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At what angle is the work done minimised?

  1. Oct 24, 2013 #1
    The problem as it appears on the problem sheet: A block of weight [itex]mg = 500 N[/itex] is sitting on a horizontal floor. The coefficient of kinetic friction is [itex]\mu_{k} = 0.4[/itex]. In class, we determined at what angle from the horizontal I should pull on the block if I'm weak and need to minimise the force [itex]F[/itex] with which I am pulling. Since then, I've been working out, and now I'm very strong - but also very lazy. At what angle [itex]\theta[/itex] from the horizontal should I pull the block in order to minimise the total work I need to do in order to move the block by [itex]10 m[/itex] in a straight horizontal line at a constant, non-zero velocity, and what will that work be?



    Relevant equations:
    (1)$$N+F\sin(\theta)= 500$$
    (2)$$F\cos(\theta)=friction=0.4N$$
    (3)$$N=\frac{F\cos(\theta)}{0.4}$$
    Plugging (3) in (1) gives: $$\frac{F\cos(\theta)}{0.4}+F\sin(\theta)=500$$ which gives force [itex]F[/itex] as a function of [itex]\theta[/itex]:
    (4)$$F=\frac{500}{\frac{\cos(\theta)}{0.4}+\sin(\theta)}$$
    (5)$$W=F\cos(\theta)\times10$$


    In a previous lecture, the teacher derived (4), and showed that to minimise [itex]F[/itex], the denominator in (4) has to maximised. So he took the denominator, differentiated it with respect to [itex]\theta[/itex], and set the derivative to equal to 0 (to determine the maxima):
    (4.5)$$\cos(\theta)-\frac{\sin(\theta)}{0.4} = 0 ∴ \theta = \arctan(0.4) = 21.8^{o}$$

    My attempt: (5) is the definition of work done, and I need to find the angle at which [itex]W[/itex] is the minimum. Clearly, [itex]W[/itex] is a multivariable function since not only does [itex]10\cos(\theta)[/itex] vary as [itex]\theta[/itex] varies, but [itex]F[/itex] also varies since it is also a function of [itex]\theta[/itex]. So substituting (4) in (5) gives (6): $$W=\frac{500}{\frac{\cos(\theta)}{0.4}+\sin(\theta)}\times 10\cos(\theta) = \frac{5000}{2.5+\tan(\theta)}$$
    For [itex]W[/itex] to be minimised, the denominator of (6): [itex]2.5+\tan(\theta)[/itex] must be maximised, so I did what my teacher did, and simply differentiated the denominator and set the derivative to equal to 0: $$\sec^{2}(\theta) = 0$$ However, this does not give me a physically meaningful value for the angle. Where did I go wrong, and what do I do? I've considered various alternative forms of (5), but none seem to give any real solutions for the angle.

    Any help would be appreciated.
     
  2. jcsd
  3. Oct 24, 2013 #2

    Simon Bridge

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    The result is[/i physically meaningful, and, describing the possible meanings in words may help you discover what happened.

    You can see, without differentiating, that a+tanθ will be maximized when ##\small \theta=n\pi /2## ... the equation is also satisfied if the denominator is minimized notice ;)

    If you push at +90deg to the horizontal, then you are pushing directly down on the box ... which means the box does not move so the work is zero - well done: minimum work.

    See what happened?
     
  4. Oct 25, 2013 #3


    Thanks for the reply.

    When I said "not physically meaningful", I meant that I was expecting an angle [itex]90^{o}>\theta>0^{o}[/itex] to the horizontal. Yes, pulling directly up on the box is going to minimise the work, however, I'm trying to find a angle such that the box moves by [itex]10m[/itex] in a straight horizontal line, at a constant non-zero velocity.

    By the way, is the equation (5) appropriate for this question, given that the force varies with [itex]\theta[/itex]?
     
  5. Oct 25, 2013 #4

    Simon Bridge

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    equ (5) is applicable - but that's not the only work done on the system is it?

    You need an expression for the total work done on the system.

    You should not substitute in the numbers until after all the algebra is complete.
    Leave the variables in as long as you can.
     
  6. Oct 25, 2013 #5
    Thanks for the reply.

    I asked my professor for clarification and he replied: "you only need to consider the work done by me since I'm the lazy one, not friction."

    Then I asked him why I'm not getting any acute angle for my [itex]\theta[/itex] and he replied: "If you think that your math is right, then maybe it's trying to tell you something?"

