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Can someone see what I've done wrong with this M1 question. watch

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    I was doing the May 2003 Edexcel M1 paper, but I seem to have got a really stupid answere for this question (no.1).

    A uniform plank AB has a mass 40kg and length 4m. It is supported in a horizontal position by two smooth pivots, one at the end A, the other a the point C of the plank where AC = 3m, as shown in Fig. 1. A man of mass 80kg stands on the plankwhich remains in equilibrium. The magnitudes of the reactions at the two picots are each equal to R newtons. By modelling the plank as a rod and the man as a particle, find

    a) the value of R.
    b) the distance of the man from A.

    This is my working out:

    a)2R = 80g + 40g
    R = 1176/2
    = 588N

    b)Taking Moments about A
    (80d) + (40g * 2) = 3R
    80d + 784 = 1764
    d = 12.25m ------the plank is only supposed to be 4m though.

    I probably done something really stupid so if anyone can spot what it is I'll be thankfull.


    I was also stuck on this question, I don't know what to do.

    A particle P moves with constant acceleration (2i - 3j) ms^-2. At time t seconds, its velocity is V ms^-1. When t=0, V= -2i + 7j

    a) Find the value of t when P is moving parallel to the vector i.
    b) Find the speed of P when t = 3
    c) Find the angle between the vector j and the direction of motion of P when t=3
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    call x length A to man:
    About A: 80gx + 3R = 80g
    About C: 40g = (3-x)70g + 3R

    add these:

    40g + 80gx = 240g - 80gx + 80g
    160x = 280
    x = 1.75m

    3R = 80g - 80gx
    3R = 80[g(1-x)]
    = 80(-0.75g)
    = -60gN

    so 60gN, or 588N upwards.

    I haven't checked these but they seem quite likely numbers.
    I'll do the vectors one after lunch because they take a while.
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    (Original post by Bebop)
    a) the value of R.
    b) the distance of the man from A.

    This is my working out:

    a)2R = 80g + 40g
    R = 1176/2
    = 588N

    b)Taking Moments about A
    (80gd) + (40g * 2) = 3R
    80gd + 784 = 1764
    d = 1.25m
    you just forgot to put the 'g' in for the man
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    For the vector question...

    a)

    First you need to write an expression for velocity at time t.

    You know u=(-2i+7j) and a=(2i-3j). Seeing as you don't need to know a value for t and you are looking for v you can use v=u+at. Rearrange so you have a expressions for i and j in terms of t.

    When the particle is parallel to i the j component of it's vector will be 0.

    Therefore work out at what value of t the j component is equal to 0.

    b)

    Use the expression you already found to find the velocity vector. Then use Pythagoras to find the speed.

    c)

    The j vector is vertically upwards. Draw this and the velocity vector at t=3. Then use tanx=opp/adj and add the answer to 90.

    Hope that helps! If you need some more explicit answers just ask.
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    (Original post by nooneknows)
    For the vector question...

    a)

    First you need to write an expression for velocity at time t.

    You know u=(-2i+7j) and a=(2i-3j). Seeing as you don't need to know a value for t and you are looking for v you can use v=u+at. Rearrange so you have a expressions for i and j in terms of t.

    When the particle is parallel to i the j component of it's vector will be 0.

    Therefore work out at what value of t the j component is equal to 0.

    b)

    Use the expression you already found to find the velocity vector. Then use Pythagoras to find the speed.

    c)

    The j vector is vertically upwards. Draw this and the velocity vector at t=3. Then use tanx=opp/adj and add the answer to 90.

    Hope that helps! If you need some more explicit answers just ask.

    When the particle is parallel to i the j component of it's vector will be 0. Why is this? :confused:
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    (Original post by Illuvatar)
    When the particle is parallel to i the j component of it's vector will be 0. Why is this? :confused:
    i is the x-direction, j is the y-direction. When the j-component is 0, the vector is not moving in the y-direction and is therefore parallel to the x-direction, i.
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    (Original post by Illuvatar)
    When the particle is parallel to i the j component of it's vector will be 0. Why is this? :confused:
    because vector i is a unit vector along the x-axis, and the unit vector j is along the y-axis. Therefore, for a vector to be parallel to i, then it must not have a vertical ( j ) component
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    (Original post by ZJuwelH)
    i is the x-direction, j is the y-direction. When the j-component is 0, the vector is not moving in the y-direction and is therefore parallel to the x-direction, i.
    ah ofcourse thanks!
 
 
 

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