'Fraid so. I can't see how a triangle with two radii as sides would help.(Original post by HashBrowns)
I used the sine rule for the triangle with two radii as sides two show the length of the longer side is 1.6 cos a. From there I used another triangle to show x = 0.8. Is this wrong?
See attached: x=0.8 tan a.
OCR M3: Rigid Objects Question 4 Watch
- Study Helper
- 17-05-2013 19:19
- Thread Starter
- 17-05-2013 19:46
Could I see your solution?
- Study Helper
- 17-05-2013 21:38
Let the weight be W, the angle of the rod to the horizonal be , and the distance of the pin to the point of contact with the cylinder "x", as we had before.
We know there is no vertical component at the pin, so resolving vertically for one of the rods we have:
Then taking moments about the pin for one of the rods we have:
Refering to the diagram, and considering the triangle BOM, we have
So, substituting into our 2nd equation we have:
And substituting for W from your first equation we get:
R cancels and multiplying through by 10 we have:
If we now convert the cos^2 into 1/sec^2 = 1/(1+tan^2) and multiply up and re-arrange we have:
Which on the face of it looks a bit of a pig!
If we divide through by 8, we can perhaps see that one root is tan a = 1/2.
Now the quadratic has no real roots, so we are left with
Hence we find alpha, and hence the half angle at the pin, and hence the whole angle.
Last edited by ghostwalker; 17-05-2013 at 21:39.