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    (Original post by Nirm)
    For M2 Jan 2013 Question 3. I do not understand why the R(b) component is multiplied by sin(a). Can someone give me a break down with triangles in M2 with components or when it is sin(x)/cos(x) times by whatever? I thought I had it figured but I guess not, I always get them wrong when in moments or slopes. I was watching examsolutions (https://www.youtube.com/watch?v=FDo1ix2Kjuc) but he just stated it and didn't go much into detail.
    You multiply by the perpendicular force. Lots of ways to get this, but if you just look at the 'z-angles' then you can see the perpendicular force is sin(alpha).

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    (Original post by target21859)
    Hello I don't understand how 0.9a=0.5(a+acostheta) in this question. Could someone explain it to me please?
    Here you go, the centre of mass is on the vertical line as the two rhombuses are symmetrical. Then we are given that it is 0.9a up the line, and therefore we can work out the centre of mass in terms of a and compare it to the real centre of mass (giving us 0.8 when you cancel).

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    (Original post by ImJared)
    Here you go, the centre of mass is on the vertical line as the two rhombuses are symmetrical. Then we are given that it is 0.9a up the line, and therefore we can work out the centre of mass in terms of a and compare it to the real centre of mass (giving us 0.8 when you cancel).

    Ah thanks but how do we know y bar is halfway up the line? Is there a formula for the centre of mass of a rhombus?
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    Name:  image.png
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Size:  387.9 KB@ImJared
    I've drawn a diagram(not as good as yours! ) to help you understand what I'm asking. Why can't the vertical go through the red point instead of the green point? That would change the distances between each mass to the vertical so the answer would be different.
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    (Original post by target21859)
    Ah thanks but how do we know y bar is halfway up the line? Is there a formula for the centre of mass of a rhombus?
    So we know that the centre of mass is directly in the centre of the rhombus as all lengths are the same and it is uniform. From this we can say that the hypotenuse (AC) of the big triangle (in green) goes through the centre of mass. If we divide AC by 2, we have a new triangle which has an adjacent side/height of the centre of mass.



    Don't worry, this question was a harsh one - took me a little bit to figure it all out.

    (Original post by target21859)
    @ImJared
    I've drawn a diagram(not as good as yours! ) to help you understand what I'm asking. Why can't the vertical go through the red point instead of the green point? That would change the distances between each mass to the vertical so the answer would be different.
    Ah! I see what you're asking. The answer is simply because the mass was added to the right of O (they tell you it was added exactly on P) and therefore the centre of mass also has to be shifted to the right. Forget about rotation and just imagine if it wasn't hanging, the centre of mass would shift to point P if you added an infinite mass at point P.
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    (Original post by ImJared)
    So we know that the centre of mass is directly in the centre of the rhombus as all lengths are the same and it is uniform. From this we can say that the hypotenuse (AC) of the big triangle (in green) goes through the centre of mass. If we divide AC by 2, we have a new triangle which has an adjacent side/height of the centre of mass.



    Don't worry, this question was a harsh one - took me a little bit to figure it all out.



    Ah! I see what you're asking. The answer is simply because the mass was added to the right of O (they tell you it was added exactly on P) and therefore the centre of mass also has to be shifted to the right. Forget about rotation and just imagine if it wasn't hanging, the centre of mass would shift to point P if you added an infinite mass at point P.
    Thanks I understand the rhombus one now! But I still don't get the second one. Sorry if I'm coming across as really stupid. Couldn't the centre of mass just shift over to the right slightly so it isn't on the right hand side of O?
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    (Original post by ImJared)
    So we know that the centre of mass is directly in the centre of the rhombus as all lengths are the same and it is uniform. From this we can say that the hypotenuse (AC) of the big triangle (in green) goes through the centre of mass. If we divide AC by 2, we have a new triangle which has an adjacent side/height of the centre of mass.



    Don't worry, this question was a harsh one - took me a little bit to figure it all out.