    So if there are no real solutions for [itex]\sec^{2}\theta=0[/itex], then the problem is basically a trick question, and the work done cannot be minimized? That seems odd, considering that this problem is worth 7 marks.
     
  7. Oct 25, 2013 #6
    Of course it can be minimized. The value of zero is the minimum value.
    Think about it: if there is no friction, what is the work necessary to move something horizontally with constant speed?
     
  8. Oct 25, 2013 #7
    Ohhh! I think I get it now. Since work done is equal to change in kinetic energy, and since there is NO change in kinetic energy when the velocity remains constant, the work done must be zero, and that ultimately means that [itex]\theta = 90^{o}[/itex], right?

    EDIT: Oops, I meant the total work must be zero, not the work done by force [itex]F[/itex].
     
    Last edited: Oct 25, 2013
  9. Oct 26, 2013 #8

    Doc Al

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    Don't think in terms of total work, but of the work done by the force F (which is the work done by you). (As long as the speed is constant, the total work will be zero at any angle.)

    Realize that 90° is the limiting value. Crank out some numbers to see the trend. For θ = 70, 80, 85, 89 degrees, for instance. What happens to the value of the friction force and thus the value of the work done?
     
  10. Oct 26, 2013 #9
    Thanks for the reply.

    As θ goes to 90°, the value of the friction force, and thus the value of the work done by force [itex]F[/itex] goes to [itex]0[/itex]. So, the solution to this problem is simply that to minimise the work done, the angle at which I should pull on the block is 90° from the horizontal, and therefore the work done is 0. Right?
     
  11. Oct 26, 2013 #10

    haruspex

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    Arguably there is an infimum but it is not a realisable minimum.
     
  12. Oct 26, 2013 #11
    Yes, you need to do some extra work to accelerate to some non-zero speed. But this work can be as small as you want it to be.
    If you are saying that this is not a "mathematical" minimum of the work as a function of angle, I agree but I don't quite see the relevance for this problem.
     
  13. Oct 26, 2013 #12

    haruspex

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    It's the consideration which resolves the paradox that was bothering subzero0137. The question asks for the minimum work to move it a specified distance. If you answer zero, that's clearly wrong because there will be no horizontal movement. The strictly correct answer is that there is no such minimum; end of paradox.
     
  14. Oct 27, 2013 #13

    Doc Al

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    Yes. But I think a better way to describe it is that you can make the work done by you (in this idealized arrangement) as small as you want by making the angle closer and closer to 90°. Of course, this ignores the additional work you needed to get to some non-zero velocity (or you can just assume that it's already moving).

    I also like haruspex's response.
     
  15. Oct 27, 2013 #14
    Thanks for the replies. I can't believe the professor gave us a trick question on an assessed problem set :devil:
     
  16. Oct 27, 2013 #15

    Simon Bridge

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    If the box is already moving at a steady speed, no additional work is needed to get it from A to B.
    But, since there's friction, don't you need to do some work to keep it moving?

    I think this is the main point where confusion can set in.
     
  17. Oct 27, 2013 #16

    haruspex

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    Yes, but it isn't.
    In the limit solution, the force is vertical and equal to the weight of the mass, so there's no normal force, so no friction. In the real world, of course, surfaces are three dimensional and springy. A small amount of work would be needed to lift the mass clear of the surface.
     
  18. Oct 27, 2013 #17

    Simon Bridge

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    ....... no, it isn't.
    The question is quite open about the strategy involved in getting the block from A to B.
    However, the initial analysis looks like it assumes a constant velocity.

    Accepting the idealization yeah... if ##F = mg## then you can get minimum work just lifting and walking along with the block however slowly from the lazy nudge given to get it moving.

    This does not take into account the muscle-strain in holding the block up - but we don't seem to be asked to.

    For ##F>mg## then you can accelerate the box as well - that does not sound all that lazy though since this means that some work goes into extra kinetic energy.

    For ##F<mg## then you still have some friction - so you need a component of F in the direction of motion to keep the box moving right?

    I've been reading things that F can be adjusted as well as the angle.


    Thing is, I don't think it's a trick question - just very open-ended.
    There's two parts that interlock - what needs to be considered is what strategy to take to get the block from A to B.

    IRL the limiting factor is usually the rate that work can be done (the power) rather than the amount of work itself. A lazy person may do a lot of work slowly so there's little effort involved.
    That changes the strategy employed and the way students will intuitively think about work.
     
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