculus bit of finding the centre of mass of a lamina? Surely if you just have the equation of the graph you just plug that stuff into the xbar and ybar formulae, integrate then get the answer?

Nearly every example in the M3 textbook involves taking a strip out of the graph but I really am struggling to understand the logic behind this :/ Like would we actually need to do this in an exam question?
somebody help pls

Reply 1

Student403

Original post by physicalgraffiti

In the exam, you'll be fine just using the equation of the graph into the xbar and ybar formulae. Those formulae are generalised, derived from taking those strips you're talking about and summing them.

If you're curious as to why the approach to the derivations in the book uses strips, that's because it's how we derive things from calculus (essentially summing up all the infinitesimally small strips or pieces of your graph or whatever).

Reply 2

physicalgraffiti

Original post by Student403

Ahh ok, I thought so. However there's a particular scenario that confuses me and that's when you have two graphs to deal with i.e. question 12 on Exercise 5A in the M3 textbook (I'll link the pdf solution, it's on page 13). Yet again, they're using a dreaded strip XD I just don't know how I'd go about doing it with the xbar and ybar formulae?

http://pmt.physicsandmathstutor.com/download/Maths/A-level/M3/Solutionbank-Heinemann/M3%20Chapter%205.pdf

Reply 3

Student403

Original post by physicalgraffiti

Lol don't be scared of the strip! The strip is your friend (or you can ignore it if you're about that jumping in to the xbar/ybar equation lyf :tongue:

).

When you have 2 curves, it's the exact same process as with 1 curve except you're taking one away from the other (usually which gets taken away is pretty obvious, the closer one). Think of it like how you'd integrate to find the area between 2 curves, taking one away from the other before integrating. Sound familiar?

Reply 4

physicalgraffiti

Original post by Student403

Lol don't be scared of the strip! The strip is your friend (or you can ignore it if you're about that jumping in to the xbar/ybar equation lyf :tongue:

The strip is honestly terrifying XD
So if I'm to use the xbar/ybar thing, do I just take away one curve from the other to get one single equation that I can then integrate? I understood I had to do this for some questions like where one graph is explicitly over another one but for this particular question the beginning of one graph is attached onto the end of the other so I didn't know if I'd take them away or what :P

Reply 5

Student403

Original post by physicalgraffiti

Pretty much. If you tilt the graph 90 degrees, it's basically like "one graph is over the other". You account for this by choosing your limits

Reply 6

physicalgraffiti

Original post by Student403

Pretty much. If you tilt the graph 90 degrees, it's basically like "one graph is over the other". You account for this by choosing your limits

Ahh ok, so would I do 2x-(24-4x)^1/2 then integrate between 0 and 6 or?

I feel so hopeless at this stuff ;__;

Reply 7

Student403

Original post by physicalgraffiti

Ahh ok, so would I do 2x-(24-4x)^1/2 then integrate between 0 and 6 or?

I feel so hopeless at this stuff ;__;

Okay, disregard my previous message as I think it might be a bit misleading (sorry about that)

For one, the solution bank has some small errors as you might have spotted (top of the final page of Q12 where they go from (x1+x2)(x1+x2) to (x2^2 - x1^2)

What you're better off doing is what they've done in the solution bank, which is find x as a function of y and integrate with bounds of x (0 to 4). You can rearrange both eqn (1) and eqn (2) quite easily to give you x in terms of y, then subtract one from the other.

I've tried to draw a diagram here to show why you can't do the 2x-(24-4x)^1/2 thing here: See what you get when you integrate (1)-(2)? It's definitely not what you're looking for

Spoiler

What you want to be doing is shown in this picture (Hope this isn't confusing, but i've switched the axes to make it look more clear which equation you should be taking away from which). See how the plain orange area is the shape you want, but to get it you have to integrate to find the total orange area and take away the purple area? So your bounds would be 0 to 4. Can you see what your equations would be?

Spoiler

(edited 6 years ago)