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Math Help - Unsolvable Fricton Problem

  1. #16
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    Quote Originally Posted by topsquark View Post
    Physics problem solving 101:
    It is a civil engineering class.

    Actually, I've noted Mathematicians using this one too. They simply redefine the problem into one they know how to solve.
    I think I know what you are talking about. For example, there is a famous problem called the 4 colour problem. It states that any region that has been decomposed into smaller regions can be coloured with at most 4 colours such that no two adjacent ones has the same color. When mathemations where working (or worked) on this problem they completely changed the problem. Because the original problem was not mathematical enough, I am myself not sure how you even begin to define something like that but that is what they did.

    Physicists, on the other hand, tend to use "spherical horses," as the joke goes.)
    I know that joke too.
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  2. #17
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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 P were acting parallel to the plane this problem would be relatively easy...since an angle x is involved too, maybe the solution is to find an expression for the angle?? cos(X)=P??
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  3. #18
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by anthmoo View Post
    If P were acting parallel to the plane this problem would be relatively easy...since an angle x is involved too, maybe the solution is to find an expression for the angle?? cos(X)=P??
    In that case the angle X is zero so we have a specified angle and can solve the problem. This turns out to be what the professor was after. But we can do the problem if X is any specified angle.

    -Dan
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  4. #19
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    Quote Originally Posted by topsquark View Post
    This turns out to be what the professor was after.
    If you want to know what happened. The professor apologized directly to me in front of the whole class saying , but he also said the problem was not solvable. Different editions of the book had the problem said differently. (I miss that class).
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  5. #20
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If you want to know what happened. The professor apologized directly to me in front of the whole class saying , but he also said the problem was not solvable. Different editions of the book had the problem said differently. (I miss that class).
    It's nice to have a professor that will do that rather than trying to create a cloud of confusion to cover his/her error. I haven't run into a lot that would apologize for a mistake.

    -Dan
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  6. #21
    Senior Member TriKri's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    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".
    But I don't understand how that can be right, if what I have calculated is true,
    P_y = W + F/ \mu _s - \frac{1}{\mu_s} \cdot P_x
    the angle wouldn't be zero when the force is minimixed. If the force shall be minimixed, \sqrt{P_x^2\ +\ P_y^2} should be minimized. The equation abowe describes a line sloping downwards, crossing the P_x axis when P_x is W + F/\mu_s. Since the force should be in a right angle to this line, the tangent of the angle should equal \mu_s if I have got this rigt.
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  7. #22
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    Quote Originally Posted by TriKri View Post
    But I don't understand how that can be right, if what I have calculated is true,
    P_y = W + F/ \mu _s - \frac{1}{\mu_s} \cdot P_x
    the angle wouldn't be zero when the force is minimixed. If the force shall be minimixed, \sqrt{P_x^2\ +\ P_y^2} should be minimized. The equation abowe describes a line sloping downwards, crossing the P_x axis when P_x is W + F/\mu_s. Since the force should be in a right angle to this line, the tangent of the angle should equal \mu_s if I have got this rigt.
    What you can do is express the force equation in terms of the angle. And then treat it as a Calculus optimization problem. Meaning find derivative make it zero, it will turn out the angle is zero.
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  8. #23
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    Quote Originally Posted by ThePerfectHacker View Post
    What you can do is express the force equation in terms of the angle. And then treat it as a Calculus optimization problem. Meaning find derivative make it zero, it will turn out the angle is zero.
    Ok I will have to check that to see what's wrong in my previous calculation... :/
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  9. #24
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    Quote Originally Posted by ThePerfectHacker View Post
    What you can do is express the force equation in terms of the angle. And then treat it as a Calculus optimization problem. Meaning find derivative make it zero, it will turn out the angle is zero.
    I read this thread this morning before I went to work. No time to post a reply here, I waited until now that I'm home again, some 11 hours after I read this. But I made calculations at the jobsite. And I found a "P" with an "x". Meaning, the minimum P is not when angle x is zero.

    No time still to post the soulution---I type really slow, because I am a one-finger typist---I will do it in maybe 2 hours from now. [Household chores first.]
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  10. #25
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    Quote Originally Posted by ticbol View Post
    I read this thread this morning before I went to work. No time to post a reply here, I waited until now that I'm home again, some 11 hours after I read this. But I made calculations at the jobsite. And I found a "P" with an "x". Meaning, the minimum P is not when angle x is zero.

    No time still to post the soulution---I type really slow, because I am a one-finger typist---I will do it in maybe 2 hours from now. [Household chores first.]
    Uh-oh.

    I am very sorry I just found out now that I really made a mistake in my fast calculations this morning. I said I found an answer where x is not zero, but I had a nagging question in my mind almost all day long---because my reasoning without computations said for minimum P, the angle x cannot be more than zero.

    Without P, the block will slide down the incline by a force of [60sin(30deg) -(0.25)(60cos(30deg)] = [30 -13] = 17 lbs.
    So with P, and angle x is zero, P = 17 lbs going up the incline to stay the block.
    If the P is raised at an angle x with the incline, then the total normal force on the incline will be decreased by P*sin(x). Hence the friction will be lesser than 13 lbs. So the sliding force will increase, will be more than 17 lbs. So the component of P parallel to the incline will be more than 17 lbs also to stay the block. Thus, the P, which is be larger than this parallel component, will be even more than 17 lbs. Therefore there is no way that P will be less than 17 lbs if angle x is more than zero. That is why for minimum P, the angle x must be zero.

    That was what I had in mind while driving to the jobsite. But to be sure, when I had a chance, I did a quick calculation, and I found a P less than 17 lbs with an angle x greater than zero. I said, "Wow, really?" No time to check the computations, I was a little bothered by the written answer the whole day. Until a while ago before I finally had time to sit down before this laptop.

    Then I again quickly solved the problem on a sratch paper. And there I saw what went wrong this morning.
    Along the incline,
    (0.25)[51.96 -P*sin(x)] +P*cos(x) = 30
    -(0.25)P*sin(x) +P*cos(x) = 30 -12.99
    P[(0.25)sin(x) +cos(x)] = 17 --------------there was the mistake.
    The (0.25)sin(x) should have been negative!

    Of course, if it were negative, then
    P[-(0.25)sin(x) +cos(x)] = 17 ----this will lead to an optimization:
    P = [17] / [sqrt(1 -sin^2(x)) -(0.25)sin(x)] -----(i)
    Get the derivative of P with respect to sin(x), then equate that to zero.
    To avoid confusion, let s = sin(x),
    P = 17 / [sqrt(1 -s^2) -(0.25)s]
    dP/ds = 0 will lead to
    -s/[sqrt(1 -s^2) -0.25 = 0
    -s = (0.25)sqrt(1 -s^2) -------------and here I could have found the mistake. And I should have not written in this thread here a while ago as I arrived home that I found a P and an x.

    Oh, well, sorry again.
    It always pay to review your computations first before stopping in your calculations. , .
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