* Applied Maths * predictions / discussion / aftermath (1 thread only please)

@geographic - thanks for the support, its needed in this question!!! Wow, that part (b) is a proper humdinger!! That’s one that I’d have spent way too much time on in the exam back in the day trying to prove it – and still not getting it right most likely Statics was never my strong point!!




I started by drawing the force diagram and then equating forces up and down. I then took moments about point B and equated the clockwise with the anticlockwise. This was the part I spent loads of time on – I couldn’t figure what the perpendicular distance of the weight line of force from point B was. Eventually (after a long time!!) I saw that the angle around B is (180 – 2theta) and so the angle I used for finding the perpendicular for the weight force was (90-2theta). So that moment was WCos(90-2theta) = WSin(2theta) = 2WSin(theta)Cos(theta) (you really must know your trig well ). Then I had 3 equations to work with:


TSin(theta) = R


W – mew(R) = TCos(theta) and


T = WCos(theta)


Eliminating T and R from the second equation I ended up with 1/Cos(theta) – mew(Sin(theta)) = Cos(theta). Mutiplying across by Cos(theta) and rearranging and noting that 1-sqr(Cos(theta)) = sqr(Sin(theta)) gave me



–mew(Sin(theta))Cos(theta) = sqr(Sin(theta))


Isolating mew gave me mew = -tan(theta). tan(theta) can be negative, in which case mew is positive and so mew > tan(theta). Likewise tan(theta) can be positive, in which case mew is negative. I'm chancing my arm here but I'm going to say that mew must always be positive and so we can ignore this "absurd" possibility. Therefore it must always be the case that mew >= tan(theta) [note that I may have taken the friction force in the wrong direction to begin with, if I had taken the other direction I would end up with mew = tan(theta) and again, because mew must always be positive mew >= tan(theta).


I could easily have made an error here though so if anyone sees any error with it, you might let me know.



Thats it until LC 2012 I guess!! Hope ye all have a great summer break - I'll check back every so often just to see if anyone has posted solutions to the questions, especially this one, as it especially tricky.

:pac:

Thanks for the feedback geographic. I've been checking out that last part again and I believe the answer I should have got was mew = tan(theta) (I took the direction of F to be wrong in the original case).

This corresponds to the general definition of mew, which is the ratio of the frictional force F to the normal reaction R i.e. F/R. If you think of it in terms of a diagram, then theta would be the angle of friction (the angle between the resultant of F and R and R itself), so in the traditional diagram illustrating mew, F, R and the angle of friction (theta), tan(theta) would actually be equal to F/R. Now (assuming the simple case of a book on a rough table, where mew > 0) as long as the force acting on the book remains below the value of the limiting friction force (the max amount of frictional force that can be called into play) then the book will not move (it is in equilibrium). When the value of the force acting on the book equals the frictional force F, then the limiting frictional force has been reached and the book is just at the point of moving (in limiting equilibrium). And finally, when the force acting on the book exceeds the limiting frictional force then the book moves. Remembering that tan(theta) = F/R then, if the frictional force called into play is less than the limiting frictional force, tan(theta) is smaller (as F is smaller) and mew will be > tan(theta) in this case. When tan(theta) = Limiting Frictional Force/R then mew = tan(theta) and the body is at the point of limiting equilibrium.

So, after all that, to answer the question, once we have that mew = tan(theta) [the limiting frictional force case], we can automatically say that mew >= tan(theta) for all cases where the system is in equilibrium.

One other piece of useless (?) information on coefficient of friction - it is actually possible for the coefficient of friction to be > 1. It just means that the frictional force exceeds the normal reaction force for the 2 surfaces in contact. Apparently, the coefficient of friction between cast-iron and copper is 1.05 and and the coefficient of friction between rubber and some other materials can be easily between 1 and 2!! I'm still learning all these years later