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M2 question

Hi,

I need help in M2 June 2014 R Q4 (b) I don't understand how they found the angle between the side AB and the downward vertical.

Which examination board is this question from?

Edexcel

I presume you mean Q3? When we freely suspend a lamina from A the downward vertical will pass through A and our centre of mass. At this point it would be useful to draw a diagram of this line and the angle we require, call it alpha. We have that from part a) that the distance from the midpoint of AC and the centre of mass is 2.8a. By division by 2 we note that the distance between A and the midpoint of AC is 6a. Now we can draw a triangle with angle say beta. We can evaluate Beta by trigonometry (i.e. tan beta = opp/adj ). We get Beta is 32.3 ish. Now consider the angle gamma, defined as alpha+beta. We can use the same side of the triangle of length 6a but extend the other side to be length from the midpoint of AC to B which is 9a. This gives us another triangle and using the same trig formula we get gamma is 56.3. now gamma-beta=alpha therefore we get the required angle alpha to be 24. If theres a particular stage of this problem you're not quite getting then I can explain it in more detail if you like :)

Sorry but I actually meant question 4 (b), where they ask to calculate the distance between O and DC

No problem. The vertical will go through A and G. Right now it would be useful to draw a diagram. By a) we found that the distance between G and the side AD was 3cm (call this point N). We were given that the side AB (therefore also AD and AN) are 25 degree inclined from the vertical. We will get a triangle now and we can calculate that AN=3/(tan25) giving 6.43 ish. Now the distance from G to the line DC is the same as the distance DN which is also AD-AN. From looking at the original diagram we get AD is 6. Therefore the distance is 6.43-6 which is 0.43. Now we return to a moments calculation done a bit differently, about line DC. Here we have y bar which is 0.43 and we want the distance from O to the line DN. therefore as the centre of mass of ABC is on the line DN its distance is 0 therefore we get moment calculation 16z=32(0.43) giving us z=0.867 ish. If you look at the official mark scheme for this they use terrible notation so thats maybe what threw you?

Thanks

If you don't mind can you also explain Q2 (b) of June 2015 paper, where they ask for the value k.

Perhaps you could specify what part of the question you don't quite get?

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