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Simple statics problem?

  1. Dec 13, 2011 #1
    Hi PF

    I have a statics problem that i solve. BUT I’m insecure rather or not it is solved correctly.

    The mechanical system is shown in the link below

    A weight less wire is supsended from the wall in point C and attached to a heavy bar at point A. The other end of the bar is attached to the wall but can move in the vertical direction.

    The weight of the bar is F_g. All other dimensions known are given in the figure.

    The aim is to find the force P required to push the bar down, so that the bar is in a horizontal position.

    My solution:
    First, the outer forces are noted

    A force balance in the x direction gives

    In the y diection

    Moment balance around point C, anti clockwise being positive gives,

    R_bx=F_g*(h-sin(alpha)*l*1/2)/ (h-sin(alpha)*l) .. (eq1)

    The main system is cut into two subsystems as shown in the link

    A force balance on the wire gives

    And then a force balance on the bar.
    in the x direction:
    In the y direction:
    And moment around point A gives:

    P= (-F_g*1/2*l*cos(alpha)+R_bx*l*sin(alpha))/(l*cos(alpha)) .. (eq2)

    Using (eq1) in (eq2) gives an expression for P as:
    P= -(1/2)*F_g*(-cos(alpha)*h+cos(alpha)*sin(alpha)*l+2*sin(alpha)*h-sin(alpha)^2*l)/(cos(alpha)*(-h+sin(alpha)*l))

    Inserting reasonable values for h, l and alpha and F_g in the expression for P also gives reasonable results. However, I'm very much in doubt rather or not the moment in the main system is possible to do due to the boundary conditions (eq1). I on the other hand don’t see any solution

    Any help is appreciated.
  2. jcsd
  3. Dec 15, 2011 #2

    My main concern is if the moment balance marked as eq.1 is even possible with the boundary conditions.

  4. Dec 15, 2011 #3


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    The diagram is not clear, though I think I understand it. It seems a precariously unstable support and surely it is not be drawn to scale? I think angle CAB would need to be very small, ~10 degrees. But eqn 1 does seems okay to me. What don't you like about it?

    I think this sentence is not expressed correctly: "The other end of the bar is attached to the wall but can move in the vertical direction." I suspect that you worded it that way, and it's not from the textbook.
    Last edited: Dec 15, 2011
  5. Dec 15, 2011 #4


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    kbka: First, h is not a constant, nor a given known dimension; h changes whenever angle alpha changes. Are you sure you drew h correctly? Is h drawn this way in your text book? Or did you add h yourself?

    If h is correct, then it is merely an initial condition, h1, used with the initial condition alpha1, to determine cable AC length L2. And let's rename L to L1. Therefore, the known dimensions are L1, L2, and alpha. L1 and L2 are constants. And alpha is a variable.

    Your force summation equations (first two equations) look fine. Your moment balance equation (eq. 1) currently looks completely wrong. Try it again.

    We are not allowed to give you the relevant equations for your homework. We can only check your math.

    Label cable length L2, change L to L1, use L2, delete h, draw angle alpha more clearly on your diagrams, and try the moment summation again more carefully. There is no problem with the given problem statement wording.
  6. Dec 15, 2011 #5

    Thank you both for your replies

    First of all this isn't a homework problem. It's a simplified version of something I'm designing. long story.
    It was placed here in the homework section by an admin. I do, however, see that in my attempt to explain and simplify the problem, I made it sound like a homework problem from back in school :)

    The reason for choosing h, L and alpha are because these are design parameters. What I need is the absolut minimum force to push down the bar. As the force required is highest when alpha is small I only need to solve the boundary condition.

    NVN – you’re right it’s completely wrong. At least the math is. Can’t believe I messed up the trigonometry :/

    Alpha is the angle between the bar and the vertical wall the system is mounted on.

    So I’ve re-calculated eq1 to

    R_bx=F_g*(h-cos(alpha)*l*1/2)/ (h-cos(alpha)*l)

    Eq2 was also incorrect. It should be

    My concern about this problem lies in the moment balance due to the boundary condition in B(the rolling support) and the fact that forces are transferred to point C by wire.

    So my question basically is
    Can the force R_bx create a moment in point C?
    Can the force F_g create a moment in point C?

    My intuition says yes to both, but I want to make sure.

    Thank you,
  7. Dec 15, 2011 #6


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    kbka: The answer to your last two questions in post 5 is, yes to both.

    Eq. 2 is now correct; but you still got eq. 1 wrong. Perhaps one problem is you are not drawing and labeling the moment arms on your second free-body diagram (FBD); and so you are perhaps having trouble visualizing and computing the moment arms. Hint 1: The first term of eq. 1 is now correct; the second term of eq. 1 is incorrect. Try again.
    Last edited: Dec 15, 2011
  8. Dec 15, 2011 #7


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    Try doing this graphically, and then sort out the math.

    You have 3 forces acting on the bar. Two of them are
    1. The weight of the bar acting downwards at the mid point
    2. The tension in the wire, acting along the direction of the wire.

    Those two forces intersect at a point, which is at the mid point of the wire.

    Now take moments about that point. For equliibrium, the moment must be zero.
    so the thrird force, at the wall, must also pass through the mid point of the wire.

    The reaction at the wall has two components, the horizontal reaction from the moving support, and the vertical force P that you apply.

    Now you know the directions of all three forces, you can make a scale drawing of the triangle of forces to find their magnitudes - or calculate them with trig, if you prefer. But the drawings are useful, because they give you a picture of how the forces vary as the end of the bar moves along the wall.
  9. Dec 16, 2011 #8
    Hi again,

    thank you both. I see now that I'm confusing myself with all the trig. Messing up the distances.


  10. Dec 16, 2011 #9


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    kbka: That is now correct. You can now try developing an expression for P. By the way, if you are interested in solving for P when alpha = 90 deg, then you already intuitively know P = -0.5*F_g when alpha = 90 deg. (The reason for the negative sign is because you drew vector P downward, whereas P is upward when alpha = 90 deg.)
  11. Dec 16, 2011 #10


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    In the days before pocket calculators, everybody was taught how to solve this sort of problem by drawing (or "graphical statics", to make it sound more impressive). IMO it's a shame that people can now go through a whole engineering course without ever doing that. At the stage of designing something, you don't necessarily want super-accurate answers, you want to understand how the structure "works" and what happens if you change things. A rough drawing can often show you that faster than setting up equations and punching numbers into a calculator, or making a computer model with a CAD system.
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