Recent content by sassora

Yes I agree about the average force. I'd like to express this as: Based on the triangle formed on the F-e graph: Work done = ½Fmax.emax The area on the σ-ε graph will be the same but rescaled. σ will be 1/A of F and ε will be 1/L of e, the area under the σ-ε graph will be the area under the...

I see from the diagram that the area is ½σmaxεmax. If we assume A,L are constant then we have (1/2AL)(Fmaxemax) For the work done are we then considering the average force applied is Fmax/2 and so the work done is (Fmax/2).emax?

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That makes sense - just because ε and e are correlated won't mean that the both give the work done, but something related to the work done.

You haven't explicitly, I was trying to work out the reason for the change in variable. I understand the reason why my attempt wasn't quite right was by treating F as a constant although I'm not completely clear on that.

σ varies with ε and I treated it as a constant? So you've rewritten the integral using ε = e/L to give ##\frac {\int F.de}{AL}##. I guess that shows that the area is work done over volume but how did you know to make that transformation? I would have thought that a F depends on ε but by...

∫ σ dε = ∫ F / A dε = F ε / A = F (e / L) / A = Fe / LA Fe is the work done and LA extends the cross-sectional area to the volume. So the area under the line is work done / volume. If the cross-sectional A wasn't constant then I guess we'd be doing a second integral, which sounds quite...

Aren't we looking for the area rather than the gradient? Not sure why I would go straight to dividing stress by strain?

Homework Statement Deducing what the area under the stress-strain curve shows. There are four option in the attached image. I can discount work done by considering the units. The remaining ones seem plausible, but only one is true. Homework Equations stress = force / area; strain =...

Thanks it's good to have that confirmation

Homework Statement Centre of gravity - the point at which: 1) gravity acts on a body or 2) weight of a body may be considered to act. The answer is 2) and I understand why - because gravity acts all over but it is easier to calculate a single point, an average point of where the mass is...

I agree. I've met the requirements of the question, we have helpfully analysed the situation a little further. I think @CWatters point was important about the assumption that they must have assumed the climber or at least his legs are horizontal (and level with his centre of mass) so that there...

I've taken the question from here: http://qualifications.pearson.com/content/dam/pdf/A-Level/Physics/2013/Exam-materials/6PH01_01_que_20150519.pdf (it's question 1). The only plausible answer is C (the one attached in my first post), I'd get the mark but I'm trying to learn this inside out...

Yes, I was quibbling with the details to have a better understanding of what I am assuming. Do you mean that the normal reaction force is due to the horizontal component of tension? I can see that these would cancel out in equilibrium but the wall cannot have a normal reaction if no force is...

Homework Statement Free body force diagram question, where I have selected the best answer but the diagram doesn't ring completely true (see diagram uploaded). I think mathematically the diagram is sufficient but it doesn't properly represent the situation. Do you agree? Homework EquationsThe...