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Homework Help: Tension of a rope

  1. Dec 4, 2017 #1
    1. The problem statement, all variables and given/known data
    A box weighs 12.6kg and it is dragged with a rope(it weighs 1.4kg and the length is 60cm).The force dragging the box is F=80N.The coefficient of kinetic friction is 0,5.
    What is the tension of the rope in the middle?(Answer:17N)

    2. Relevant equations
    No revelant equations

    3. The attempt at a solution
    (F-T=m2 (1.4kg)*a;T-md (kinetic friction)*m1(12.6kg)*g)
    T=78N but it is wrong.
  2. jcsd
  3. Dec 4, 2017 #2
    What is the acceleration? Have you included the mass of the rope in determining the acceleration?
  4. Dec 4, 2017 #3
    Accelleration a: F-m1*g*md=(m1+m2)*a=> a=(F-m1*g*md)/(m1+m2)=1.29m/s^2
    Book answer a=1.3m/s^2
  5. Dec 4, 2017 #4
    Do you think that the tension in the rope is constant from end to end?
  6. Dec 4, 2017 #5
    I don't know this is why I am asking help.
  7. Dec 4, 2017 #6
    One end of the rope is at tension F and the other end is st tension T. If you do a firce balance on the rope, what do you get?
  8. Dec 4, 2017 #7
    How do i do that?
  9. Dec 4, 2017 #8
    The tension on the rope will be maximum at point B, because the point B pulls both the block and the rest of the rope. It will be minimum at point A, as point A only pulls the mass of the block. You found the acceleration of the block + rope system to be ##a_{system} = 1.3 ms^{-2}##. What will be the mass of the section of the rope that is 0.3 m long?
  10. Dec 4, 2017 #9
    Thank you for the drawing.The mass of that section is 0.7kg?Am i right.But how can the tension in the middle 17N?
  11. Dec 4, 2017 #10
    Correct , it will be 0.7 kg. Now take the block and the section of the rope as a system( A to the black cut). What will be the acceleration of the system?
  12. Dec 4, 2017 #11
    I think that I have the solution:T (middle)=F-0.7kg*1.3m/s^2-md*m1*g=17.287N
    Thank you.
  13. Dec 4, 2017 #12


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    There's a serious flaw in the question.

    Which way is the 80N acting? In the absence of any statement on that we should normally assume it is horizontal. So what is holding the rope up? The rope has mass, so it has weight.
    If the 80N is horizontal then the rope forms an arc which is horizontal at the right hand end and angles up to the block at the left end. Thus, the weight of the rope is being taken by the block and must be included in the normal force.
    The setter may have overlooked that this has a signicant affect on the acceleration. It is barely more than half that calculated by ignoring the rope's weight.
    A further complication, though less severe, is that the rope's weight will also contribute to its tension at its mid point.

    A compromise would be to take the rope arc to be symmetric, but then the 80N must be angled upwards, making it hard to calculate the horizontal component.
    Edit: no, it's quite easy to calculate the horizontal component.
    Last edited: Dec 4, 2017
  14. Dec 4, 2017 #13
    Why does this include the frictional force?????? Let's see your free body diagram on the half-section of the rope to the right.
    Last edited: Dec 4, 2017
  15. Dec 4, 2017 #14
    I think the student is supposed to assume that the rope is perfectly horizontal, and to neglect the weight of the rope. I know that this isn't precisely correct, but it is a reality of student homework problems.
    Last edited: Dec 4, 2017
  16. Dec 4, 2017 #15


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    Asuming horizontal is one thing, we could think of it as a rod. But why should the weight be neglected? It would certainly be reasonable even for a novice student to count half the rope as contributing to normal force.
  17. Dec 4, 2017 #16
    Hi haruspex.

    I think this is a good point. How about you volunteer to solve it both ways and then report the comparison back to us? The comparison should be pretty interesting.

  18. Dec 4, 2017 #17
    Please tell me you really don't think the answer for the tension at the center of the rope is anything near 17 N. There is no way this is correct.
  19. Dec 5, 2017 #18
    As the acceleration of the block + the segment of rope is ##1.3 ms^{-2}## , and the mass of the system is ##13.3 kg##, Newton's second law should give a force of ##17 N##. ##F_{x} = Ma_{x} \Longrightarrow F_{X} = (13.3)(1.3) \approx 17N##.
    Now I ignored the normal force due to the rope , but I think the problem wants us to ignore it.
  20. Dec 5, 2017 #19
    Your force balance on the rope is incorrect. The frictional force is not acting on the rope. Draw a free body diagram of just the right half of the rope. The horizontal force balance on this section should read $$F-T_{1/2}=0.7a$$where ##T_{1/2}## is the tension at the mid point of the rope. This gives a tensile force of 79.1 N at the center of the rope. Does a value of 17 N really make sense to you if, for a massless rope, the tension at the center would be 80 N?
  21. Dec 5, 2017 #20
    Me too got the same value.17 N at mid doesn't make sense if mass +half rope will be accelerated 17 N then how 63 N will responsible for acceleration only 1.3 m/s^2 of mass just 0.7kg mass.
  22. Dec 5, 2017 #21
    I think I messed up by not including the frictional force on the block + segment of rope system. Let me try again. So the system consists of the block and the rope A. Total mass of the system is ##M_{system} = 13.3 kg##. The external forces on the system are friction and the tension.

