# Finding a moment

1. Feb 26, 2013

### United

1. The problem statement, all variables and given/known data
http://imgur.com/gedpNxN

2. Relevant equations
M=force times perpendicular distance

3. The attempt at a solution
I tried multiplying 10*6, 10*3, 10*9 and plugging in their respective negatives as well, but I still don't have a correct answer.

2. Feb 27, 2013

### SteamKing

Staff Emeritus
First, stop guessing.

Second, figure out what is the total load produced by the distribution.

Third, if this load were concentrated at one point along the beam, where would that point have to be located? (Hint: what is the center of gravity of the distributed load?)

Fourth, calculate the moment about point A.

3. Feb 27, 2013

### United

I found the center to be 4 m (1/3 from the right side) into the triangle.

I assumed the total load was 45N, based on (10*9)/2 (Weight * total length/2)

based on this, I used my moment equation, and still got the wrong answer (noting that my force is in the negative direction)

I feel my total load is incorrect. If it is, how can I correct it?

Last edited: Feb 27, 2013
4. Feb 27, 2013

### SteamKing

Staff Emeritus
The load acts only over 6 m of the length of the beam. It is 0 N for the first 3 m.

5. Feb 27, 2013

### United

To account for that I changed my 9 to a 6, used my moment equation, and still got a wrong answer.

(10*6)/2 times 4, noting that the answer is negative because of direction

What am I missing here?

6. Feb 27, 2013

### SteamKing

Staff Emeritus
For one thing, the c.g. of the distributed load is not 4 m from point A.

7. Feb 27, 2013

### United

I thought you calculated cg by calculating the value 1/3 to the left of your total "triangle" distance.

On that basis is what I attained the answer 4m. What am I missing there?

8. Feb 27, 2013

### SteamKing

Staff Emeritus