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1. Please can someone help me with question 8 (ii) and (iii) on OCR (not MEI) Mechanics 2 June 2007 paper?
For (ii) - I thought to take moments about O you're meant to resolve the frictional force so that you can use the component that's perpendicular to the line of O.
So I got 1.2T = Fcos(30)*0.8 instead of 0.8F = 1.2T

The paper = http://www.ocr.org.uk/Images/64988-q...echanics-2.pdf

Thank you for the help!
2. (Original post by BlueSi)
Please can someone help me with question 8 (ii) and (iii) on OCR (not MEI) Mechanics 2 June 2007 paper?
For (ii) - I thought to take moments about O you're meant to resolve the frictional force so that you can use the component that's perpendicular to the line of O.
So I got 1.2T = Fcos(30)*0.8 instead of 0.8F = 1.2T

The paper = http://www.ocr.org.uk/Images/64988-q...echanics-2.pdf

Thank you for the help!
No resolving is required.

Take a look at this diagram:

https://gyazo.com/00cc80b84a50a3e7531fd046c40a1ccd

Note that the slope is a tangent to the hemisphere. What does this tell us? Well, according to a well known circle theorem, this means that OX and the slope meet at a right angle, as OX is a radius of the hemisphere. Friction acts along the slope and through the point X. Can you finish this off?
3. Ah okay I get it now. I was thinking in terms of getting the perpendicular distance from the vertical line through O .
I see how it's 1.2T - 0.8F = 0.
Thank you so much for the reply!
4. (Original post by BlueSi)
Ah okay I get it now. I was thinking in terms of getting the perpendicular distance from the vertical line through O .
I see how it's 1.2T - 0.8F = 0.
Thank you so much for the reply!
Once you see it, it's almost a gift of 4 marks
5. (Original post by Ayman!)
Once you see it, it's almost a gift of 4 marks
Yeah I can't believe it was that simple

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