UKH

I've been teaching A Level maths for a few years and have never been stumped by a question like this one before.

It is from a Level 3 (A Level standard) Engineering Maths paper (Design and Mechtronic Engineering).

First part is easy, 2nd part, I haven't got a clue what it is wanting, even when I've been through the mark scheme.

a) Friction and Load are connected by F=aL+b, where a and b are constants to be determined,  when F=100, L=70.   When F=80, L=50.

b) Confirm your values for a and b, by plotting a graph where a and b are values to be determined.

The axis given for (b) have 'a' on the x axis from 0-1.4 and F on the y-axis from 0-140.

100 points for anyone who can guess what they are expecting for part b!

In reply to ablackett:Annotated graph showing how intercept and gradient determine, confirm the earlier calculated values?

 

In reply to ablackett:

Simultaneous equations surely?

a=1

b=30

It's very badly written but I assume that's what they're after. However you'd need a law degree to be certain.

In reply to Removed User:

They are the values of a and b, but the answer given states that the graph would plot a and F, That isn't what you would plot - you would plot F and L. Given that a is a constant, the suggested answer seems very odd. 

In reply to ablackett:

Simultaneous equation for a.

For b, F=aL+b is in the form of a straight line. Plot the 2 given points and draw a line between them. From this graph work out the y intercept and gradient to confirm these give the same answers as in a.

Edit: with a on the x axis, rearrange the equation so that it is of the form y=mx+c where y=L and x=a, and m and c constants in terms of F and L

In reply to ablackett:

Yes - part a is easy simultaneous equations to give a=1 and b=30.

I initially thought it was an obvious question to plot F against L and read off 'a' as the gradient and 'b' as the intercept.  But that isn't what it wants. 

Plotting F=100 and a=1, then plotting F=80 and a=1 and joining both lines to a=0, F=30 gives two lines with a gradient of 50 and 70, so does answer the question.  But it seems such an odd way of doing things I can't help thinking I have missed something.

In reply to ablackett:

Jas128 - what exactly are you then plotting? You're left in a situation where you have two known points in terms of F and L, but plotting in a-F space. So without using the whole of the method of part (a) too, which defeats the object of part (b), you can't really plot two points. Unless I'm wrong here???

But, you can plot some lines... here's my convoluted stab at it (though it's a pretty poor question and answer...).

Let b=30. Let's assume nothing about a for now. So, with b=30, we can calculate the values of F given the two l-values (might help to put this info in three columns (1) a from 0-1.4 (2) calculated F-values for l=70 (3) calculated F-values for l=50). Now plot on the same graph (2) vs (1) and (3) vs (1). These will be upward sloping. Now ask yourself, given we know that F = 100 when L=70 and F=80 when L=50, what is a? I.e. what single a-value satisfies (a) being on the (2)-(1) line when F=100 and (b) being on the (3)-(1) line when F=80? If helpful, whack a vertical line to show this is satisfied at a=1.

In reply to ablackett:

I think you can find a and b from an a vs F graph without needing to measure gradient:

- Think of the equation as F = La + b (so L is the constant and a the variable)

- Draw 2 lines from the origin, one with gradient 70, one with gradient 50

- Slide a vertical ruler along the lines until you find where they are 20 apart (because we are interested in 2 values of F that are 20 apart).

- Read off the a (x) axis at this point - it should give 1

- Read off the F (y) axis for either line at this position

- Subtract your reading from either 100 or 80 depending on which line was used - this should give b

In reply to tjhare1:

no you're right, I hadn't worked through the whole question explicitly and hadn't appreciated the axes given! 

In reply to ablackett:

Do you have to use the axes given? The question just says "a graph" so would one in the F-L space be allowed?

In reply to ablackett:

That's very perverse thing to do. Three points and two straight lines to represent F=aL+b. What is the extra engineering insight or educational value of that? It actively obscures any insight and reduces the educational value.

It is not specified how a & b are to be determined from graph. Hence F vs L is a valid choice of graph and you have two points and one line to represent F=aL+b.

Do you have to use the graph paper supplied?

In educational terms it seems really bad.

In engineering terms I can't see any advantage, we need an expert opinion but there's never a tribolist when you want one.

