@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.