# Rigid equilibrium question involving a ladder

1. Nov 9, 2010

### LizzleBizzle

1. The problem statement, all variables and given/known data

"Three people are carrying a horizontal ladder 4.00 m long. One of them holds the front end of the ladder, and the other two hold opposite sides of the ladder the same distance from its far end. What is the distance of the latter two people from the far end of the ladder if each person supports one-third of the ladder's weight?"

2. Relevant equations

T=F*l
ET=0

3. The attempt at a solution

I drew a picture of the ladder with one person in the front, and two people (x) distance away from the opposite end of the ladder. So, x is the distance from the two people to the far end, and 4-x is the distance of the front person from the two people.

Each holds one-third of the ladder's weight.

I decided I'd set the axis of rotation as the far end of the ladder and set up my equilibrium equation, so...

ET = (2/3w)(x) + (1/3w)(4-x) = 0

And then I kind of got stuck after fiddling around with my equation and getting, uh, nowhere. Tried to throw in a center of gravity equation and realized that wasn't really helping me. Any ideas?

Thanks,
Liz

2. Nov 9, 2010

### p21bass

Try setting the person at the front of the ladder at 0.00m, and the other two people at x. Then take the moment about the front of the ladder.

Also, don't forget to take into account the weight of the ladder in your FBD.

3. Nov 9, 2010

### PhanthomJay

Hi, Liz, and welcome to Physics Forums!

Yes, but what's the distance of the front person from the far end?
Yes.
good idea
If you are summing moments about the far end, then you must determine the distance of each force from the far end. What is the distance from the front person to the far end? It is not (4-x). And you forgot to include the moment from the ladder's weight, the resultant force of which acts at the center of gravity of the ladder.

4. Nov 9, 2010

### PhanthomJay

That's an even better idea

5. Nov 9, 2010

### LizzleBizzle

Ah ha. After a second cup of coffee, I gained a little more inspiration...thank you for the hints. So here's what I have so far, but maybe I have missed something:

I changed my point of rotation to the front end, as suggested. I accounted for the weight of the ladder.

I found the center of gravity to be (2/3w)(x) + (1/3)w(4) / w , which boils down to 2/3x + 1.3.

In this case, the force of the two guys in the back would rotate clockwise around the axis of rotation, so that value is negative, and the weight, which would rotate counterclockwise, is positive.

So...
ET = 0 = (2/3w)(4-x) - [(w)(2/3x + 1.3)]
0 = 2.6666w - 2/3wx - 2/3wx - 1.3w
1.3w = 1.3wx
x = 1 m?