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    (Original post by GarlicBread01)
    When can you assume that the reaction forces at two supports are the same? Is it only when the questions tell you they are, otherwise always assume they are not the same magnitude?
    Only when they tell you they are, definitely

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    (Original post by 83457)
    What are some of the hardest past papers people have come across for M1? Doesn't have to be Edexcel (and I'd prefer non-Edexcel ones, since I think I've done the hard ones from that )
    My favourite M1 problem ever!!! And home made of course.
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    (Original post by Cryptokyo)
    My favourite M1 problem ever!!! And home made of course.
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    only 6 marks?
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    (Original post by GarlicBread01)
    When can you assume that the reaction forces at two supports are the same? Is it only when the questions tell you they are, otherwise always assume they are not the same magnitude?
    Yes exactly! Or the reaction forces are equidistant from the centre of mass of the rod and the weight of the rod is the only force acting downwards.
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    (Original post by KloppOClock)
    only 6 marks?
    Well for the tension... [please elaborate]
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    (Original post by KloppOClock)
    here
    Minor update to certain wording of the questions but otherwise the same stuff!
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  1. File Type: pdf Mechanics 1 Practice Paper 4 UPDATE.pdf (510.5 KB, 65 views)
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    (Original post by KloppOClock)
    if its is an isoceles triangle, and sinA=3/4, then drawing a triangle gives cosA as 3/4 as well i think?

    cosa^2+sina^2 = 1.125 but i thought that was supposed to be equal to one
    \sin(a)=\frac{3}{4} and \cos^{2}(a)=1-\sin^{2}(a)=\frac{7}{16} so \cos(a)=\frac{\sqrt{7}}{4}
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    (Original post by Cryptokyo)
    \sin(a)=\frac{3}{4} and \cos^{2}(a)=1-\sin^{2}(a)=\frac{7}{16} so \cos(a)=\frac{\sqrt{7}}{4}
    yh i forgot that was just for right angle triangles
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    (Original post by KloppOClock)
    yh i forgot that was just for right angle triangles
    Well you can split into two right angled triangles.
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    (Original post by Cryptokyo)
    My favourite M1 problem ever!!! And home made of course.
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    Do yoy have the answer for a? Not sure ive done it right
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    (Original post by metrize)
    Do yoy have the answer for a? Not sure ive done it right
    yeah same, i got (18-3root7)/4
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    [QUOTE=metrize;65449145]Do yoy have the answer for a? Not sure ive done it right[/QUOTE

    Answer in spoiler
    Please say if anything wrong.
    Spoiler:
    Show
    Set up equations:
    T-g\sin(\alpha)-\frac{1}{2}g\cos(\alpha)=a
    6g\sin(\alpha)-3g\cos(\alpha)-T=6a
    Add equations together.
    6g\sin(\alpha)-3g\cos(\alpha)-g\sin(\alpha)-\frac{1}{2}g\cos(\alpha)=7a
    Sub in values known to get
    a=2.008954644...
    Back into first equation:
    T=a+g\sin(\alpha)+\frac{1}{2}g \cos(\alpha)=12.6
    Tension is 12.6N

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    [QUOTE=Cryptokyo;65450051]
    (Original post by metrize)
    Do yoy have the answer for a? Not sure ive done it right[/QUOTE

    Answer in spoiler
    Please say if anything wrong.
    Spoiler:
    Show
    Set up equations:
    T-g\sin(\alpha)-\frac{1}{2}g\cos(\alpha)=a
    6g\sin(\alpha)-3g\cos(\alpha)-T=6a
    Add equations together.
    6g\sin(\alpha)-3g\cos(\alpha)-g\sin(\alpha)-\frac{1}{2}g\cos(\alpha)=7a
    Sub in values known to get
    a=2.008954644...
    Back into first equation:
    T=a+g\sin(\alpha)+\frac{1}{2}g \cos(\alpha)=12.6
    Tension is 12.6N

    Ah i messed up because for the 1kg particle i didnt realise it would have moved up the plane
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    [QUOTE=Cryptokyo;65450051]
    (Original post by metrize)
    Do yoy have the answer for a? Not sure ive done it right[/QUOTE

    Answer in spoiler
    Please say if anything wrong.
    Spoiler:
    Show
    Set up equations:
    T-g\sin(\alpha)-\frac{1}{2}g\cos(\alpha)=a
    6g\sin(\alpha)-3g\cos(\alpha)-T=6a
    Add equations together.
    6g\sin(\alpha)-3g\cos(\alpha)-g\sin(\alpha)-\frac{1}{2}g\cos(\alpha)=7a
    Sub in values known to get
    a=2.008954644...
    Back into first equation:
    T=a+g\sin(\alpha)+\frac{1}{2}g \cos(\alpha)=12.6
    Tension is 12.6N

    oh i thought that as it says the rod remains in equilbrium, then the blocks remained in equilbrium, which isnt true
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    [QUOTE=KloppOClock;65450303]
    (Original post by Cryptokyo)

    oh i thought that as it says the rod remains in equilbrium, then the blocks remained in equilbrium, which isnt true
    The force that the tension exerts on the pulley keeps the rod in equilibrium as it produces a moment.
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    [QUOTE=Cryptokyo;65450607]
    (Original post by KloppOClock)

    The force that the tension exerts on the pulley keeps the rod in equilibrium as it produces a moment.
    so the resultant force of the tension acting on the rod is equal to the weight of the particle on the rod?
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    Yes
    (Original post by KloppOClock)

    so the resultant force of the tension acting on the rod is equal to the weight of the particle on the rod?
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    Can someone help me urgently please? <3

    LOOK IN DIAGRAM!


    So in this question where it says 2.5 -x

    I labelled that part as x and the other part where it says x as 2.5 - x, so essentially I switched them around

    I did the calculation and got x = 2 MY WAY, but the mark scheme way (not switched around) got x = 0.5

    Am I not aloud to label the end part as 2.5 - x or something?
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    (Original post by KINGYusuf)
    Can someone help me urgently please? <3

    LOOK IN DIAGRAM!


    So in this question where it says 2.5 -x

    I labelled that part as x and the other part where it says x as 2.5 - x, so essentially I switched them around

    I did the calculation and got x = 2 MY WAY, but the mark scheme way (not switched around) got x = 0.5

    Am I not aloud to label the end part as 2.5 - x or something?
    No yours is right, you just have to mention that it is 2 metres from M to B or something like that.

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    (Original post by SaphBruh)
    No yours is right, you just have to mention that it is 2 metres from M to B or something like that.

    Posted from TSR Mobile
    Thanks, I was so stressed
 
 
 
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