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Math Help - What is the weight of the beam?

  1. #1
    Member looi76's Avatar
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    What is the weight of the beam?

    Question:
    A long uniform beam is pivoted at one end. A force of 300N is applied to hold the beam horizontally.

    What is the weight of the beam?

    Attempt:
    Need Help... Don't know how to start
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  2. #2
    Junior Member shinhidora's Avatar
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    I can't really help you exactly.. though doesn't it have something to see with the strenght of gravity? You need an amount of strenght equal to the gravity to hold the beam in air.

    This is just a gamble... so you best wait 'till someone proves me wrong (which 'll probably be the case but anyway...)

    The strenght of a gravity field: g= 9.81 N/kg

    As they apply 300N to hold the beam in the air...

    300/9.81 = 30.58... kg

    I could be terribly wrong though...

    Edit: and 'topsquark' has proven me wrong so ignore me ^^
    Last edited by shinhidora; June 11th 2008 at 04:15 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    A long uniform beam is pivoted at one end. A force of 300N is applied to hold the beam horizontally.

    What is the weight of the beam?

    Attempt:
    Need Help... Don't know how to start
    I don't have time to fully analyze this, but the CM of the beam will be 1.25 m from the pivot, so that's where the weight acts from. If we take the sum of the torques using the pivot as our axis of rotation we will get an equation for w. (The net torque is, of course, 0 Nm.)

    -Dan
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  4. #4
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    Quote Originally Posted by looi76 View Post
    Question:
    A long uniform beam is pivoted at one end. A force of 300N is applied to hold the beam horizontally.

    What is the weight of the beam?

    Attempt:
    Need Help... Don't know how to start
    Since the beam is uniform the center of gravity C is 1.25 m from the pivot.
    The weight W of the beam creates a right-turning momentum

    m_r = 1.25 \cdot W \ m

    while the force F = 300 N creates a left-turning momentum of

    m_l = 2.0 \cdot 300 \ Nm = 600\ Nm

    Both momenti must be equal that means the beam doesn't turn in any direction:

    1.25 \cdot W \ m = 600\ Nm~\implies~ \boxed{W = 480\ N}
    Attached Thumbnails Attached Thumbnails What is the weight of the beam?-drehmoment_balken.gif  
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shinhidora View Post
    I can't really help you exactly.. though doesn't it have something to see with the strenght of gravity? You need an amount of strenght equal to the gravity to hold the beam in air.

    This is just a gamble... so you best wait 'till someone proves me wrong (which 'll probably be the case but anyway...)

    The strenght of a gravity field: g= 9.81 N/kg

    As they apply 300N to hold the beam in the air...

    300/9.81 = 30.58... kg

    I could be terribly wrong though...

    Edit: and 'topsquark' has proven me wrong so ignore me ^^
    I want to quickly mention why this doesn't work. There is a reaction force at the pivot point and we don't know what size that force is. The reason this isn't a difficulty using the torque method is that by taking the axis of revolution as the pivot the torque due to this reaction force is 0 Nm because it has a 0 moment-arm.

    -Dan
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  6. #6
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by topsquark View Post
    I want to quickly mention why this doesn't work. There is a reaction force at the pivot point and we don't know what size that force is. The reason this isn't a difficulty using the torque method is that by taking the axis of revolution as the pivot the torque due to this reaction force is 0 Nm because it has a 0 moment-arm.

    -Dan
    Just to make sure

    Torque = the force required to rotate something around a certain axis?
    And could you define pivot please?

    English not being my mothertongue makes this a bit hard sometimes ^^
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  7. #7
    Super Member
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    Quote Originally Posted by shinhidora View Post
    Just to make sure

    Torque = the force required to rotate something around a certain axis?
    And could you define pivot please?

    English not being my mothertongue makes this a bit hard sometimes ^^
    If you speak French then the meaning should be clear: In both languages (English and French) the word pivot means the same.

    If you speak Vlaams then a pivot is een ashals.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shinhidora View Post
    Just to make sure

    Torque = the force required to rotate something around a certain axis?
    And could you define pivot please?

    English not being my mothertongue makes this a bit hard sometimes ^^
    Torque is the rotational version of force. \vec{\tau} = \vec{r} \times \vec{F}. A "pivot" is a point around which a beam (or other object) would rotate about since it is attached to that point. (This differs slightly from "axis of rotation" because an axis of rotation could be anywhere on the object. A pivot is a point that is attached to the object and forces the object to rotate around it.)

    -Dan
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