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Math Help - uniform plank walk

  1. #1
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    uniform plank walk

    A uniform plank of length 4.9 m and weight 210 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 453 N walk on the overhanging part of the plank before it just begins to tip?
    in meters
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by rcmango View Post
    A uniform plank of length 4.9 m and weight 210 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 453 N walk on the overhanging part of the plank before it just begins to tip?
    in meters
    (since no one answers it yet and it is placed in the Urgent HW, i'll try it..)
    i can't imagine the drawing.. maybe something like this..

    Code:
     o                      o
     o                      o
     o                      o
    ''''''''''''''''''''''''''''''''''''''''''
    do you have the answer.. actually, i got 0.6258 m from the right support.. well, not sure on this that is why i don't put the solution.. can i see what you're doing?
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  3. #3
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    Your answer is correct, I could not figure this one out.

    looks alot like this, plank - View Image - Post Your Image

    can you please explain how i am supposed to set this up.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by rcmango View Post
    Your answer is correct, I could not figure this one out.

    looks alot like this, plank - View Image - Post Your Image

    can you please explain how i am supposed to set this up.

    good.. now, since the plank has uniform weight w_p, let's assume that it is concentrated at the middle of the plank..

    so, if the length of the plank is L, then the weight is located at \frac{L}{2}..

    the distance of the left support from the middle of the plank is \frac{L}{2} - 1.1m..

    now, if w_{man} is the weight of the man, then the product of w_p and \frac{L}{2} - 1.1m (the distance of the middle of the plank from the right support) must be equal to the product of w_{man} and the distance, x, of the man from the right support..

    that is, w_p \cdot \left( \frac{L}{2} - 1.1 \right) = x \cdot w_{man}..

    now, substitute the values..
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    Your answer is correct, I could not figure this one out.

    looks alot like this, plank - View Image - Post Your Image

    can you please explain how i am supposed to set this up.
    This is just going to mimic kalagota's response, but let me put it in "Physics" terms for you.

    The plank is not rotating, so it is in static equilibrium. This denotes not only a = 0 for the system, but also means the system doesn't rotate.

    So let's pick a coordinate system where +y is upward. According to the diagram we have a force S_L by the left support acting straight upward, a force S_R by the support on the right acting straight upward. Both of the support forces act on the board at the "point" of the support. We also have the man's weight w_m acting straight down from where the man is standing (x meters to the right of the right support), and a weight w_p of the plank acting straight down from the geometric center of the plank.

    Newton's 2nd tells us
    \sum F_y = S_L + S_R - w_m - w_p = 0

    or
    S_L + S_R - 453 - 210 = 0

    Now, we have two unknown forces and one unknown x. So we need two more equations.

    Let's pick an axis of rotation at the point where the support on the right meets the plank and call this point A. The system is not rotating about point A, so
    \sum \tau _A = 0

    \sum \tau _A = -S_L(4.9 - 1.1) - w_p(2.45 - 1.1) + w_mx = 0
    (Recall that a counterclockwise torque is negative by default.)

    -3.8S_L - 283.5 + 453x = 0

    We still need one more equation. The nice thing about static equilibrium is that the system doesn't rotate about any point, so we are free to pick whatever axis of rotation we desire, whether it is a physical axis of rotation or not. So I'm going to pick another axis of rotation B located where the support on the left meets the plank. Thus
    \sum \tau _B = w_p(2.45) - S_R(4.9 - 1.1) + w_m(4.9 - 1.1 + x) = 0

    or
    514.5 - 3.8S_R + 453(3.8 + x) = 0

    So we now have three equations in three unknowns:
    S_L + S_R - 453 - 210 = 0
    and
    -3.8S_L - 283.5 + 453x = 0
    and
    514.5 - 3.8S_R + 453(3.8 + x) = 0

    Solving the system I get x = 0.625828~m

    -Dan
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  6. #6
    MHF Contributor kalagota's Avatar
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    you're really a physics maestro..
    i was about to that solution before but when i realized that there was in need to solve a system of equation, i became passive, so i tried the simplest way to solve it..

    (whoa, it made me remember topics of the first of the two physics course i had..)
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