In reply to elsewhere:

I'm a mechanical engineer and haven't got a clue what they're trying to get the student to demonstrate either mathematically or in terms of an engineering principle.

In reply to tlouth7:

Further to my previous post, here is the graph I would plot:

https://www.wolframalpha.com/input/?i=plot+y+%3D+50+x+and+y+%3D+70+x+betwee...

Reading off the F axis for a = 1 we get F = 50 and 70

Subtracting these from 80 and 100 respectively gives 30 which is the value of b

Edit to add that I think the question is pretty daft and I would absolutely plot F against L given the choice.

In reply to ablackett:

You could be over-analysing and they just mistakenly mislabelled the axes. I don't think I've ever encountered a textbook that didn't slip up somewhere.

In reply to Luke90:

But it’s an exam question, not a text book!

 

i will work through the suggestions tomorrow but I’m tending towards it’s just a bloody daft question, or perhaps a printing error which they have made the mark scheme fit to cover up the error.

In reply to ablackett:

The problem is trivial: simultaneous equations quickly show that F = L + 30.

But the final part of the question:

"b) Confirm your values for a and b, by plotting a graph where a and b are values to be determined. The axis given for (b) have 'a' on the x axis from 0-1.4 and F on the y-axis from 0-140"

is more or less gobbledegook.

Is this really how the sentence was written, or has it somehow got mangled in it's transfer to UKC? As it stands, it's a disgrace, because it doesn't make sense from either an English grammar or Maths/Science point of view! Luckily, this last ghastly sentence is completely redundant, because the problem is easily -much more easily - solved without it.

In reply to ablackett:

One hundred what?  Bananas?

Solve simultaneous equations for a and b, then plot a graph of F vs L and read a and b off as the gradient and intercept.

This makes no sense and is I think an error considering the earlier wording, with the plot supposed to be F and L.

In reply to ablackett:

The F=30 result for L=0 lies on the a=1 vertical line along with the F=80 and F=100 data. 

It might be what the examiners want but F=30, a=0 does not represent a physical reality consistent with F=aL+b.

What physics conveniently says F=30, a=0 at L=0 rather than F=30, a=2 at L=0? 

If the examiner is saying you can arbitrarily make a=0 when L=0 because the product aL is zero then 2 is an equally valid arbitrary choice. a=2 maintains aL=0 resulting in different gradients so the graph confirms nothing.

Unless it is a standard pure mathematical tool, expecting the student to plot a point with a=/=1 is unfair as it contradicts the question. 

Is this an engineering maths exam question written without showing it to an engineer?

Ask the examiners to circulate it round some mech eng departments for comment? 

In reply to John Stainforth:

I think the last sentence must be OP describing the axes, rather than part of the question text. I read the first sentence to mean that the student cannot use the values found in part a to plot the graph; they must derive them from the graph.

Solving graphically may not be redundant in as much as this is probably part of the syllabus.

In reply to wintertree:

While I agree that these axes are not the most logical, it is possible to solve part b as posed using them.

In reply to tlouth7:

You are right - going back and looking at the axis range for ‘a’ in the OP.  The language is then very ambiguous.

No matter how it’s looked at it is an awful question in terms of science (units), language and clarity, methodology (nobody does it this way) and generally typifies what elsewhere said - questions written with “flavour” by someone without a clue.

In reply to wintertree:

Yes it is pretty bad isn't it. I had been thinking of it as a pure maths question so wasn't fussed about the lack of units, but given it is framed as being about friction (where has mu disappeared to?) they should absolutely be included (though it is possible OP skipped them in the interest of clarity).

Isn't plotting straight line graphs and finding gradient and intercept a little basic for A-level? I would expect a subject with "mechatronic" in its name to include some control theory such as Bode plots or PID time response, both of which are considerably more complex than y = mx + c

In reply to tlouth7:

To answer one point the original wording was as follows.

 

"A law connecting friction F and load L is of the form F=aL+b were a and b are values to be determined.

Determine the values for a and b:

F=100 when L=70 and

F=80 when L=50                                   (7 marks)

Confirm your values for a and b by plotting a graph on Figure 2 where a and b are values to be determined.                                       (6 marks)"