Calculate distribution of mass

  • Thread starter Mikey-D
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  • #1
11
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this post also appears in the mechanics physics section, but thought I may get some results here...

I asked a similar question a while back, but am still unclear on something.

If I have some arbitrary object (a box, for instance) with known centre of mass that is held up at four (or more) planar points, is it possible to calculate exactly what portion of the mass of the box is at each of the points?

I know it can be done with three points (using three equations: sum of forces, and both components of the sum of torques about an arbitrary point), but am unsure whether it is even possible with more than three points. Seems to me it must be possible to calculate, but my searching has turned up nothing thus far.

Can anyone point me in the right direction here??

Thanks a lot!

Edit: Assume zero elasticity in the supports and object.
 

Answers and Replies

  • #2
83
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you have 6 equations to work with. its not torque but a moment
sum of moments in the one direction equal zero. Should be solvable with only these. regardless of how many so called planar points your have
 
  • #3
11
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I found a workaround, so no longer need to make these calculations, but I'm still curious.

I'm not sure I follow...What six equations are there, and even with six equations how can we solve for more than six points. I should also note that all forces are in the vertical direction. So our sum of forces equation has no horizontal (x- and y-components). Given that all forces are exerted at on a plane, perpendicular to that plane, all components of torque (moment of force?) are also in the z-direction.
 
  • #4
7
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I found a workaround, so no longer need to make these calculations, but I'm still curious.

I'm not sure I follow...What six equations are there, and even with six equations how can we solve for more than six points. I should also note that all forces are in the vertical direction. So our sum of forces equation has no horizontal (x- and y-components). Given that all forces are exerted at on a plane, perpendicular to that plane, all components of torque (moment of force?) are also in the z-direction.

The problem is statically indeterminate. The six equations are the six equilibrium equations: 3 force directions (in x, y, z) and 3 moment directions (about x,y,z).

If, as you suggest, we assume that the reactions are pinned rollers then the number of equilibrium directions reduces to three: 1 vertical direction (z) and 2 moment directions (x, y). (Your statement about all components of torque (moment) being in the z-direction is incorrect.) We can therefore solve for the case where we only have three supports.

To solve for the case where we have four or more supports we need to assume a deflected shape of the body. This will generate another equation at each of the supports assuming there is no vertical deflection at the supports.
 

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