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Math Help - Unsolvable Fricton Problem

  1. #1
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    Unsolvable Fricton Problem

    How do I solve this,

    "Given that the static fricton is \mu_s =.25 between the object weighing 60 lbs and the inclined plane. Find the force P and angle x such that motion is impending (about to happen).

    (Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

    I resolve everything into component form.
    \vec{W}=<0,-60>
    \vec{F}=<-.87F,-.5F>
    \vec{N}=<-.5N,.87N>
    \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>
    We also know that since motion is about to happen,
    F=\mu_s N=.25N
    But that is not sufficent to solve this problem.
    Attached Thumbnails Attached Thumbnails Unsolvable Fricton Problem-picture14.gif   Unsolvable Fricton Problem-picture15.gif  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    How do I solve this,

    "Given that the static fricton is \mu_s =.25 between the object weighing 60 lbs and the inclined plane. Find the force P and angle x such that motion is impending (about to happen).

    (Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

    I resolve everything into component form.
    \vec{W}=<0,-60>
    \vec{F}=<-.87F,-.5F>
    \vec{N}=<-.5N,.87N>
    \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>
    We also know that since motion is about to happen,
    F=\mu_s N=.25N
    But that is not sufficent to solve this problem.
    Resolve everything into components normal to the plane and parallel to
    the plane (then ignore the components normal to the plane)

    To me this looks like it leaves you with the equation if the block is about
    to move up the plane (frictional force act down the plane):

    P*cos(x)-mu*sqrt(3)*30g-30g=0

    and if about to move down the plane (friction acts up the plane):

    P*cos(x)+mu*sqrt(3)*30g-30g=0

    (you will have to check that these are the equations that you get
    also check that you dont lift the block off of the plane).

    RonL
    Last edited by CaptainBlack; November 21st 2006 at 10:57 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How do I solve this,

    "Given that the static fricton is \mu_s =.25 between the object weighing 60 lbs and the inclined plane. Find the force P and angle x such that motion is impending (about to happen).

    (Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

    I resolve everything into component form.
    \vec{W}=<0,-60>
    \vec{F}=<-.87F,-.5F>
    \vec{N}=<-.5N,.87N>
    \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>
    We also know that since motion is about to happen,
    F=\mu_s N=.25N
    But that is not sufficent to solve this problem.
    Your second diagram troubles me. Either you rotated by 30 degrees and have the direction of the weight incorrect, or you didn't rotate anything and have the directions of the normal force and friction incorrect.

    Just to be clear: the weight vector is always directed toward the center of the Earth. The normal force is always perpendicular to the plane formed by the two surfaces that meet. The friction force is always parallel to the plane formed by the two surfaces that meet.

    I would highly recommend CaptainBlack's coordinate system, but you can use any you like. Using what you appear to be using, for example:
    \sum F_{horizontal} = Fcos(30) + Nsin(30) - Pcos(x + 30) = 0
    \sum F_{vertical} = -W - Fsin(30) + Ncos(30) + Psin(x + 30) = 0

    Given that we are looking for the maximum static friction force we have (as you correctly stated)
    F = \mu N

    We now have three equations in 4 unknowns (P, x, F, and N) so I agree, you can't solve this system.

    -Dan
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    Forum Admin topsquark's Avatar
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    I have posted what I laughingly call a diagram. I hope it comes out clear!

    -Dan

    (Sigh) I blew the image size up by 300% and it looks small as ever. Anyway, the force pointing down is W, the force down the slope is F, the force up and to the right is P and the force up and to the left is N. (I gotta get a new scanner.)
    Attached Thumbnails Attached Thumbnails Unsolvable Fricton Problem-force-diagram.jpg  
    Last edited by topsquark; November 21st 2006 at 12:39 PM. Reason: Pitcher too small. Waaa!
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  5. #5
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    So maybe I copied the diagram incorrectly from my text. See what my Professor says tomorrow. If he says there is a solution I will tell him a physics professor told me it cannot be solved.
    -------------------------------------------
    I had class today and my professor had to apologize for giving an unsolvable problem. His edition stated (the one I used) how it is worded. It should have said "What is the minimum value of the force...". He wanted us to solve an optimization problem.
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  6. #6
    Senior Member TriKri's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How do I solve this,

    "Given that the static fricton is \mu_s =.25 between the object weighing 60 lbs and the inclined plane. Find the force P and angle x such that motion is impending (about to happen).

