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Thread: Statics help: Rigid body equilibrium

  1. #1
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    Statics help: Rigid body equilibrium

    Got a problem that i have no idea on how to start working at. Idea is that im supposed to figure out the lenght of h with respect to b so the water can flow. Any tips? Thanks in advance.

    Image of problem:

    pressure-problem.gif

  2. #2
    Administrator Kelly Bramble's Avatar
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    The pressure along length h needs to be greater than the pressure along length b unless there is an external force (torque) applied to the mechanism.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  3. #3
    Administrator Kelly Bramble's Avatar
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    For b, see Hydrostatic Pressure vs Depth -> https://www.engineersedge.com/fluid_...c-pressure.htm

    Also,

    https://www.engineersedge.com/fluid_...h_pressure.htm

    The pressure is variable along h, increasing as the depth becomes greater. You are going to need the width of the perspective lengths b and h to complete your calculations.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  4. #4
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    Yes, the principle of the solution is clear, but the description of the problem is stated in the original post. What im saying is that there are no perspective lenghts and the solution is supposed to be sqrt3 *b, and i have come to a dissapointing conclusion that the answer lies somewhere in a right triangle where b=1 and the angles respectively 60 30 and 90 degrees. Dissapointing because i would like to know why. Might be wrong tho. does anything come to mind after reading this?



    Quote Originally Posted by Kelly Bramble View Post
    For b, see Hydrostatic Pressure vs Depth -> https://www.engineersedge.com/fluid_...c-pressure.htm

    Also,

    https://www.engineersedge.com/fluid_...h_pressure.htm

    The pressure is variable along h, increasing as the depth becomes greater. You are going to need the width of the perspective lengths b and h to complete your calculations.

  5. #5
    Technical Fellow jboggs's Avatar
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    First - is this homework? Sure looks like it to me. If that is the case, there are some forum rules to which Kelly can refer you.

    Second - if this is indeed homework, then my compliments go out to the professor that dreamed it up. In one example he is requiring an understanding of how fluid pressure works in an open volume, how to analyze forces and sum moments, and how to analyze rigid bodies, as well as the understanding of math, variable, and solution of multiple equations. He or she is to be congratulated.

    Third - despite all that, your analysis must begin by just staring at the diagram and thinking about some details for a few minutes before you try to write down anything. You must understand the mechanics involved. The lever will move only when the sum of the moments does not equal zero. What moments? The moments created by forces acting in certain directions and along certain vectors. Let's assume a positive moment tends to rotate the body counterclockwise and a negative moment rotates it clockwise. Or put another way, a positive moment closes the bottom and a negative moment opens it.

    So, your mission is to resolve the question of what ratio of h/b will result in a total negative moment on the body? (Hint: when I don't know what the effect will be of missing variable, like the depth or width of the tank, I just make some up, figure it out and figure out how to apply them in the equations.)

    Your task also involves an understanding the fluid pressures on bottoms and sides of tanks.

    As for the side wall, you know that fluid pressure in an open volume is directly related to depth. The pressure at 2 ft below the surface is twice the pressure at 1 ft. That means that pressure on the side wall varies from zero at the top to a maximum at the bottom. That means the resultant force from that pressure will also vary linearly along the wall. Since it varies linearly the total force acting on the wall will be the same as if half of the maximum pressure was acting on the whole wall surface. If that total force were applied as a point load rather a distributed load, where would it be applied? Where would the effective center of that linearly varying load be? That line and its distance from the pivot point determines the moment arm for the force acting on the wall.

    As for the bottom surface, you know that pressure acts equally in all directions. That means that the pressure acting on a horizontal flat bottom does not vary with position. It is the same as the maximum pressure at the bottom of the side wall, and it is applied equally to the entire bottom surface. Equally distributed. You know how to determine the total force on the bottom, and the position of the equivalent point force relative to the pivot point. (Hint: 1/2b)

    So the moment resulting from the force acting on the bottom surface can be derived from: (max pressure * b) * 1/2b.

    A similar statement can be made for the side wall moment, accounting for its linear variation.

    I know there is a number, a ratio, at which the lever will flip. How do I know? Simple. Assume b = 100 and h = 1. Is it going to open up? I doubt it. Want verification? Crank out the math. Now, assume b = 1 and h = 100. I think it will open up. Do you?

  6. #6
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    Quote Originally Posted by jboggs View Post
    First - is this homework? Sure looks like it to me. If that is the case, there are some forum rules to which Kelly can refer you.

    Second - if this is indeed homework, then my compliments go out to the professor that dreamed it up. In one example he is requiring an understanding of how fluid pressure works in an open volume, how to analyze forces and sum moments, and how to analyze rigid bodies, as well as the understanding of math, variable, and solution of multiple equations. He or she is to be congratulated.

    Third - despite all that, your analysis must begin by just staring at the diagram and thinking about some details for a few minutes before you try to write down anything. You must understand the mechanics involved. The lever will move only when the sum of the moments does not equal zero. What moments? The moments created by forces acting in certain directions and along certain vectors. Let's assume a positive moment tends to rotate the body counterclockwise and a negative moment rotates it clockwise. Or put another way, a positive moment closes the bottom and a negative moment opens it.

    So, your mission is to resolve the question of what ratio of h/b will result in a total negative moment on the body? (Hint: when I don't know what the effect will be of missing variable, like the depth or width of the tank, I just make some up, figure it out and figure out how to apply them in the equations.)

    Your task also involves an understanding the fluid pressures on bottoms and sides of tanks.

    As for the side wall, you know that fluid pressure in an open volume is directly related to depth. The pressure at 2 ft below the surface is twice the pressure at 1 ft. That means that pressure on the side wall varies from zero at the top to a maximum at the bottom. That means the resultant force from that pressure will also vary linearly along the wall. Since it varies linearly the total force acting on the wall will be the same as if half of the maximum pressure was acting on the whole wall surface. If that total force were applied as a point load rather a distributed load, where would it be applied? Where would the effective center of that linearly varying load be? That line and its distance from the pivot point determines the moment arm for the force acting on the wall.

    As for the bottom surface, you know that pressure acts equally in all directions. That means that the pressure acting on a horizontal flat bottom does not vary with position. It is the same as the maximum pressure at the bottom of the side wall, and it is applied equally to the entire bottom surface. Equally distributed. You know how to determine the total force on the bottom, and the position of the equivalent point force relative to the pivot point. (Hint: 1/2b)

    So the moment resulting from the force acting on the bottom surface can be derived from: (max pressure * b) * 1/2b.

    A similar statement can be made for the side wall moment, accounting for its linear variation.

    I know there is a number, a ratio, at which the lever will flip. How do I know? Simple. Assume b = 100 and h = 1. Is it going to open up? I doubt it. Want verification? Crank out the math. Now, assume b = 1 and h = 100. I think it will open up. Do you?
    Thank you very much. There are no words i can write to express my gratitude, so i only hope that thanks will do.

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