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How exactly is tension exerted?

  1. Jul 13, 2013 #1
    Say you have a rope looped around a frictionless pulley, and a mass m is attached to each side of the rope. When the system is released, the tension in the rope is T = mg, and the total force exerted on the pulley due to tension in the string is 2T = 2mg. We usually derive this by adding up the downward tension in the rope on either side of the pulley, but I don't buy this. There are only two points for which the rope is in contact with the pulley and the tension is directly downward. These two points can't possibly be the sole origin of force pulling the pulley down. I understand that this issue can be resolved if one considers differential portions of the rope as it loops around the pulley, but does this math always translate directly into adding up the tensions of the two sides of the rope?
     
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  3. Jul 13, 2013 #2
    I don't understand your concern: if the pulley's centre of mass is stationary, there must be a costraint force directed upward which is equal to 2mg, to equilibrate the weight forces directed downward.
     
  4. Jul 13, 2013 #3
    Yes, I understand it is 2mg, but I am considering a more detailed and involved question, how exactly the string is interacting with the pulley that causes the 2mg.
     
  5. Jul 13, 2013 #4
    I just thought this through imagining that the total vertical component of the force of the rope normal radial to the pulley must vary from a maximum at the top to a minimum of 0 at the sides +/- 90 degrees from the top... and that the total might be the sum of a trig function because of the varying force along the arc...

    But since the vertical component at +/- 90 degrees is already 1mg for each side, there does not seem to be a contribution from the force of the rope against the pulley.

    It looks like the rope laying over the pulley is the same as just fixing the rope to each side of the pulley...???
     
  6. Jul 13, 2013 #5

    tiny-tim

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    welcome to pf!

    hi greendog77! welcome to pf! :smile:
    you're correct of course, the rope exerts a force (with a vertical component) on the pulley all the way round, not just at the ends :smile:

    however, if we regard the pulley and the rope round it as being a single body, then the only external forces on that body are at the two ends, ie 2mg …

    the vertical components of all the "intermediate" forces cancel out :wink:

    (how is fairly obvious if the pulley is moving, less obvious and dependent on the internal structure of the pulley and of the rope if the pulley is stationary)
     
  7. Jul 13, 2013 #6

    WannabeNewton

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    The tension is transmitted through differential elements of the rope/string through Newton's 3rd law and the electrostatic forces between the molecules making up the rope. If the rope is massless then the tension will be transmitted exactly; if the rope is not massless then the tension will be diminished as it gets transmitted. See Kleppner and Kolenkow p.87-88
     
    Last edited: Jul 13, 2013
  8. Jul 13, 2013 #7

    sophiecentaur

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    You can look at it in a similar way to the hydrostatic force on the bottom of a round bottomed tube. In that case we don't usually worry about the fact that there are forces acting in all directions on the bottom of the tube. We ignore all the horizontal forces and just consider the sum of the vertical forces and then accept / can calculate if we're fussy / never give it a second's thought that the total force downwards is rho gh times the cross sectional area.
    The pulley problem has just one fewer dimensions to consider.
     
  9. Jul 13, 2013 #8
    Thanks for all the input. The reason I asked this was for less obvious problems where you can't simply use center of mass and one-system tricks. For instance, say you have a mass m on a frictionless table connected to a rope which is looped over a pulley on the edge of the table and connected to another mass m. As the two-mass system accelerates, what is the total force exerted on the pulley? If we simply add up the tensions on the two "end" points touching the pulley, we'd get T*sqrt(2). Is this really the correct result? I'm a little hesitant here.
     
  10. Jul 13, 2013 #9

    PhanthomJay

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    For an ideal pulley whose center of mass is stationary, there is no net (total) force acting on it. But if you mean what is the force of the rope on the pulley pin, you are correct. And the table pulley support arm exerts that same force in the opposite direction, for equilibrium (Newton 1). These forces are directed at 45 degrees from the horizontal.
     
  11. Jul 13, 2013 #10

    WannabeNewton

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    Is the pulley pin the thing that the pulley is extending out of?
     
  12. Jul 13, 2013 #11

    PhanthomJay

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    Presumably, there is a clevis eye anchored to the table extending outward from the table edge at 45 degrees connecting to a pin (axle) at the center of the pulley. The pin passes through the pulley and the holes (eyes) of the clevis on each side. The resultant rope force T *(sq. root 2) acts on the pulley quarter circumference and the pulley transmits that force to the center of the pin (axle).
     
