Force Distribution: Understanding Points A, B, C & D - Joseph

[JosephNY]

Could someone please help me understand how force is distributed across a plane as shown in the attached picture (I don't think I have the vocabulary to clearly describe the problem)?

If 100 pounds/kg, for example, were placed point 1, what would the force be at points A, B, C and D? Would the forces at A, B, C and D change if the 100 pounds were moved to points 2, 3, 4, or 5?

Thank you,

Joseph

 

[BvU]

@anorlunda : same as other thread ?

Hello Joseph,

what would the force be at points A, B, C and D

100, 0,0,0

Would the forces at A, B, C and D change if the 100 pounds were moved to points 2, 3, 4, or 5?

Yes.

 

[JosephNY]

@anorlunda : same as other thread ?

Hello Joseph,

100, 0,0,0
Yes.

Thank you!

What would the force be at A, B, C & D if the 100lbs were moved to 2, 3, 4 and 5 (assume points 2, 3, 4 and 5 were each equidistant from each other and the same distance as between 1 and 2)?

 

[JosephNY]

Would the forces at A/B/C/D be different is the force down at 1/2/3/4/5 were moved to the center between A-C and B-D, as shown?

And generally, why is the force not evenly distributed across (i.e. within) each plane (A-C and B-D)? Or evenly across between those planes (A-B and B-D)?

 

[BvU]

why is the force not evenly distributed across (i.e. within) each plane

you placed all the load on the front edge. so all the force ends up at A and B.

Would the forces at A/B/C/D be different if the force down at 1/2/3/4/5 were moved to the center between A-C and B-D, as shown?

That is a different case. The answer is 'yes'. And then the load distributes over A-C and B-D

Do you know about the laws (equations) that govern this domain ? Force equilibrium $\sum \vec F = 0$, moment equilibrium $\sum \vec \tau = 0$ ?

 

[JosephNY]

you placed all the load on the front edge. so all the force ends up at A and B.
That is a different case. The answer is 'yes'. And then the load distributes over A-C and B-D

Do you know about the laws (equations) that govern this domain ? Force equilibrium $\sum \vec F = 0$, moment equilibrium $\sum \vec \tau = 0$ ?

Unfortunately, no, I am not familiar with the laws or equations, but I'm starting to see that the location of the force does indeed have a great influence on the location of the lower force (as opposed to being fully distributed across planes in the horizontal direction).

Thank you!

 

[CWatters]

This is a class of problems called statics. Basically the structure isn't accelerating either linearly or rotationally so the net force in any direction is zero. This follows from Newton's F=ma.

Typically this allows you to write a set of simultaneous equations, each representing a sum of forces or torques equated to zero. When solved you end up with values for the individual forces.

 

[Dr.D]

To build on what CWatters has said, if the frame is supported on a flat surface as seems to be implied, then the reaction forces will be distributed and statics alone does not provide enough information to determine them all. It is called a "statically indeterminate" problem. It becomes necessary to consider the deformations of the structure to get the full picture, and this can get pretty complicated even for a simply system like you have shown.

 

[Stephen Tashi]

Could someone please help me understand how force is distributed across a plane as shown in the attached picture (I don't think I have the vocabulary to clearly describe the problem)?

If 100 pounds/kg, for example, were placed point 1, what would the force be at points A, B, C and D? Would the forces at A, B, C and D change if the 100 pounds were moved to points 2, 3, 4, or 5?

The concept of a force "at" some location is ambiguous. For example, "at" the location of force 1 in your diagram, there is a force of equal an opposite magnitude acting opposite to force 1 that keeps the situation in balance.

The analysis of forces in static situations is done by analyzing only forces exerted upon an object. This type of analysis is done with a "free body diagram".

A free body diagram of the table would show force 1 and the upward forces exerted by the floor upon the supports (and the force of gravity acting on the table if the table's weight is considered.)

When you speak of the forces "at" points A,B,C,D it isn't clear whether you are visualizing tiny legs at A,B,C,D that protrude so that only the locations A,B,C,D bear the load or whether you are thinking of AC and BD as flat surfaces that are in continuous contact with the floor.

Simple problems in statics can be solved by setting up equations that say the net (vector) force on a free body is zero and the net torque upon the body is zero. Have you studied this type of problem?

Most real life situations are so-call "static indeterminate" problems. In such situations the basic equations of balance do not have unique solutions. So the analysis must be done by considering the properties of the materials involved. The answers for a table built from a rigid material might be different that the answers for a table made of a flexible material.

 

[BvU]

Unfortunately, no, I am not familiar with the laws or equations, but I'm starting to see that the location of the force does indeed have a great influence on the location of the lower force (as opposed to being fully distributed across planes in the horizontal direction).

Thank you!

In that case the latest replies won't be of much practical help for you (or are they ?). To say something about the distribution of the load forces over e.g AC you need more information, unless you can assume that for post #4 an even distribution is an adequate estimate.

Is the red table ideal (weightless, absolutely stiff, etc...) ?

 

[JosephNY]

BvU, CWatters, Dr.D, and Stephen Tashi:

Thank you all very much for the analysis and explanation.

Clearly, 11th grade physics decades ago just didn't provide the foundation sufficient to understand this problem.

I understand the idea of "net force in any direction is zero" and solving basic simultaneous equations.

I also have a basic understanding of the difference between distributed and point/location forces (and now the knowledge of the existence of the term "statically indeterminate").

I also now know of the existence of the term "free body diagram" and how these static indeterminate problems consider the properties of the materials used (rigidity, various strengths, etc.).

So, in addition to some great concepts to be on the lookout for, and to hopefully get a chance to learn a little about, it seems that, generally, it is safe to say that a downward force applied at 3 would be evenly distributed along both places AC and BD whereas a downward force applied at 2 would result in a greater downward force on the AC plane (as it touches the floor --in grey) than the downward force on the BD plane. And, the differences in force between these planes is proportional to the how far away from 3 the force is applied.

Accurate (generally)?

Thank you all again!

 

[BvU]

Accurate, almost. It's not the differences but the forces themselves that are inversely proportional to those distances. And with that we come to the balance of moments. Just like F_net = ma describes a balance of forces F_net = 0 in equilibrium (absence of acceleration, a = 0), there is a law about rotation: (angular acceleration $\alpha$) : $\tau_{\rm net} = I\alpha$ where $\tau_{\rm net}$ is the sum of all moments (aka as torques) $\tau$ working on the object (the table in your case). In equilibrium, $\alpha = 0$ and that means $\tau_{\rm net} = \sum \tau = 0$.

And the moment about a given axis from a force is proportional to the perpendicular distance from the axis to the line of action of the force.