    (Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

    I resolve everything into component form.
    \vec{W}=<0,-60>
    \vec{F}=<-.87F,-.5F>
    \vec{N}=<-.5N,.87N>
    \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>
    We also know that since motion is about to happen,
    F=\mu_s N=.25N
    But that is not sufficent to solve this problem.
    If the angle of the plane is \alpha,
    mg = 60\ lbs \cdot g
    F = sin(\alpha) \cdot mg
    W = cos(\alpha) \cdot mg
    N is the normal force
    f is the friction
    f = \mu_s \cdot N\ \Rightarrow\ N = f/\mu_s

    Then P_x = P \cdot cos(x)
    P_y = P \cdot sin(x)
    Then F + f = P_x\ \Rightarrow\ f = P_x - F
    N = W - P_y \ \Rightarrow \ P_y = W - N
    P_y = W - f/\mu_s
    P_y = W + (F - P_x)/\mu_s
    P_y = W + F/ \mu _s - \frac{1}{\mu_s} \cdot P_x

    I didn't really understand the question, but isn't this an approach to anything?
    Last edited by CaptainBlack; November 22nd 2006 at 08:41 PM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    So maybe I copied the diagram incorrectly from my text. See what my Professor says tomorrow. If he says there is a solution I will tell him a physics professor told me it cannot be solved.
    -------------------------------------------
    I had class today and my professor had to apologize for giving an unsolvable problem. His edition stated (the one I used) how it is worded. It should have said "What is the minimum value of the force...". He wanted us to solve an optimization problem.
    Now that makes more sense.

    I was thinking about this again. There is a calculable value for the component of P along the plane, but there is a range of possible values for the perpendicular component. In the terms of the problem that you should have been given, then the angle x = 0 to give the smallest P.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    If the angle of the plane is \alpha,
    mg = 60\ lbs \cdot g
    F = sin(\alpha) \cdot mg
    W = cos(\alpha) \cdot mg
    N is the normal force
    f is the friction
    f = \mu_s \cdot N\ \Rightarrow\ N = f/\mu_s

    Then P_x = P \cdot cos(x)
    P_y = P \cdot sin(x)
    Then F + f = P_x\ \Rightarrow\ f = P_x - F
    N = W - P_y \ \Rightarrow \ P_y = W - N
    P_y = W - f/\mu_s
    P_y = W + (F - P_x)/\mu_s
    P_y = W + F/\mu_s - \frac{1}{\mu_s} \cdot P_x

    I didn't really understand the question, but isn't this an approach to anything?
    That depends.

    First, the weight is 60 lbs. That is NOT m, but W. "lbs" is the unit of force in the British system of units, not mass which is usually taken to be in "slugs" if I remember correctly.

    Second, W = mg by definition, not mg*cos(alpha). The weight component along the plane has this magnitude. Is this what you meant?

    Finally, what is F? ThePerfectHacker was using this symbol for the static friction force and you are using f for that purpose.

    I'm not saying your approach is wrong, I'm saying I don't understand what you are trying to do with it.

    -Dan
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  9. #9
    Grand Panjandrum
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    [quote=topsquark]

    That depends.

    First, the weight is 60 lbs. That is NOT m, but W. "lbs" is the unit of force in the British system of units, not mass which is usually taken to be in "slugs" if I remember correctly.[quote]

    Wonderfully system, you have two options pounds are mass and poundals are
    force, or pounds are force and slugs are mass. My experience is that the
    first of these is more commonly used (in a bizarre usage of the term common
    - we never use imperial or customary units over here any more).

    I must admit that I always want to write a tirade about the use of customary
    units in maths or science homework whenever I see it, telling the student to
    go back to their professor/lecturer/teacher and telling them off on my behalf.

    RonL
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Wonderfully system, you have two options pounds are mass and poundals are
    force, or pounds are force and slugs are mass. My experience is that the
    first of these is more commonly used (in a bizarre usage of the term common
    - we never use imperial or customary units over here any more).

    I must admit that I always want to write a tirade about the use of customary
    units in maths or science homework whenever I see it, telling the student to
    go back to their professor/lecturer/teacher and telling them off on my behalf.

    RonL
    (Shakes his head in despair) Never heard of "lb - mass" before! I have heard of poundals, though, so I should have suspected it.

    I suppose that also gives a reason for the (mis)usage of "kg" as a unit of weight.

    -Dan
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    Senior Member TriKri's Avatar
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    > Wonderfully system
    Thank you CaptainBlack!

    Haha, ok maybe I missunderstood ThePerfectHacker's use of variables names, this is how I imagined it:

    Hope that makes any sense

    And by the way, by setting out the forces in the system I meant their absolute values and nothing else. Maybe I should have stated that.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    > Wonderfully system
    Thank you CaptainBlack!

    Haha, ok maybe I missunderstood ThePerfectHacker's use of variables names, this is how I imagined it:

    Hope that makes any sense

    And by the way, by setting out the forces in the system I meant their absolute values and nothing else. Maybe I should have stated that.
    Gotcha. Yes, that is valid. We still have the problem, though, that we have 4 unknowns (Px, Py, f, and N) and only 3 independant equations to work with.

    -Dan
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    Senior Member TriKri's Avatar
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    Maybe |P| shall be minimized? I agree, the problem statement is a little bit fuzzy.
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    Quote Originally Posted by TriKri View Post
    Maybe |P| shall be minimized? I agree, the problem statement is a little bit fuzzy.
    It turns out that the force is minimized when the angle is minimized. One of my classmates said he solved the problem. When I asked him how, he simply said, "I ignored the angle".
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  15. #15
    Forum Admin topsquark's Avatar
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    Physics problem solving 101:

    When you can't solve the problem, change it to something that can be solved. (Actually, I've noted Mathematicians using this one too. They simply redefine the problem into one they know how to solve. Physicists, on the other hand, tend to use "spherical horses," as the joke goes.)

    -Dan
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