    Ah! I see what you're asking. The answer is simply because the mass was added to the right of O (they tell you it was added exactly on P) and therefore the centre of mass also has to be shifted to the right. Forget about rotation and just imagine if it wasn't hanging, the centre of mass would shift to point P if you added an infinite mass at point P.
    i still find da reason/explanation fr da first example tooo complex cant we jzt picture da rhombus of equal lengths as a square so da c of mass of da rhombus would lie in da middle of it(like in a square) so the y component of position of centre of mass if perpendicular height of rhombus divided by 2
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    can someone plz confirm whether we shuld write mass per unit area in centre of mass wuestions as k and use it in our calculations
    are marks allocated fr dat
    if so wat is da correct way of writin it i find it a bit confusing
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    Find the range of values if e( newtons law of restitution) How do you answer question like this.
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    hello could anyone help me with part a and b? i don't understand why they don't consider T when taking moments
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    (Original post by CoolDocA)
    Find the range of values if e( newtons law of restitution) How do you answer question like this.
    It will usually give you a piece of information along the lines of:
    Given that the motion of P is reversed by the collision (so final velocity is negative)
    Given that Q collides with P again (Q's final velocity is larger than P's velocity).

    If you post a specific question I can help, it's literally the one bit of M2 I can do (and I find it kinda fun too).
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    Hello. Thank you for explanations on centre of masses, but i am still confused on how you find the new centre of mass?

    @ImJared could you possibly explain this please?
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    (Original post by Craig1998)
    It will usually give you a piece of information along the lines of:
    Given that the motion of P is reversed by the collision (so final velocity is negative)
    Given that Q collides with P again (Q's final velocity is larger than P's velocity).

    If you post a specific question I can help, it's literally the one bit of M2 I can do (and I find it kinda fun too).
    June 2012 2)b)
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    (Original post by CoolDocA)
    June 2012 2)b)
    step1 find speed of P after collision.
    This is (2u-12ue)/7
    step 2 This speed must be <0 since direction is reversed
    step 3 simplify to get 1-6e<0
    this becomes e>1/6
    Since max e is 1 the inequality becomes 1/6<e<=1
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    (Original post by candol)
    step1 find speed of P after collision.
    This is (2u-12ue)/7
    step 2 This speed must be <0 since direction is reversed
    step 3 simplify to get 1-6e<0
    this becomes e>1/6
    Since max e is 1 the inequality becomes 1/6<e<=1
    Thanks so much, it makes so much sense now.
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    (Original post by alvosm)
    hello could anyone help me with part a and b? i don't understand why they don't consider T when taking moments
    The tension in the string acts through the point D, so when taking moments will equal 0. It works the same for the normal reaction at B, you don't need to include it when taking moments about D because as you can see it would pass straight through D meaning it would equal 0.

    For part b), taking moments about C will eliminate the tension when taking moments again, leaving you with the reaction force, the weight of the rod and the frictional force which you can solve for the magnitude.
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    Hey guys, in this question why does the ladder not exert any normal reaction on the person [Reece]?? (this is from Q2 gold paper 2)
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    what paper was it?
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    (Original post by ImJared)
    So we know that the centre of mass is directly in the centre of the rhombus as all lengths are the same and it is uniform. From this we can say that the hypotenuse (AC) of the big triangle (in green) goes through the centre of mass. If we divide AC by 2, we have a new triangle which has an adjacent side/height of the centre of mass.



    Don't worry, this question was a harsh one - took me a little bit to figure it all out.



    Ah! I see what you're asking. The answer is simply because the mass was added to the right of O (they tell you it was added exactly on P) and therefore the centre of mass also has to be shifted to the right. Forget about rotation and just imagine if it wasn't hanging, the centre of mass would shift to point P if you added an infinite mass at point P.
    Which year is this paper?
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    (Original post by eftio.gea)
    Hey guys, in this question why does the ladder not exert any normal reaction on the person [Reece]?? (this is from Q2 gold paper 2)
    Reece is modelled as a particle
 
 
 
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