Submitted by Justabitwearden on Fri, 03/10/2017 - 19:55
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Hey so I'm currently stuck on the last part of Q9 where you have to find a necessary and sufficient condition for d. I understand that you probably have to find an inequality of some sort to show that friction is acting upwards that also gets rid of mu but I've come stuck on what this is going to be. I have horizontal and vertical force components and have taken moments about A which I know are probably right from part (i) working but it would be difficult to use the result in (i) as that assumes limiting equilibrium
I guess you have $F + T\cos
I guess you have $F + T\cos \theta = mg$ and $mg(a\cos \phi + b\sin \phi)=Td\sin(\theta + \phi)$ from your previous parts. If friction is not limiting then you have $(F+T\cos \theta)(a\cos \phi + b\sin \phi) = Td\sin(\theta + \phi)$ which you can then solve for $F$.
If $F$ is positive then it is directed upwards and conversely for downwards. So you then have an inequality $$F = \frac{Td\sin(\theta + \phi)}{a \cos \phi + b\sin \phi} - T\cos\theta> 0$$
To make this seem a little
To make this seem a little less magical and more systematic: the way you typically approach this is to list down all the equations you have, which you did; you have your vertical resolving, horizontal resolving and moments about A. This was perfect. Now, you look at the result you want, you want it in terms of $a, b, \theta$ and $\phi$.
That means you need to use those equations you have somehow to get F in terms of everything but $mg$. In this case, eliminating $mg$ is rather easy to do, you have one thing equal to $mg$ from your vertical resolving, so you can just replace $mg$ anywhere else with that. This then gives you some ugly looking expression involving $\sin$ and $\cos$. But you want $\tan$, so you'll need to expect that somewhere along the line you'll need to
(i) Expand $\sin(\theta +\phi)$ to get it in terms of $\sin$ and $\cos$ of $\theta$ and $\phi$.
(ii) Divide/multiply the numerator and denominator of the fractions by $\sin$ or $\cos$ to get everything in the form $\frac{\sin(\cdot)}{\cos(\cdot)} = \tan({\cdot})$
After some easy algebra, you get the required result. For the very last part, thing about the obvious physical restriction on $d$. It can't exceed what...?
Okay the results have all
Okay the results have all come out now, thanks!
No problem.
No problem.