# Distribution of Weight on Two Supports

1. Feb 13, 2012

### zoobyshoe

An oblong weight of uniform shape and density is supported horizontally on two supports. The placement of the two supports can be shifted. What is the best way to determine what percentage of the weight is born by each support given any possible arrangement of the supports?

2. Feb 13, 2012

### jehan60188

sum of moments = sum of forces = 0

3. Feb 13, 2012

### FeX32

If it's a static problem. Then yes. Sum the moments about any point on the body and equate to zero. Then sum the forces in $x$ and $y$ directions and equate to zero. You are then left with enough equations to solve for the forces.
This assumes there are only 2 supports.
Statically indeterminant problems usually need to include some deflection to be able to solve (more than 2 supports.)

4. Feb 13, 2012

### zoobyshoe

I have no idea what either of you are talking about.

5. Feb 13, 2012

### FeX32

6. Feb 13, 2012

### zoobyshoe

I could be wrong but that seems to be about a different situation.

Here's an illustration of what I'm asking:

Say we have three bricks, A, B, and C. We stand bricks A and B upright and lay brick C on top of them such that A and B are beneath each end of C. In this position we should be able to assume that bricks A and B each bear 1/2 the weight of C.

Now if I move Brick A more toward the center of C, say 1/3 the length of C, then here I think brick A is bearing more of the weight of C than brick B is.

It seems there should be a simple formula to discover what percentage of the weight of C A is now bearing and what percentage B is bearing.

7. Feb 13, 2012

### jehan60188

that is exactly what i'm talking about

8. Feb 13, 2012

### jehan60188

mg -( Fna + Fnb ) = 0
mg*L2 - Fna*X1 = 0
mg*L1 - Fnb*X2 = 0

you'll know mg, L1, L2, X1 and X2
so that's three equations, and two unknowns, so you can find a solution for Fna and Fnb

9. Feb 13, 2012

### zoobyshoe

Where:

mg = mass x gravity

L = length

X = ?

Fna = Force something brick A?

Come on, guys, help me out.

10. Feb 13, 2012

### jehan60188

mg = weight of one brick
L1 = distance from axis of rotation to center of gravity of brick
L2 = distance from other axis of rotation to center of gravity of brick
X1 = distance between brick a and b
X2 = same distance
Fna = normal force exerted by brick a

if you don't understand this, you'll have to get your hands on a physics textbook, or wait for someone else to explain torque to you

11. Feb 13, 2012

### I like Serena

Suppose your brick A has a distance "a" to the left of the center of your brick C.
And suppose your brick B has a distance "b" to the right of the center of brick C.

Then the weight on A is b/(a+b) of the weight of C.
And the weight on B is a/(a+b) of the weight of C.

12. Feb 13, 2012

### zoobyshoe

No, I get this now. That was the last info I needed. Thanks for your time.

13. Feb 13, 2012

### zoobyshoe

THAT'S what I was looking for! Thanks!