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Tilting force on a frame

  1. Jan 8, 2009 #1
    I am trying to find the force necessary to tilt a vertical frame.

    I have a triangular frame, which is standing on one of its sides vertically. I am applying a force or around 5 kg's in the horizantal direction on its upper corner. how do i design the frame so that it will not tilt under this force? what should the angles, and the base width be for optimum design.

    i have a hunch that an equilateral triangle will be the best disign for anti-tilt frame. Please correct me if i am wrong.

    The weight of frame will me around 1 kg and the height 20 to 30 cm.

    (diagram attached )
     

    Attached Files:

    Last edited: Jan 8, 2009
  2. jcsd
  3. Jan 9, 2009 #2
    This is basicaly a geometry problem. The condition for not tiliting can established by observation of the torques. Now we dont want it to tilt over the intersection point of sides B and C.
    Lets figure out the torques on the system. There is a the torque resulting from the gravity acting on the three sides of the frame. This effect on the three sides can be considered as a single force acting in the center of mass of the frame, in this case the centroid or geometric center of the triangle. Denote the sum of the masses of the indivual sides of the frame as M.
    The force acting on the top point as F. The side lengths as a,b,c (as on the diagram) Then the condition for not tilting is:

    [tex]Mg\cdot r_s \geq F\cdot H [/tex]

    where [tex]r_s[/tex] is the "hand" of the force with respect to the tilting point.

    Now we only need to calculate r_s. This is:

    [tex]r_s = \frac23 s_a \cos\theta [/tex]

    Where s_a is the length of the line connecting the tilting point and the center of the side a. [tex]\theta [/tex] is the angle enclosed by s_a and c.
    Using the cosine law and the geometry of the triangle (If you need I can write down explicitly, but this really is just geometry.. :D) s_a and \theta can be calculated, so we have for r_s:

    [tex]r_s= \frac{3c^2+b^2-a^2}{6}[/tex]

    Plugging this into the torque inequality and rearranging we have for the side c:

    [tex]c \geq \sqrt{\frac{2FH}{M}+\frac{a^2-b^2}{3}}[/tex]

    So this is how you have to choose c, so that the frame doesnt tilt.
     
  4. Jan 9, 2009 #3
    Thank you very much for the precise answer.

    If i keep the frame as right angle triangle, with a=H=height, and angles as 90, 45 and 45 degrees, then b becomes the hypoteneus=28, with M=1 kg, F=3 Kg, the equation goes negative. i guess i am wrong somewhere...
     
    Last edited: Jan 9, 2009
  5. Jan 10, 2009 #4
    The unit for force is Newtons. So I guess under a force of 3kg, you mean weight of 3kg, this means that: F=3kg*10m/s^2 = 30N

    And your equation wont go negative.

    But if you say you want a right angled triangle with 45 45 90, and you know the height, then you already determine all of the sides... :D
     
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