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lads can only do this m2 question watch

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    it's 5a first question in heinemann book

    a uniform beam AB of mass 20 kg and length 2m is attached to a vertical wall by means of a smooth hinge at A. The beam is maintained in the horizontal psition by means of a light inextensible string, one end of which is attached to the beam at B and the other end is attached to the wall at a point in the wall 2m vertically above A

    a) calc T
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    I'll go... T = 139N
    Cud be horribly wrong, ill check my genitalia in a moment err!
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    (Original post by YorkshirePhoeni)
    I'll go... T = 139N
    Cud be horribly wrong, ill check my genitalia in a moment err!
    i get t'same
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    ditto;

    if you take moments about A, you have:-
    20g = 2Tsin45
    T= 10g / sin45 = 138.6N
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    wt if tension in string was 400 n n there was a mass at b
    wtn is da mass?
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    (Original post by mabs)
    wt if tension in string was 400 n n there was a mass at b
    wtn is da mass?
    Take moments about A again and call the mass at b 'M', you already know the tension so it should come out using the moments equation.
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    everyone knows the method but wots the ansa darlin???
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    (Original post by mabs)
    wt if tension in string was 400 n n there was a mass at b
    wtn is da mass?
    As Mysticmin said, using the same format; Moments about A

    20g + 2M = 400 sin 45

    2M = 400 sin 45 - 20g

    M = (400 sin 45 - 20g)/2

    M = 43.4N

    Therefore Mass M = 43.4/9.8 = 4.43 kg (Do not forget this step)
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    (Original post by YorkshirePhoeni)
    As Mysticmin said, using the same format; Moments about A

    20g + 2M = 400 sin 45

    2M = 400 sin 45 - 20g

    M = (400 sin 45 - 20g)/2

    M = 43.4N

    Therefore Mass M = 43.4/9.8 = 4.43 kg (Do not forget this step)
    I got a different answer by taking moments, ive probably gone wrong somewhere, if so , feel free to point out and **** off!...

    moments from A... letting mass at B =B

    20gm + 2mBg = 400sin45x2m

    remove 2m from all...

    10g + Bg = 400sin45
    98 + 9.8B = 282
    9.8B = 184
    B = 18.86 kg
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    Taking moments
    ===========

    2m.Tcos45 - 2m.Mg = 1m.20g
    2.400.1/√2 - 2Mg = 20g
    Mg = 200√2 - 10g
    M = 200√2/g - 10
    M = 28.86 - 10
    M = 18.86
    M = 18.7 kg (3 sf)
    ========
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    like i sed only men can do m2
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    (Original post by mabs)
    ===========

    M = 18.86
    M = 18.7 kg (3 sf)
    ========
    I liked the answer before the strange sig figs...
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    kk georgey

    we having fun

    girls arre wallowin in self pity
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    (Original post by mabs)
    like i sed only men can do m2
    Hove you done it yet? what did you get?
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    (Original post by mabs)
    like i sed only men can do m2
    You've got to be bloody joking. I'm a girl and I have never got below 65/75 in M2. :rolleyes:
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    (Original post by Mysticmin)
    You've got to be bloody joking. I'm a girl and I have never got below 65/75 in M2. :rolleyes:
    Do you get the joke or not? lol
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    (Original post by George-W-Duck)
    I got a different answer by taking moments, ive probably gone wrong somewhere, if so , feel free to point out and **** off!...

    moments from A... letting mass at B =B

    20gm + 2mBg = 400sin45x2m

    remove 2m from all...

    10g + Bg = 400sin45
    98 + 9.8B = 282
    9.8B = 184
    B = 18.86 kg
    Oh aye forgot the 2m x the 400 sin 45, else i get the same 18.86
    This is the start of my m2 revision i DO apologise!
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    (Original post by glance)
    Do you get the joke or not? lol
    Nah, I only saw that post, explaination?
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    (Original post by glance)
    Do you get the joke or not? lol
    I dont' get the joke...explain
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    Maybe there wasn't one. I didn't think it was very funny...anyway, forget that.
 
 
 
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