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    How to solve that kind of problems? I know about clock&anticlockwise forces, but can't work out solutions, especially the first one.
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    (Original post by rizamadiyar)
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    How to solve that kind of problems? I know about clock&anticlockwise forces, but can't work out solutions, especially the first one.
    What are the clockwise and anti-clockwise moments in the first one?
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    (Original post by SeanFM)
    What are the clockwise and anti-clockwise moments in the first one?
    I guess the clockwise one is to the right off the pivot due to the larger distance(40 cm)
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    noot noot
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    (Original post by rizamadiyar)
    I guess the clockwise one is to the right off the pivot due to the larger distance(40 cm)
    Correct yes, but in terms of numbers and x?
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    (Original post by SeanFM)
    Correct yes, but in terms of numbers and x?
    If only I could know... Does it really matter? The answer is 30 cm, I can't figure it out
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    (Original post by rizamadiyar)
    If only I could know... Does it really matter? The answer is 30 cm, I can't figure it out
    I am trying to guide you towards the solution.

    Okay, to go one step back, it says in the question the beam is balanced. So what can you say about the clockwise and anti-clockwise moments?

    Answer: the clockwise and anti-clockwise moments are _ _ _ _ _ (1 word)
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    (Original post by SeanFM)
    I am trying to guide you towards the solution.

    Okay, to go one step back, it says in the question the beam is balanced. So what can you say about the clockwise and anti-clockwise moments?

    Answer: the clockwise and anti-clockwise moments are _ _ _ _ _ (1 word)
    equal?
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    (Original post by rizamadiyar)
    equal?
    Correct.

    Now the anti-clockwise moment is equal to ? * ?

    And the clockwise moment is equal to ??? + ???

    And since they are equal, x = ?
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    (Original post by SeanFM)
    Correct.

    Now the anti-clockwise moment is equal to ? * ?

    And the clockwise moment is equal to ??? + ???

    And since they are equal, x = ?
    500*20=200*x+200*40, x=10 cm, but it says that the answer is 30 cm
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    (Original post by rizamadiyar)
    500*20=200*x+200*40, x=10 cm, but it says that the answer is 30 cm
    I guess it has smth to do with the weight of the beam=800 N
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    (Original post by rizamadiyar)
    I guess it has smth to do with the weight of the beam=800 N
    That was my initial thought, however the beam is uniform and the pivot is in the centre. Hence the distance of the force due to the weight is 0 from the centre, and so it has no impact if you take moments about where the weight is (which we know is the centre)

    I suspect that the answer is just incorrect, I do not see anything else that can be done,
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    (Original post by SeanFM)
    That was my initial thought, however the beam is uniform and the pivot is in the centre. Hence the distance of the force due to the weight is 0 from the centre, and so it has no impact if you take moments about where the weight is (which we know is the centre)

    I suspect that the answer is just incorrect, I do not see anything else that can be done,
    I guess so as well, what about the next one? 8000(moment to the right off the pivot)=200(50-x)+10*(what? 30?)
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    (Original post by SeanFM)
    I suspect that the answer is just incorrect, I do not see anything else that can be done,
    I agree, x=10cm.

    (Original post by rizamadiyar)
    I guess so as well, what about the next one? 8000(moment to the right off the pivot)=200(50-x)+10*(what? 30?)
    The clockwise moment of the 800N load is 8000Ncm. The anti-clockwise moment of the 200N load is not what you've written - remember, we're taking moments about the pivot.

    The uniform beam is 60cm, so we model its weight as a single force at the centre - how far from the pivot is that?
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    (Original post by RogerOxon)
    I agree, x=10cm.


    The clockwise moment of the 800N load is 8000Ncm. The anti-clockwise moment of the 200N load is not what you're wirtten - remember, we're taking moments about the pivot.

    The uniform beam is 60cm, so we model its weight as a single force at the centre - how far from the pivot is that?
    20 cm? 8000=200x+10*20, i got it. Thank you all!
 
 
 
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