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Finding Magnitude of Unkown Force on Rough Horizontal Surface

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data
    A block of mass 100g is at rest on a rough horizontal plane. It has a massless rope attached to one end inclined at 20° above the horizontal. When the tension in the string is 0.5N the object is found to be in limiting equilibrium.

    Part A: Find the coeffecient of static friction between the object and the plane.

    Part B: What would the tension in the string have to be to make the object accelerate at 1.5m/s^2.


    2. Relevant equations
    Fnet=ma
    R=mg (R is the normal reaction force)


    3. The attempt at a solution
    For clarification, F_f is the frictional force. and P is the unkown foce in part B.

    Part A:
    [itex]
    0.5cos20=0.47\\
    0.5sin20=0.17\\
    mg=R+0.17\\
    0.98=R+0.17\\
    ∴ R=0.98-0.17=0.81\\

    F_f=0.5cos20 (because \space the \space force \space applied\space by\space the\space string\space to\space the\space horizontal \space must \space equal \space the \space frictional \space force)\\
    ∴F_f=0.47\\
    μR=0.47\\
    μ0.81=0.47\\
    ∴ μ \frac{0.47}{0.81}=0.58 \\
    [/itex]

    Part B:
    [itex]
    R+Psin20=mg \\
    R+Psin20=0.98 \\
    ∴ R=0.98-0.34P \\
    \\
    Pcos20-F_f=ma\\
    Pcos20-μR=0.15\\
    ∴0.94P=0.15+0.58R\\
    0.94P=0.15+0.58(0.98-0.34P)\\
    0.94P=0.15+0.57-0.2P\\
    1.14P=0.15+0.57\\
    1.14P=0.72\\
    ∴ P=\frac{0.72}{1.14}=0.63N \\
    [/itex]

    My main issue is if I have done it right because the answer to Part B seems rather low compared to the force when it is in limiting equilibrium i.e. only 0.13N more.

    Any help/pointers is appreciated.
     
    Last edited: Feb 13, 2013
  2. jcsd
  3. Feb 13, 2013 #2

    haruspex

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    Hi Faradazed,
    As a matter of technique, it's a really good idea to do most of the work purely in algebra, using symbols even for values that are known. Then plug all the numbers in at the end. It's much easier to spot mistakes, and less prone to silly errors.
    I assume you meant Ff = .5 cos 20
    The question is not quite kosher here. It should tell you the dynamic friction.
    You know it will be roughly .15N more, since that would be the force to accelerate it that fast with no gravity or friction. You may be surprised that it's a bit less than .15. That's because some of the extra force goes into reducing the normal force, and hence the friction. That 'frees up' some of the 0.5N to produce acceleration. In general, there is an optimal angle at which to drag a load.
     
  4. Feb 13, 2013 #3
    Hi, thanks for replying :)

    Yes it is a habit I need to get out of, my math teacher has commented on it, I am trying to improve upon this :)

    Yes I think this was a copying error from my written work to typing it onto the computer
    Is that the same as kinetic friction? We have only been taught about static friction so I assume all problems given are meant to act like as if kinetic friction is nonexistent.
    I see. Does what I have done (apart from not writing it out in an algebraic way to start with) look OK to you then?
     
  5. Feb 13, 2013 #4

    haruspex

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    Yes, kinetic is the usual word, sorry.
    I would think the default assumption is that it's the same as the static friction. That seems to be what you have to assume here.
    Yes, it's fine.
     
  6. Feb 13, 2013 #5
    OK, many thanks :)
     
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