  13. Jul 13, 2013 #12

    WannabeNewton

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  14. Jul 14, 2013 #13
    But how can this be justified?
     
  15. Jul 14, 2013 #14
    You have the right idea, just missing another trig term. The tension of the rope also folows the curve of the pulley. Vertical components of the tension changes with a function of the angle of tension with respect to the tangent or the normal to the surface depending upon whether you want to use cos or sine. You thus have the tension changing its direction as you move along the pulley and the value of the vertical force from the tension changing as you move up the pulley.

    Do note that at the very top the rope exerts no vertical or radial component since at that location the tension is purely horizontal.
     
  16. Jul 14, 2013 #15
    256bits,

    OK, so I was right in my original idea - the vertical component on the circumference of the pulley varies from max to 0 from top to +/- 90 degrees, each side summing to 1?

    But I'm not sure I agree about the variation on tension about the pulley... the tension of the rope is constant throughout its length and always tangent to the pulley, although the pressure of the rope against the pulley does vary...

    I think the rope does apply a vertical radial component to the pulley, actually the maximum is there.

    The puzzle is that the vertical component goes from 0 to max from +/- 90 degrees to top. For the total to be 2, then each side must be 1, so the total is a summation on each side from 0 to X where the total is 1.

    That means the vertical component at the top can't be 1 for each side, but some lessor value.

    It looks to me like the vertical component is the sine of tension.

    The sines of the tension through the upper side quarter of the circumference is the summation of the vertical components... which is equal to 1. That seems to work...?
     
  17. Aug 5, 2013 #16
    You can always do such things, only newtonian mechanics is required.

    Since this is a system of rigid bodies, what you need are just the two "cardinal laws of dynamics" (I don't believe this is the correct name in english but I don't know how to translate it from the italian):

    dP/dt = R(e)

    dK/dt = M(e)

    wher P is the system's total momentum in an inertial frame, R(e) is the resultant of the (e)xternal forces applied on the system in that frame; K is the system's angular momentum in an inertial frame, M(e) is the resultant of the (e)xternal momentums applied on the system in that frame.

    In this kinds of problems we usually neglect the rope's mass and the pulley's moment of inertia; we only consider the two masses m1 (the one on the table) and m2 (the one which pulls down the first with its weight) then we can assert that the two forces T1, T2 that the two masses makes on the rope are the same:

    (T2 - T1)*r = I d2θ/dt2

    where θ is the pulley's angulus of rotation and r its radius.
    So if I = 0, then T2 = T1 = T.

    The first cardinal law of dynamics applied (twice) to the two masses becomes (here I have called with "T" the force the rope makes on a mass since for the third principle it's equal to the force that mass makes on the rope):

    m1 d2x/dt2 = T

    m2 d2x/dt2 = m2 g - T

    Solving: T = m1 g / (1 + m1/m2).

    Since the two forces on the rope, T1 and T2 are the same, the two forces that the rope applies to the pulley are the same = T and at 90°, so the resultant force applied on the pulley from the rope is sqrt(2)*T and directed at 45°:

    sqrt(2)*T = sqrt(2)*(m1 g) / (1 + m1/m2).

    So, if the pulley's centre of mass stays stationary, the same force, with opposite direction, have to be applied to the pulley as costraint force from its mechanical support.
     
    Last edited: Aug 5, 2013
  18. Aug 5, 2013 #17
    If the pulley's moment of inertia I cannot be neglected, the equations becomes:

    1. m1 d2x/dt2 = T1

    2. m2 d2x/dt2 = m2g - T2

    3. I d2θ/dt2 = (T2 - T1) r (this is the second cardinal law of dynamics applied to the pulley)

    4. x = rθ

    They are 4 independent equations in the 4 unknowns x, θ, T1, T2 and can be easily solved finding:

    d2x/dt2 = m2g/(m1 + m2 + I/r2)

    → x(t) = (1/2) m2g t2/(m1 + m2 + I/r2) + v0t + x0

    where v0 is the velocity of m1 (or m2) and x0 its position, at t = 0.


    T1 = m1m2g/(m1 + m2 + I/r2)

    T2 = m2g(m1 + I/r2)/(m1 + m2 + I/r2)

    → T1/T2 = m1/(m1 + I/r2) < 1

    Now we can see exactly in which case we can neglect I and assert that T1 = T2:

    when I/r2 can be neglected with respect to m1.
     
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