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    (Original post by fpmaniac)
    I dont know how to draw the force exerted on B by the wall so i havent gotten very far.
    This is how I've done it:Name:  1465833966906.jpg
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    (Original post by pineneedles)
    This is how I've done it:Name:  1465833966906.jpg
Views: 174
Size:  34.8 KB

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    Thanks, is one of the forces friction or in questions like this there is always 2 forces like that acting on the rod attached to the wall.
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    (Original post by TheMarshmallows)
    Here's my solution to 11

    Attachment 549129
    Attachment 549131


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    Thanks, I dont get what you've done with T. Shouldnt T be acting vertically and horizontally from A? if so the moments of T about A will be 0
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    (Original post by TheMarshmallows)
    Here's my solution to 11

    Attachment 549129
    Attachment 549131


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    Could you explain what you did in the second step, where you've written m(a)?
    At that point I was trying to take moments about a, but I wasn't sure how to because the vertical component of thrust from the rod is acting into a, and if you take moments anywhere else you can't ignore the force from the hinge. Sorry for hijacking this question by the way, OP, hahah

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    (Original post by fpmaniac)
    Thanks, is one of the forces friction or in questions like this there is always 2 forces like that acting on the rod attached to the wall.
    Always two forces, you have a vertical component and a horizontal component. The hinge is smooth so there's no friction there. It doesn't matter if you model the forces in the wrong direction either, in my drawing I could have had x going right and y going down, it would have just become negative when I do the calculations. The other poster demonstrates that in their working out for your other question 😊

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    (Original post by fpmaniac)
    Thanks, I dont get what you've done with T. Shouldnt T be acting vertically and horizontally from A? if so the moments of T about A will be 0
    One part acts in line with A, the other part acts perpendicular to A. There's two places to consider T: at the bottom or on the rod.

    (Original post by pineneedles)
    Could you explain what you did in the second step, where you've written m(a)?At that point I was trying to take moments about a, but I wasn't sure how to because the vertical component of thrust from the rod is acting into a, and if you take moments anywhere else you can't ignore the force from the hinge. Sorry for hijacking this question by the way, OP, hahahPosted from TSR Mobile
    m(a) means moments about a. The vertical component is acting in line with A, but the horizontal component is acting perpendicular to it. (or the other way around depending on where you want to consider T).

    I drew a thing that hopefully makes it clearer:

    Name:  ImageUploadedByStudent Room1465835189.379063.jpg
Views: 142
Size:  135.8 KB
    Name:  ImageUploadedByStudent Room1465835244.898151.jpg
Views: 130
Size:  125.1 KB


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    (Original post by TheMarshmallows)
    One part acts in line with A, the other part acts perpendicular to A. There's two places to consider T: at the bottom or on the rod.


    m(a) means moments about a. The vertical component is acting in line with A, but the horizontal component is acting perpendicular to it. (or the other way around depending on where you want to consider T).

    I drew a thing that hopefully makes it clearer:

    Name:  ImageUploadedByStudent Room1465835189.379063.jpg
Views: 142
Size:  135.8 KB
    Name:  ImageUploadedByStudent Room1465835244.898151.jpg
Views: 130
Size:  125.1 KB


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    O didnt know you could do that lol Thanks. I find right side easier to understand
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    (Original post by pineneedles)
    Always two forces, you have a vertical component and a horizontal component. The hinge is smooth so there's no friction there. It doesn't matter if you model the forces in the wrong direction either, in my drawing I could have had x going right and y going down, it would have just become negative when I do the calculations. The other poster demonstrates that in their working out for your other question 😊

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    If you model the forces in the wrong direction will the final answer be negative? Also would questions about rough hinges come up in the exam
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    (Original post by fpmaniac)
    If you model the forces in the wrong direction will the final answer be negative? Also would questions about rough hinges come up in the exam
    Yes in some cases, I demonstrate this in my workings for part b here:

    (Original post by TheMarshmallows)
    Attachment 549129
    Where I drew Y going the wrong direction.
    If you draw more than one force going the wrong way though you might have problems (depends on the question really).

    As for rough hinges, I don't think so, at least I've never seen a question ask it
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    (Original post by TheMarshmallows)
    Yes in some cases, I demonstrate this in my workings for part b here:


    Where I drew Y going the wrong direction.
    If you draw more than one force going the wrong way though you might have problems (depends on the question really).

    As for rough hinges, I don't think so, at least I've never seen a question ask it
    Oh crap. If a question like that comes up im gonna mess up :laugh:
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    Guys, considering the difficulty of last year's and this year's Maths papers, how hard do you think M2 will be?
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    I'm really not looking forward to this exam, I plainly cannot do M2.
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    (Original post by Craig1998)
    I'm really not looking forward to this exam, I plainly cannot do M2.
    me too! but i need to do well to make up for the shambles that was fp2
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    (Original post by Tizzydag)
    me too! but i need to do well to make up for the shambles that was fp2
    I'm relying on FP3 to make up for this exam.
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    Guys can someone help me, on the January 2008 paper for Q6c, i used v^2=u^2+2as and got the answer, but why does it in the mark scheme say 'watch out for incorrect use of v^2=u^2+2as' ????
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    (Original post by eftio.gea)
    Guys can someone help me, on the January 2008 paper for Q6c, i used v^2=u^2+2as and got the answer, but why does it in the mark scheme say 'watch out for incorrect use of v^2=u^2+2as' ????
    as long as you handle the x and y directions separately you should be ok (depending on what you did)
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    (Original post by eftio.gea)
    Guys can someone help me, on the January 2008 paper for Q6c, i used v^2=u^2+2as and got the answer, but why does it in the mark scheme say 'watch out for incorrect use of v^2=u^2+2as' ????
    As the question contains a variable u and the motion equations also involve a u, that notice is simply to ensure that students don't substitute the incorrect value.

    Here's the solution (using the equation noted):

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    (Original post by candol)
    as long as you handle the x and y directions separately you should be ok (depending on what you did)
    (Original post by ImJared)
    As the question contains a variable u and the motion equations also involve u as the initial velocity, it is simply warning examiner that some students will use the incorrect value of u in their equations.

    Here's the solution for others, relatively simple:

    Ohhhh alright yes that's what I did, i was worried i thought we couldn't use v^2 = u^2 +2as
    Thanks!!!!!!
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    They are equal in the sense that work done against resistance is the same same as the force you put in to do work against that resistance
    (Original post by fpmaniac)
    I have a question regarding work done. Sometimes they use the total resistance force x distance to calculate work done but other times they use forward force x distance to calculate work done. How do u know which one to use. E.g. for question 5 june 2013 they used resistance force x distance
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    Can someone help with M2 2015 2b please.

    I don't understand what the mark scheme is saying.
 
 
 
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