OCR Physics AS Resit G481 - 24th May 2016

6) i) Torque is one of the forces multiplied by the distance between the forces, so first 3cm = 0.03m, So Torque = 0.03 x 4 = 0.12 Nm
ii) Moments about a point, mean you ignore any forces acting at that point (not relevant here), and every other moment is given by the magnitude of the force multiplied by the distance it acts from point (that you are taking the moments about).
So the upwards (anti-clockwise is better incase you get a question with moments on both sides of the point) moment is given by distance ( x cm ) times force ( 4 N ). ( x cm = x/100 m) therefore anti-clockwise moment = 4x/100 Nm.
Clockwise moment is given by distance ( 3cm + xcm ) times force ( 4N ), ( 3cm + xcm = (3 + x)/100 m), so clockwise moment = 4(3 + x)/100 = 12/100 + 4x/100
Total moment in clockwise direction (since this will be a positive value) is given by the [clockwise moment] minus the [anticlockwise moment],
i.e. total moment in clockwise direction = [ 12/100 + 4x/100 ] - [ 4x/100 ] = 12/100 = 0.12Nm

lol i just realised theres a whole other section to that question so tell me if thats the part you got stuck on and ill go through that

How did everyone find it?

Not bad but my answer for part (a) of 7 i know i got wrong because i didnt put the area in mm^2 i got an answe of 93mm the rest of the paper was fine the moments question was free marks

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i did the same mistake plus you had to divide it by 4 to find the extension of one of the string

Because i got 23mm? I think the cross sectional area was like 4.5^-4m so i left my answer in mm as 23 which makes sense as its 4 times as small as the 93mm

I had 23 aswell wasn't that confident on it

It makes sense though because it was the weight divided by 4 then calculate stress and extension and so on and then leave your answer in mms which was 23

Yeah that's true, what do you think the grade boundary is looking like?im thinking 45/46
For the past 3 years it's been 43 to get an A