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Angle to lift a body off the surface

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A rectangular box is being pulled by a rope with a force T on a horizontal surface, with a friction coefficient μ. What is the minimum angle α at which it will lift of the surface? See attached image.

    2. Relevant equations
    \begin{equation*}
    \tau = rF\sin\alpha
    \end{equation*}
    \begin{equation*}
    F_y=F_g-Tsin\alpha
    \end{equation*}
    \begin{equation*}
    F_x=F_f-Tcos\alpha
    \end{equation*}
    3. The attempt at a solution
    OK, I am slightly confused here. Should I look at this as a torque problem where the y component of force T will cause rotation around the lower left corner until the angle the box makes with the horizontal is equal to the angle of the force T? In the key the answer is expressed as
    \begin{equation*}
    \alpha=tan^{-1}(\frac{a}{b}+\frac{1}{\mu})
    \end{equation*}
     

    Attached Files:

  2. jcsd
  3. Mar 20, 2015 #2

    Suraj M

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    Gold Member

    This is a nice question :)
    I solved it using moments and i got your answer.
    Start by framing equations.
    The body should not move horizontally, so get 1 equation involving friction.
    then
    you just need one more equation, which you can get by equating the moments(3), give it a try,
     
  4. Mar 20, 2015 #3
    Okay, so if the body is not moving horizontally, the cosine component of the force T has to be equal to the force of friction, meaning
    \begin{equation*}
    \mu mg=Tcos\alpha
    \end{equation*}
    The moments of inertia of the rectangle at the corner has to be
    \begin{equation*}
    I=\frac{m(a^2+b^2)}{3}
    \end{equation*}
    The other moment of inertia I can see is where the (rope?) connects to the box but I don't have the length of the rope so I'm assuming I can use the moment of inertia of the box at the edge, which would be
    \begin{equation*}
    I=[\frac{mb^2}{3}+\frac{ma^2}{12}] sin\alpha
    \end{equation*}
    Is sinα correct here? I don't see how else moments would be equal
    The force in y direction is converted to torque (have alpha already, I'll let angular acceleration be varphi) so
    \begin{equation*}
    F_g-Tsin\alpha=I\varphi
    \end{equation*}
    Now do I try to define T from the friction equation in x direction?
     
  5. Mar 20, 2015 #4

    haruspex

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    You don't need moments of inertia. There's no acceleration here, it's effectively a statics question. Moment, as mentioned by Suraj, is just force times perpendicular distance.
     
  6. Mar 20, 2015 #5
    Ah, he meant torque, as moment of force :sorry:

    So I have my equation to keep things stationary in x direction
    \begin{equation*}
    \mu mg=Tcos\alpha
    \end{equation*}
    And the torque
    \begin{equation*}
    \tau=bTsin\alpha
    \end{equation*}
    (Am I correct in saying r=b?)
    Is torque in this case acting against Fg? I've tried using that and then plugging it in the friction equation, but, naturally I'm missing a and it looks quite different:
    \begin{equation*}
    tan\alpha=\frac{1}{\mu b}
    \end{equation*}
     
  7. Mar 20, 2015 #6

    NascentOxygen

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    Staff: Mentor

    The force T has vert and horiz components, and both components contribute a moment about the lower left corner of the block.
     
  8. Mar 20, 2015 #7
    :sorry:If I understand Nascent correctly, then I wasn't supposed to add sin component to the torque. I was under the assumption that displacement vector has to be perpendicular to the force, but now that I think about it, it simply makes the torque maximum. If I don't have to worry about 90°, then the displacement vector from axis of rotation to the force applied will be
    \begin{equation*}
    r={\frac{1}{2}a}^2+b^2
    \end{equation*}
    and torque will be left with just
    \begin{equation*}
    \tau=rT
    \end{equation*}
    acting against mg?
     
  9. Mar 20, 2015 #8

    NascentOxygen

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    When evaluating the moment of a force (or a component of a force) about a point you need the perpendicular distance from the point in question to the line of action of that force (or component).
     
  10. Mar 20, 2015 #9
    I don't see how can I make a right angle here between the force along the rope and the point, unless I'm imagining extension of a rope, in which case I'm lost at how I would calculate it only knowing the hypotenuse. What am I missing here?
     
  11. Mar 20, 2015 #10

    NascentOxygen

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    That is exactly what you must do, extend the line of action of the force until you can form the right-angle. If you do it this way, the problem becomes an exercise in geometry because you need to determine that perpendicular distance in terms of a, b, and angle alpha.

    Alternatively, you can resolve the force T into its vert and horiz components and calculate moments about that corner point due to these two components.


    I worked this problem both ways, the answers agree, but the second method was less complicated.
     
  12. Mar 20, 2015 #11

    Suraj M

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    it should be ( mg - vertical component of the force) µ
    The other equation is through moments.
    Use the Centre of mass, vertical and horizontal component of T at the point of attachment. All these can be got by the perpendicular distance in terms of a and b.
     
  13. Mar 21, 2015 #12
    I've got my r expression, seems to check out with random values if I check by graphing
    \begin{equation*}
    r=b\tan\alpha-\frac{a}{2}
    \end{equation*}
    If I'm doing it this way, where does the friction play into this? Is there "frictional torque" around the corner?
     
  14. Mar 21, 2015 #13

    NascentOxygen

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    No. Friction acts at the corner, and by taking moments about that point any force whose line of action passes through that point contributes no torque. So your equation ΣM does not involve friction.

    However, the equation for ΣF still does.
     
  15. Mar 21, 2015 #14
    x and y components of T are Tcosα and Tsinα. As haru said, no acceleration, which yealds
    \begin{equation*}
    F+Tsin\alpha-mg=0
    \end{equation*}
    Which gives maximum static friction as
    \begin{equation}
    \mu(mg-Tsin\alpha)
    \end{equation}
    Consistent with Suraj, yay.
    Then in x
    \begin{equation*}
    Tcos\alpha-\mu(mg-Tsin\alpha)=0
    \end{equation*}
    And the torque
    \begin{equation}
    \tau=[b\tan\alpha-\frac{a}{2}]T
    \end{equation}
    What else am I missing?
     
  16. Mar 21, 2015 #15

    NascentOxygen

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    The leading edge of the block is about to lift off the surface, indicating the net torque about the trailing edge = ??
     
  17. Mar 21, 2015 #16

    NascentOxygen

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    Draw a large figure, check this carefully.
     
  18. Mar 21, 2015 #17
    If there is no angular acceleration, then the net torque is zero; its used to overcome friction and mg, right?

    That is embarrassing, but seems I fixed it

    \begin{equation*}
    r=b\sin\alpha-0.5a\cos\alpha
    \end{equation*}
     

    Attached Files:

    • r.png
      r.png
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  19. Mar 21, 2015 #18

    NascentOxygen

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    Is that image you attached really 600kB?!! I'm not about to click on it.

    You can probably reduce it to under 20kB without significant deterioration, and that way it won't chew through mobile reader's download allowance.
     
  20. Mar 22, 2015 #19

    Suraj M

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    Gold Member

    Totally, i honestly feel you're complicating this.
    I highly recommend, you try the other method, of you have not.
    The T you have, you could have got the components which you have
    then gét their distance from the point of contact (left down). then get the equation for moments.
     
  21. Mar 22, 2015 #20

    NascentOxygen

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    Totally has got through the hard part, the remainder is no more complex than the alternative, so may as well finish this, now.
     
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