Rigid equilibrium question involving a ladder...by LizzleBizzle Tags: homework, ladder, rigid equilibrium, torque 

#1
Nov910, 02:32 PM

P: 3

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 onethird 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 4x is the distance of the front person from the two people. Each holds onethird 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)(4x) = 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
Nov910, 02:58 PM

P: 137

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
Nov910, 03:05 PM

Sci Advisor
HW Helper
PF Gold
P: 5,959

Hi, Liz, and welcome to Physics Forums!




#4
Nov910, 03:09 PM

Sci Advisor
HW Helper
PF Gold
P: 5,959

Rigid equilibrium question involving a ladder... 



#5
Nov910, 05:27 PM

P: 3

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)(4x)  [(w)(2/3x + 1.3)] 0 = 2.6666w  2/3wx  2/3wx  1.3w 1.3w = 1.3wx x = 1 m? This answer COULD make sense. 



#6
Nov1010, 12:24 PM

P: 137

Huh? The center of gravity of the ladder would be at the midpoint  don't overthink it. And again, make the location of the twoperson support as x. Then solve the equation for sum of moments.



Register to reply 
Related Discussions  
Rigid Body Equilibrium Question  Introductory Physics Homework  6  
Equilibrium Style Question involving a Weight  Introductory Physics Homework  3  
Equilibrium of a Rigid Body question  Introductory Physics Homework  7  
Ladder Question  Force of friction required to prevent ladder from slipping  Introductory Physics Homework  4  
Ladder question (Static equilibrium)  Introductory Physics Homework  2 