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Moment and torque physics problem

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=20478&stc=1&d=1252419897.jpg

    2. Relevant equations

    Equilibrium- Moment/Torque

    3. The attempt at a solution

    The answer given is x = L/2, y = L/4, z = L/6.
    What should I do to find these answer? (Those answers may or may not be correct)
    :confused:
     

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  3. Sep 8, 2009 #2

    kuruman

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    Re: Moment/Torque

    The statement of the problem did not come through. You may have to type it in. Also you have to make a better effort. Equilibrium- Moment/Torque is not an equation. If you don't know how to get the answer, at least tell us what you know about the problem, what it is about and what you think might be relevant in your effort to get to the answer.
     
  4. Sep 8, 2009 #3

    kuruman

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    Re: Moment/Torque

    Ah! The picture came through.

    Suppose that you had only one block resting at the edge of a table so that x meters hang over the edge. Do you understand why x = L/2? Why not larger? Why not smaller? Answer this question and you will understand why the top block was placed as shown. Then worry about the other blocks.
     
  5. Sep 9, 2009 #4
    Re: Moment/Torque

    Actually my main problem is I don't know why x = L/2. Can you show me the force acting on the top block that cause it to be in equilibrium?
    Thanks.
     
  6. Sep 9, 2009 #5

    kuruman

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    Re: Moment/Torque

    The forces acting on the top block are gravity and the normal force exerted by the body supporting it. Where, i.e. at what point, does the force of gravity act on the top block?
     
  7. Sep 9, 2009 #6
    Re: Moment/Torque

    Hi stpmmaths,

    i have another approach. Think about the http://en.wikipedia.org/wiki/Center_of_mass" [Broken] of each block! That means, that each block can be imagined as a point mass, at a certain Point.

    So the x-component (see sketch below) of the center of mass of a bunch of blocks must be on the edge of the block below (acutally an infinitesimal piece forward the edge, to be in stable equilibrium, otherwise the "point mass" falls down!)

    Cause my english is not so well, i've made a sketch:

    http://go-krang.de/physicsforums.com%20-%20thread%20-%202339260.png [Broken]


    So considered from the origin of the coordinate system i have chosen, i declare the x component of the center of mass of the blocks with
    • Block 1 : [tex]r^{^{(cm)}}_{x,1}[/tex]
    • Block 2 : [tex]r^{^{(cm)}}_{x,2}[/tex]
    • Block 3 : [tex]r^{^{(cm)}}_{x,3}[/tex]

    with [tex] R_x = \frac{1}{\sum_i m_i} \, \sum_i m_i \cdot r^{^{(cm)}}_{i,x} [/tex] it yields

    [tex] R_x = \underbrace{L}_{\mbox{edge of the lowest block}} = \frac{1}{3\cdot m_{block}} \cdot m_{block} \cdot \left( \underbrace{\left(z+\frac{L}{2}\right)}_{ \mbox{center of mass B3} } + \underbrace{\left(y+z+\frac{L}{2}\right)}_{ \mbox{center of mass B2} }+ \underbrace{\left(x+y+z+\frac{L}{2}\right)}_{ \mbox{center of mass B1} } \right) = \frac{L}{2} + \frac{x}{3} + \frac{2y}{3} + z[/tex]

    if you put the answers for x,y,z in the above equation it results L, like i've promised ;)

    But my equation above has infinity results (Cause my equation treats the 3 blocks as one continuum....) , so you must think about what constraints are on x and on y!

    TIP: work top down through the blocks


    with best regards and i hope i could help
     
    Last edited by a moderator: May 4, 2017
  8. Sep 11, 2009 #7
    Re: Moment/Torque

    thanks
     
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