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    (Original post by thebrahmabull)
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    Two things:

    First off they asked for a geometrical method, so no points, even if you get it right.

    Second, when taking moments about A, your distance for the force S is incorrect. Have another go.
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    (Original post by ghostwalker)
    Two things:

    First off they asked for a geometrical method, so no points, even if you get it right.

    Second, when taking moments about A, your distance for the force S is incorrect. Have another go.
    Whoops! I am not sure how to use a geometric method though.
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    (Original post by thebrahmabull)
    Whoops! I am not sure how to use a geometric method though.
    What do you know about three forces in equilibrium?
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    (Original post by ghostwalker)
    What do you know about three forces in equilibrium?
    Hmmm well they can be used in a triangle of forces, but that would not involve the distances between them, just the magnitude and direction of them. Well that's all the geometry I know that is applicable in here but which wouldn't solve the problem.
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    (Original post by thebrahmabull)
    Hmmm well they can be used in a triangle of forces, but that would not involve the distances between them, just the magnitude and direction of them. Well that's all the geometry I know that is applicable in here but which wouldn't solve the problem.
    You should have come across, the lines of action of three forces in equilibrium act through a common point.
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    (Original post by ghostwalker)
    You should have come across, the lines of action of three forces in equilibrium act through a common point.
    I see. Thanks. Okay so will this same method as the one you mentioned work for more than 3 forces? When will not work in a system in equilibrium?
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    (Original post by thebrahmabull)
    I see. Thanks. Okay so will this same method as the one you mentioned work for more than 3 forces? When will not work in a system in equilibrium?
    Method won't work if there are more than three forces, except when you can reduce the extra forces so there are three in total.

    Don't understand your last sentence.
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    (Original post by ghostwalker)
    Method won't work if there are more than three forces, except when you can reduce the extra forces so there are three in total.

    Don't understand your last sentence.
    Thanks! Ok the last sentence doesn't matter. For understanding: Name:  IMG_20160410_191947.jpg
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Size:  244.2 KBis it possible to reduce the two downward forces here into 1 since they are both downward? I tried this question to find the centre of mass of rod using the above method using the downward force as 7g and either my working was wrong or this is one of those systems where the extra force cannot be reduced to 3 forces.
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    (Original post by thebrahmabull)
    Thanks! Ok the last sentence doesn't matter. For understanding: Name:  IMG_20160410_191947.jpg
Views: 86
Size:  244.2 KBis it possible to reduce the two downward forces here into 1 since they are both downward? I tried this question to find the centre of mass of rod using the above method using the downward force as 7g and either my working was wrong or this is one of those systems where the extra force cannot be reduced to 3 forces.
    Yes, I believe it's feasible.

    But note that the line of action of the combined downward forces is not the centre of gravity of the rod.

    Post working if you'd like me to check it through.
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    (Original post by ghostwalker)
    Yes, I believe it's feasible.

    But note that the line of action of the combined downward forces is not the centre of gravity of the rod.

    Post working if you'd like me to check it through.
    Ok thanks! By saying feasible did you mean this method?

    And this must mean that centre of gravity of the rod for this question cannot be found using this method, right? I did try it and it came wrong
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    (Original post by thebrahmabull)
    Ok thanks! By saying feasible did you mean this method?
    No, I was answering your question regarding considering three concurrent forces.

    The method in your photo is fine for finding the tension.
    You could then just take moments about A or B to find the CofG of the rod.

    And this must mean that centre of gravity of the rod for this question cannot be found using this method, right? I did try it and it came wrong
    As I said previously, this will be the position of the line of action of the combined forces, NOT the CofG of the rod. It's fine so far, but you then need to do further work to find the CofG of the rod. If you want to find the CofG from here, take moments about A, say, solely for the vertical forces.
    The sum of the moments (rod + weight) will equal the moment of the combined force.
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    I meant "solely for the foces being combined", not "solely for the vertical forces"

    For some reason, the previous post won't edit properly.

    Edit: As you can see, combining forces is not the best way forward with this question, although it can be done.
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    (Original post by ghostwalker)
    No, I was answering your question regarding considering three concurrent forces.

    The method in your photo is fine for finding the tension.
    You could then just take moments about A or B to find the CofG of the rod.



    As I said previously, this will be the position of the line of action of the combined forces, NOT the CofG of the rod. It's fine so far, but you then need to do further work to find the CofG of the rod. If you want to find the CofG from here, take moments about A, say, solely for the vertical forces.
    The sum of the moments (rod + weight) will equal the moment of the combined force.
    Thanks! I solved it using moments.But i don't know how to do it with the 3 concurrent forces. Are both a and b solvable using concurrent forces ?
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    (Original post by thebrahmabull)
    Thanks! I solved it using moments.But i don't know how to do it with the 3 concurrent forces. Are both a and b solvable using concurrent forces ?
    Huh! a and b are different methods for the same problem. "a" triangle of forces, "b" concurrent forces. You could also resolve horizontally, and vertically.

    I've explained how to do it with 3 concurrent forces, and get the impression you're not reading what I've posted.
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    (Original post by ghostwalker)
    Huh! a and b are different methods for the same problem. "a" triangle of forces, "b" concurrent forces. You could also resolve horizontally, and vertically.

    I've explained how to do it with 3 concurrent forces, and get the impression you're not reading what I've posted.
    No, I did read.. I was too slow to understand though. Thanks for all your help
 
 
 
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