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Tension at a point in a slab

  1. Mar 8, 2015 #1
    Two opposite forces F1 = 120 N and F2 = 80 N act on an elastic plank of modulus of elasticity Y = 2 x 10^11 N/m2. and length l = 1 m placed over a smooth horizontal surface. The cross-sectional area of the plank is S = 0.5 m2. The change in the length of the plank is x *10-11m Find the value of x.

    I understood most of the solution but I didn't get one part of it, which says:
    Tension at any point x in the plank T(x)= F2 + ((F1 - F2)/l)*x

    Why do we include the term F2? Shouldn't T(x) without the term F2 be the tension at the point?
     
    Last edited: Mar 8, 2015
  2. jcsd
  3. Mar 8, 2015 #2
    Anybody there?
     
  4. Mar 8, 2015 #3
    I am not quite sure from the question what are the orientations of the forces. Are they or are they not perpendicular to the cross section? On the face of it, the plank is not in static equilibrium, and so my favourite definition of the axial force (the algebraic sum of all the forces on one side of a section - or the other side) is problematic to apply.
     
  5. Mar 8, 2015 #4
    Yes, the forces are perpendicular to the cross section.
     
  6. Mar 8, 2015 #5
    Can someone please help me on this? One possible explanation I could think of was that when we rearrange the expression for tension it becomes as follows-
    T(x) = F2(l-x)/x + F1x/l. This can be explained as - Force the on the differential element at a length x from F1 would be F1x/l because force on the differential element right next to the application of F1 (at length x = 0) would be F1, which is essentially at one end of the plank (length = l). So for unit length it would be F1/l and for x length it would be F1x/l. Similarly, for F2, the force is being applied on the other end of the plank. And so, its distance from the differential element in consideration would be l-x. And using the same method as before, the force on this element due to F2 would be F2(l-x)/l.

    But my question is that why do we add the two forces together? Shouldn't we subtract them as they are in opposite direction? Or does this have something to with the definition of tension here?
     
  7. Mar 8, 2015 #6
    Let F1 be the tensile force at x = l, and let F2 be the tensile force at x = 0 (pulling in the opposite direction from F1). Is the plank going to be in equilibrium? If not, write an overall force balance on the plank. After you do this, I'll get back to you with the next step.

    Chet
     
  8. Mar 8, 2015 #7
    The net force would be F1 - F2, so that the block would be moving in the direction of force F1.
     
  9. Mar 9, 2015 #8
    Hello! Is anybody there? Am I screaming on a deserted island or something? :(
     
  10. Mar 9, 2015 #9
    Chill man. Some of us have to sleep too.

    ##F_1-F_2=ma##

    where m is the mass of the plank.

    Now, what I would like you to do is draw a free body diagram of the portion of the plank between the cross section at x = 0 and the cross section at arbitrary location x (< l) along the plank. Call the tension at this cross section T(x). What fraction of the mass of the plank lies between these two cross sections? From this result, what is the mass of plank material between the cross sections at x = 0 and the arbitrary location x (in terms of the total mass of the plank m)? Are all portions of the plank accelerating at the same rate (as given by the equation above)?

    Chet
     
  11. Mar 9, 2015 #10
    Yep. That would be ##mx/l##. Yes, all the portions are accelerating at the same rate (so that they are moving the same distance per unit time). If this wouldn't have been true, then the plank would have broken apart (since some particles are moving faster than others, so the distance between them shouldn't remain constant). But wait, shouldn't that be the case only when the body is a rigid body? The question clearly says that the body is elastic, so it is not a necessity that the particles have a constant distance between each other. So the particles 'can' have different accelerations. Whether they have different accelerations or not is another question, something which I am not sure about at present. The force that the element at ##x## would be experiencing would be ##(F_1-F_2)x/l##.
     
    Last edited: Mar 9, 2015
  12. Mar 9, 2015 #11
    To do this problem, you need to assume that somehow a steady state has been established so that all portions of the bar are moving with the same acceleration. This is what they expect you to assume.
    Whoa. Slow down pardner. This result is not correct.

    In your free body diagram, you know that the mass of your body is mx/l, you know that the tensile force at one end is T(x), you know the tensile force at the other end is F2 in the negative x direction, and you know that the acceleration for the section of plank being considered is the same as that for the overall plank a. So now please write a force balance equation using Newton's second law for the section of your body between x = 0 and x.

    Chet
     
  13. Mar 10, 2015 #12
    Whoa. Okay. It would be: T(x) - F2 = mx/l*a. Is that correct?
     
  14. Mar 10, 2015 #13
    But what about F1? Shouldn't F1 also act at the body between x = 0 and x?
     
  15. Mar 10, 2015 #14
    Yes! Yes! Yes!

    But now substitute ma from the overall force balance into this equation and see what you get. I think you will be pleasantly surprised.

    Chet
     
  16. Mar 10, 2015 #15
    That would make it as follows-
    ##F_1-F_2 = ma ⇒ a = (F_1-F_2)/m ##
    So, ##T(x) - F_2 = mx(F_1-F_2)/l * m = x/l *(F_1-F_2)##
     
  17. Mar 10, 2015 #16
    Nice job.

    Chet
     
  18. Mar 11, 2015 #17
    But why isn't F1 also included at the end which is being pulled by tensile force T(x)? Shouldn't it then be T(x) + F1 - F2 = (mx/l)*a?
     
    Last edited: Mar 11, 2015
  19. Mar 11, 2015 #18
    No. The force F1 is only being applied to the right face of the overall plank, and nowhere else. What we are doing here is equivalent to breaking the plank into two separate segments and asking "what tension would we have to manually apply to the right hand face of the segment on the left in order for that segment to have the same acceleration as it did in the intact plank (i.e., if we had not made the break)?" This is the same as the tension that the right hand segment exerts on the left hand segment at the cross section under consideration in the intact plank.

    Chet
     
  20. Mar 11, 2015 #19
    Then even F2 is being applied to the left face of the overall plank. Why do we include it in for a mass element at x from the right while computing its overall balanced force equation so as to apply Newton's Second law?
     
    Last edited: Mar 11, 2015
  21. Mar 11, 2015 #20
    We are analyzing the segment of plank between x=0 (where F2 is applied) and location x (where we want to determine the tension).

    Chet
     
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