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Static Equlibrium involving a weird object

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A uniform beam of weight w is inclined at an angle [tex]\theta[/tex] to the horizontal with its upper end supported by a horizontal rope tied to a wall and its lower end resting on a rough floor. (a) If the coefficient of static friction between the beam and floor is [tex]\mu_s[/tex], determine an expression for the maximum weight W that can be suspended from the top before the beam slips (W is hanging from a rope at the top of the beam). (b) Determine an the magnitude of the reaction force at the floor and the magnitude of the force exerted by the beam on the rope at P in terms of w, W, and [tex]\mu_s[/tex]. (w is the weight of the beam, W of the mass hanging from the top of the beam...point P is located at the top of the beam where the top of the beam is both connected to the wall and the hanging mass)

    2. Relevant equations

    [tex]\Sigma F=0[/tex]

    [tex]\Sigma\tau=0[/tex]


    3. The attempt at a solution

    This problem really confuses me...I suppose the thing that confuses me the most is the location and direction of the reaction forces. I first assumed that the reaction force exerted by the floor would be straight up, but I'm not sure this is the case since the beam is at an angle. Also, would the tension in the upper beam where it is connected to the wall be the reaction force exerted on the rope by the beam?

    I also cannot for the life of me determine how the force equations and torque equations are used to determine all of this information. It is in static equilibrium, so the net external force and torque must be zero. But again, I'm having trouble even setting up a force diagram...and I have the system drawn in front of me! The answer given in the book is very complex, but I have no clue how to reach it. Any help would be GREATLY appreciated.
     
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2
    To clear confusion, the reaction force exerted by the floor is straight up even if it is angled. The x-component of the floor on the beam is friction force given by [tex]\mu_s[/tex]*normal force. The horizontal rope can be cut and replaced with a force in tension.
     
  4. Mar 9, 2008 #3

    tiny-tim

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    Hi XxBollWeevilx!

    (btw, if you type alt-m, it prints µ)

    Since you aren't told exactly how the rope is supporting the beam, and since you aren't asked to find the forces at that point of support, I suppose you'd better take moments about that point.

    The forces not through that point are the weight w of the beam (through the centre of mass), and the reaction from the floor.

    The horizontal component of that reaction will be equal to the vertical component times µ.
     
  5. Mar 9, 2008 #4
    Hmm...the answers that my book gives are as follows:

    (a) [tex]W=\frac{w}{2}(\frac{2\mu_ssin\theta-cos\theta}{cos\theta-\mu_ssin\theta})[/tex]

    (b) [tex]R=(w+W)\sqrt{1+\mu_s^2}[/tex]

    [tex]F=\sqrt{W^2+\mu_s^2(w+W)^2}[/tex]

    Thanks for the help...so it would be best to take my torque sum around the point P at the top of the beam? I assume that the horizontal is completely tight from the beam to the wall. And the other thing I wasn't sure about...torque requires me to know the length of the lever arm for each force...none is given in the problem and none is in the solution. So is using toque an appropriate approach, or do I need to use more of a force analysis? Thank you!
     
  6. Mar 9, 2008 #5
    Anyone?
     
  7. Mar 9, 2008 #6
    well for part (b)[tex]F=\sqrt{W^2+\mu_s^2(w+W)^2}[/tex]
    if you do sum of forces in y direction n-w-W=0 n=w+W and Fs=[tex]\mu_s[/tex]*n
    then sum of forces in x-direction you get T=[tex]\mu_s[/tex]*n
    so resultant force at point P = magnitude of the components W and Fs. which is the same as the book.

    for resultant force at the normal force... R you get magnitude of Fs and n which you know... the answer is the same as the book

    for part a i think you want to consider the conditions when it slides... which would be when n=0 and use theta to express lengths
     
    Last edited: Mar 9, 2008
  8. Mar 9, 2008 #7
    Thanks a lot. I managed to get both halves of part B...but part A is giving me trouble, especially regarding the angles. If I try to use the BOTTOM of the beam as my axis of rotation for torque, I get

    [tex]\Sigma\tau=Wcos\theta+wcos\theta-Fsin\theta=0[/tex]

    F is the force by the beam on the rope. I am not sure my cosines and sines are correct...the angle theta is measured from the horizontal to the beam (not from the same side that the rope is connected to the upper beam...it is an angle less than 90 degrees, in other words. If I take torques around the top of the beam, I get

    [tex]\Sigma\tau=ncos\theta-wsin\theta-fcos\theta=0[/tex]
    where f is the friction force, [tex]\mu_sn[/tex]

    Am I on any sort of right track?
     
    Last edited: Mar 9, 2008
  9. Mar 9, 2008 #8
    Those lower case sigmas should be uppercase, but for some reason it won't show them that way.
     
  10. Mar 10, 2008 #9
    [tex]\Sigma\tau=Wcos\theta+wcos\theta-Fsin\theta=0[/tex] using right hand rule should all be positive... or all "into the page" your angles seems to be correct.

    [tex]\Sigma\tau=ncos\theta-wsin\theta-fcos\theta=0[/tex] signs are correct, fcos(theta) for friction force, sin theta... and w cos theta for beam weight... however, to solve and expression for W I am not quite sure...
     
  11. Mar 10, 2008 #10
    Alright...well I'll play with it and see if I can come up with the book's solution. Thanks!
     
  12. Mar 10, 2008 #11

    tiny-tim

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    Hi XxBollWeevilx!

    Yes, you do need to know the lengths - but you can just call the length of the beam L, and do all the lengths as multiples of L.

    In the end, all the Ls will cancel out - that's why there's no length in the answer.

    Which is what you'd expect, isn't it? The forces angles etc should be the same for different lengths, provided the weights are the same! :smile:

    New hint: remember, using µ, you know the direction of the total force at the bottom, don't you? …

    Take torques about the top, and have another go! :smile:
     
  13. Mar 10, 2008 #12
    OK...but if I take the torques about the top, then the moment arm for W is 0. That's just what I need to find...how would this work? Thank!
     
  14. Mar 10, 2008 #13

    tiny-tim

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    Hi XxBollWeevilx!

    I think the book answer [tex]R=(w+W)\sqrt{1+\mu_s^2}[/tex] is wrong - it shows the whole weight of the beam-plus-weight, w + W, being supported at the bottom, when we know some of it is supported at the rope. :frown:
    It would work … one step at a time!

    Find the floor-reaction force strength from that torque equation (you already know the direction - it's (µ,1)). That won't have W.

    Then forget torque and just draw in all the forces and take components along the beam to get W, and horizontally to get F. :smile:

    (I think you're supposed to assume the rope is cylindrical and frictionless, so that the reaction is perpendicular to the beam.)

    Hey, tongpu, mentors … I need a second opinion here … is the book answer wrong, or am I missing something? :redface:
     
  15. Mar 10, 2008 #14
    Well my test is over, I got part B but I didn't get all of part A correct. It's alright though, I appreciate the help a lot. I tried to take the torques about the center, meaning only the weight of the beam would not be significant. I almost got there! Thanks again.
     
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