M3
https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202014%20(IAL)%20QP%20-%20M3%20Edexcel.pdf
question 5 b
I don't understand what they did in the mark scheme. I get that they resolve forces and thats fine, but they take moments about the point of contact, i dont get how 2rsina = w(9r/4sina-rcosa)
question 5 b
I don't understand what they did in the mark scheme. I get that they resolve forces and thats fine, but they take moments about the point of contact, i dont get how 2rsina = w(9r/4sina-rcosa)
Original post by beta_tester
https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202014%20(IAL)%20QP%20-%20M3%20Edexcel.pdf
question 5 b
I don't understand what they did in the mark scheme. I get that they resolve forces and thats fine, but they take moments about the point of contact, i dont get how 2rsina = w(9r/4sina-rcosa)
question 5 b
I don't understand what they did in the mark scheme. I get that they resolve forces and thats fine, but they take moments about the point of contact, i dont get how 2rsina = w(9r/4sina-rcosa)
I presume the lefthand side is OK - though it's missing a P.
For the right, you want the perpendicular distance of the line of action of the centre of mass. Refering to the diagram, this is BC.
BC = AC - AB
= 9r/4 sin a - r cos a
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