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    Why is the angle at O equal to the angle a C?

    And why would there be a reaction force at O?
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    (Original post by creativebuzz)
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    Why is the angle at O equal to the angle a C?
    If you were to extend the vertical line at O you could draw a right angle triangle and the angle to the left of B is 90-\theta. The angle the the right of B if you were to draw a horizontal line is 180-(90-\theta+90)=\theta
    The angle to the right of the B vertex is \theta and it's alternate to the angle at C.
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    (Original post by Kvothe the arcane)
    The angle to the right of the B vertex is theta
    Why is the angle at the right theta? I tried drawing it out but I can't seem to get it
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    Bad drawing



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    (Original post by Kvothe the arcane)
    Bad drawing

    Ohh I see, thank you!! Is that a theorem that I've just completely forgotten or?
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    (Original post by creativebuzz)
    Ohh I see, thank you!! Is that a theorem that I've just completely forgotten or?
    I used the GCSE? geometry principles that the sum of the internal angles in a (right angled) triangle is 180^{\circ} and similarly, that angles on a straight line add up to 180^{\circ}. And lastly, the idea that alternate angles two angles, formed when a line crosses two other lines, that lie on opposite sides of the transversal line and on opposite relative sides of the other lines are equal when when the lines crossed are parallel.

    Internal angles of triangle



    Angles on a straight line



    Alternate angles




    As for your 2nd question about reactive force, I'm not sure. I haven't yet covered M2 moments and even my M1 moments were weak.

    I've tagged a couple members in who may be able to help .
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    (Original post by creativebuzz)
    And why would there be a reaction force at O?
    Because it's hinged at O so there'd be a reaction force from the hinge.
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    (Original post by creativebuzz)
    And why would there be a reaction force at O?
    Something's got to support the whole structure; without it the whole construction would fall down, i.e. no equilibrium.
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    (Original post by ghostwalker)
    Something's got to support the whole structure; without it the whole construction would fall down, i.e. no equilibrium.
    I see!

    In this question I did sin45 instead of cos45 (I put a purple line in so you can see why I put sin45 instead of cos45) but the mark scheme says it should be cos45, why is that?

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    (Original post by creativebuzz)
    I see!

    In this question I did sin45 instead of cos45 (I put a purple line in so you can see why I put sin45 instead of cos45) but the mark scheme says it should be cos45, why is that?
    Does it matter? \sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}
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    (Original post by creativebuzz)
    I see!

    In this question I did sin45 instead of cos45 (I put a purple line in so you can see why I put sin45 instead of cos45) but the mark scheme says it should be cos45, why is that?

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    As Zacken said - it doesn't matter.

    However, since they've marked an angle as 45, I would assume they're using that same angle in their calculation. Sin45 would have been the better choice.

    If the angle had been 30 degrees, say, then cos30 would be incorrect.
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    (Original post by ghostwalker)
    As Zacken said - it doesn't matter.

    However, since they've marked an angle as 45, I would assume they're using that same angle in their calculation. Sin45 would have been the better choice.

    If the angle had been 30 degrees, say, then cos30 would be incorrect.
    Oh yeah I knew that it didn't matter!

    I just wanted to make sure that I didn't get the answer right because of a fluke/my method was legit
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    (Original post by ghostwalker)
    As Zacken said - it doesn't matter.

    However, since they've marked an angle as 45, I would assume they're using that same angle in their calculation. Sin45 would have been the better choice.

    If the angle had been 30 degrees, say, then cos30 would be incorrect.
    Also, why did they give the lamina's mass as 12 (+15 of course) because isn't the mass of a lamina usually found by doing the area (so 5x8 and then add the particles' masses) in centre of mass questions? This is literally the only thing I don't understand about CoM.
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    (Original post by creativebuzz)
    Also, why did they give the lamina's mass as 12 (+15 of course) because isn't the mass of a lamina usually found by doing the area (so 5x8 and then add the particles' masses) in centre of mass questions? This is literally the only thing I don't understand about CoM.
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    The area by itself is insufficient. Assuming there is some thickness to the lamina, you'd need to know the mass per unit area, to then work out the mass of the lamina. 5x8 is not a mass.
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    (Original post by ghostwalker)
    The area by itself is insufficient. Assuming there is some thickness to the lamina, you'd need to know the mass per unit area, to then work out the mass of the lamina. 5x8 is not a mass.
    Ah thickness of course, that makes sense! But how would I know when finding the area is sufficient because in quite a lot of CoM questions (particularly composite shapes) tend to involve finding the area!
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    (Original post by creativebuzz)
    Ah thickness of course, that makes sense! But how would I know when finding the area is sufficient because in quite a lot of CoM questions (particularly composite shapes) tend to involve finding the area!
    If you're given any masses, then you need to work in terms of mass.

    If no masses are given, then it's areas; assuming density is constant.
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    (Original post by ghostwalker)
    If you're given any masses, then you need to work in terms of mass.

    If no masses are given, then it's areas; assuming density is constant.
    Gotcha.

    I managed to get full marks on part a, but in part b what do they mean by "give the direction"?
    Also, is there some reason/theory behind how we can find what direction the X and Y reaction forces are working in without plugging in your values and just seeing if your answer is negative or not?

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    (Original post by creativebuzz)
    I managed to get full marks on part a, but in part b what do they mean by "give the direction"?


    Force is a vector quantity, it has a magnitude and direction.

    E.g. a force represented by i+j has magntude of root(2), and a direction 45 degrees to i vector going anticlcockwise.


    Also, is there some reason/theory behind how we can find what direction the X and Y reaction forces are working in without plugging in your values and just seeing if your answer is negative or not?
    Logic and common sense. Can't really give a detailed description, but take your example.

    If the rod wasn't attached at A, would it move to the right or left? Force at A must be in the opposite direction to hold it in equilibrium.

    Then consider D. If the rod wasn't attached at A, would it rotate clockwise or anticlockwise about D? Vertical force at A must act to stop that, hence....
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    (Original post by ghostwalker)
    [/b]

    Force is a vector quantity, it has a magnitude and direction.

    E.g. a force represented by i+j has magntude of root(2), and a direction 45 degrees to i vector going anticlcockwise.


    [b]

    Logic and common sense. Can't really give a detailed description, but take your example.

    If the rod wasn't attached at A, would it move to the right or left? Force at A must be in the opposite direction to hold it in equilibrium.

    Then consider D. If the rod wasn't attached at A, would it rotate clockwise or anticlockwise about D? Vertical force at A must act to stop that, hence....
    If the rod wasn't fixed at A it would fall clockwise, yes?
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    (Original post by creativebuzz)
    If the rod wasn't fixed at A it would fall clockwise, yes?
    Yes.

    These things will be clearer when you've covered moments.
 
 
 
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