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    So I understand resolving forces when they are all on the one line but how do you resolve when the point you take moments about is not on the line?

    Questions like this, where you take moments about P, could someone please explain it to me?
    Name:  moment.jpg
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    Any help is greatly appreciated
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    *cough*
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    Name:  moments.png
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    This is the Physics and Maths tutor solution but I don't know where they got the numbers from
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    (Original post by Music With Rocks)
    Name:  moments.png
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    This is the Physics and Maths tutor solution but I don't know where they got the numbers from
    i'm not sure either >.>

    Zacken
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    (Original post by Music With Rocks)
    So I understand resolving forces when they are all on the one line but how do you resolve when the point you take moments about is not on the line?

    Questions like this, where you take moments about P, could someone please explain it to me?
    Name:  moment.jpg
Views: 135
Size:  131.3 KB
    Any help is greatly appreciated
    Maybe try drawing the rod horizontal like it would be in m1, then it'll be easier to see.
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    (Original post by BBeyond)
    Maybe try drawing the rod horizontal like it would be in m1, then it'll be easier to see.
    Sorry not quite sure what you mean? I drew it like this

    Name:  moment 2.jpg
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    (Original post by Music With Rocks)
    Sorry not quite sure what you mean? I drew it like this

    Name:  moment 2.jpg
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    Try turning your page so your dotted line which I assume is a rod or something is horizontal, then you want the vertical components.
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    (Original post by BBeyond)
    Try turning your page so your dotted line which I assume is a rod or something is horizontal, then you want the vertical components.
    Would that work? The question asks you to take moments about P? (the point in the bottom right corner)
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    (Original post by Music With Rocks)
    Would that work? The question asks you to take moments about P? (the point in the bottom right corner)
    Soz I'm awful at explaining things ahaha I'll leave it to someone else
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    (Original post by BBeyond)
    Soz I'm awful at explaining things ahaha I'll leave it to someone else
    Fair enough, thank you for trying anyway
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    So I think I have found a way to get the two outer components by using trig to get the lengths then multiplying by the force. I end up with
    3*4sin35
    5*4sin55

    but still unsure of how to get the moment of the middle 5N component?
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    (Original post by Music With Rocks)
    So I think I have found a way to get the two outer components by using trig to get the lengths then multiplying by the force. I end up with
    3*4sin35
    5*4sin55
    but still unsure of how to get the moment of the middle 5N component?
    You should resolve the distances rather than the force. Remember that torque = force x perpendicular distance from the pivot. Find the perpendicular distance from the point P to the line of action of each force using right angled trigonometry.
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    (Original post by Ayman!)
    You should resolve the distances rather than the force. Remember that torque = force x perpendicular distance from the pivot. Find the perpendicular distance from the point P to the line of action of each force using right angled trigonometry.
    I think that is what I did? but I am not entirely sure haha

    I drew the lines to the forces (perpendicular distance) then used trig to get the lengths of 4sin35 & 4sin55, although (and I may just be thick) I can't seem to find the length for the middle force?
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    (Original post by Music With Rocks)
    I think that is what I did? but I am not entirely sure haha

    I drew the lines to the forces (perpendicular distance) then used trig to get the lengths of 4sin35 & 4sin55, although (and I may just be thick) I can't seem to find the length for the middle force?
    You've 2/3rd of the work done then :borat:

    Extend the line of action of the 5 N force like I've done in the diagram (the white dotted line):

    Diagram


    The red line is the perpendicular distance from the pivot P to the line of action of 5 N.
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    (Original post by Ayman!)
    You've 2/3rd of the work done then :borat:

    Extend the line of action of the 5 N force like I've done in the diagram (the white dotted line):

    Diagram


    The red line is the perpendicular distance from the pivot P to the line of action of 5 N.
    I was using completely the wrong technique, this has cleared everything up!

    THANK YOU SO MUCH

    I have been struggling with this for the entire day, you are an absolute life saver, thank you!
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    (Original post by Ayman!)
    ...
    Dem paint skillz. :adore:
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    (Original post by Music With Rocks)
    I was using completely the wrong technique, this has cleared everything up!

    THANK YOU SO MUCH

    I have been struggling with this for the entire day, you are an absolute life saver, thank you!
    No worries! Once you see it, it can't be unseen :ninja:

    (Original post by Zacken)
    Dem paint skillz. :adore:
    The funniest part is that I don't even have paint on my computer :laugh:
 
 
 
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