#### Archived

This topic is now archived and is closed to further replies.

# Weight distribution question

This topic is 5286 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hello. I have a question about some simple force equations. Say you hav a table that measures 5x10 cm. It only has 90 degrees corners. You place a scale under each leg. All the scales show the same, say 100N. Then you add 100N to one of the corners, what will happen to all the scales? The table is totaly stiff and doesn't bend or anything. [edited by - Niklas2k2 on September 2, 2003 10:19:36 AM]

##### Share on other sites
See, that''s not really a simple question, you need to know the shape of the mass and do some integration to get the distributed mass of the object at each scale.

##### Share on other sites
Oki... I didn''t realise how complex the question wass. But any guidance or recomendation where to find help or what to "google" would på much appreciated =).

/Niklas

##### Share on other sites
I think that if we could exploit some symmetry of your problem, it could be easy to solve. We already have two constraints: sum of the forces is 0, and sum of the torques is 0. However, we have four unknows, which is the big problem. If your force was applied in the middle of an edge, I think that we could solve it, since the symmetry would eliminate two unknowns.

But as it is, I don''t think that there is anything we can do that doesn''t involve integrals, as jperalta said.

Cédric

##### Share on other sites
supposing that the weight is a point weight that act on the table at a given point and since the table is static (balanced) then you can find the sum of the moments about each point of intersection between the legs and the ground and equating it with zero to find the reaction at each leg.for example if the legs of the table intersect the ground at the points [(1,0,1)(-1,0,1);(-1,0,-1);(1,0,-1)] ,the weight of the table=100N acting in its middle and the added weight is 100N acting at the point (1,1,1).the reactions are R1,R2,R3,R4 for the legs 1,2,3,4 respectively then we will find that from the symmetry of the figure that R2 and R4 are equal in magnitude.the weight of the table and the added weight both of them equal (0,-100,0).since the body is static then the sum of forces acting on it is zero ,you have the equation R1+R2+R3+R4=200N .since R2 and R4 are equal then R1+2*R2+R3=200N.Then find the moment about the first and second points using the vector product and equate it with zero.so you have 3 equations in 3 Unknowns so you can find R1,R2 and R3.

[edited by - mohamed adel on September 2, 2003 3:01:24 PM]

##### Share on other sites
Oh yeah... A vector equation is three constraints.

*Repeats it mentally*

Mohamed is entirely right (on principle, at least).