    Using Newton's second law:

    ##\Sigma F = ma##
    ##T_{1/2} - f_{kinetic} = M_{system}a##
    ## T_{1/2} = M_{system}a + f_{kinetic}##
    ##T_{1/2} = (13.3 kg)(1.3 ms^{-2}) + (0.5)(126) = 80.29 N##

    Please correct me if I am wrong.
    Last edited: Dec 5, 2017
  23. Dec 5, 2017 #22
    You have taken g=10m/s^2 that's why you are getting more than 80N put g=9.8 result will be 79.1 N
  24. Dec 6, 2017 #23


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    Here are the results for five different views of the problem:
    1. the simple view, in which the weight of the rope (but not its inertial mass) is ignored
    2. the rod view, so half the rope's weight contributes to frictional force
    3. the rope hangs down from the block but is horizontal at the 80N end; all of the rope's weight contributes to friction; the midpoint tension has an x and a y component
    4. the rope hangs down at each end symmetrically (sort of.. the horizontal tension varies a bit along its length); the horizontal component of the applied 80N is reduced because some of that force is now vertical; half the rope's weight contributes to friction
    5. the rope is horizontal at the block; the horizontal component of the 80N is reduced further as it takes the whole weight of the rope, while the midpoint tension again has a vertical component
    Code (Text):

    79.087 T, simple view

    79.259 T, rod view

    79.430 right hor, Tx
    06.860 right hor, Ty
    79.726 right hor, T

    79.705 mid hor, Fx
    78.979 mid hor, T=Tx

    78.815 left hor, Fx
    77.961 left hor, Tx
    06.860 left hor, Ty
    78.262 left hor, T
  25. Dec 7, 2017 #24
    This is not exactly what I had in mind. What I was envisioning was the following:

    If we do a differential force balance on the increment of rope between s and s + ds, we obtain:
    $$\frac{d(T\sin{\phi})}{ds}=\rho g$$
    $$\frac{d(T\cos{\phi})}{ds}=\rho a$$where s is the distance along the rope contour measured from the left end, ##\phi## is the angle that the tangent to the rope contour makes with the horizontal at location s, T is the tension in the rope at location s, ##\rho## is the linear density of the rope, and a is the horizontal acceleration.

    If we integrate these equations from location s to the right end of the rope at s = L, we obtain:
    $$T\sin{\phi}=T_L\sin{\phi_L}-\rho g (L-s)\tag{1}$$
    $$T\cos{\phi}=T_L\cos{\phi_L}-\rho a (L-s)\tag{2}$$where ##T_L## is the tension at s = L (80 N in our problem) and ##\phi_L## is the contour angle at s = L.
    In addition, the shape of the rope is described by:
    We can eliminate the tension T from our equations by dividing Eqn. 1 by Eqn. 2 to yield:
    $$\tan{\phi}=\frac{(dy/ds)}{\sqrt{1-(dy/ds)^2}}=\frac{T_L\sin{\phi_L}-\rho a(L-s)}{T_L\cos{\phi_L}-\rho g(L-s)}\tag{5}$$
    Next, solving for dy/ds and reducing the equation to dimensionless form yields:
    $$\frac{dY}{dS}=\frac{\sin{\phi_L}-\xi_g(1-S)}{\sqrt{(\sin{\phi_L}-\xi_g(1-S))^2+(\cos{\phi_L}-\xi_a(1-S))^2}}\tag{6}$$where Y=y/L, S=s/L, and where the dimensionless constants ##\xi_g## and ##\xi_a## are given by:
    $$\xi_g=\frac{\rho g L}{T_L}$$and$$\xi_a=\frac{\rho a L}{T_L}$$
    If we integrate Eqn. 6 from S = 0 (the left end of the rope) to S = 1 (the right end of the rope), we obtain the difference between the dimensionless rope elevations at the two ends:
    If the two ends of the rope are at the same elevation, then we must have that:
    $$\int_0^1{\frac{\sin{\phi_L}-\xi_g(1-S)}{\sqrt{(\sin{\phi_L}-\xi_g(1-S))^2+(\cos{\phi_L}-\xi_a(1-S))^2}}dS}=0\tag{7}$$This equation will be satisfied if the proper value is employed for the contour angle at the right hand end of the rope ##\phi_L##. Thus, Eqn. 7 provides a relationship for determining ##\phi_L##. Once ##\phi_L## is known, the value of the contour angle and tension at all locations along the rope are established, including the values at the left end where the cord is attached to the block.
  26. Dec 7, 2017 #25


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    Ah, you wanted the full catenary treatment. I was thinking just in terms of getting good approximations in the present context so that we could compare the consequences of the different feasible assumptions.
    I believe the difference the catenary analysis would make is very small, since it depends principally on how the horizontal mass distribution of the rope differs from uniform. In the present problem that would be tiny.
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