1. Nov 17, 2008

### jromega3

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

A ladder is sitting on a frictionless surface as shown in the figure. Each half has a mass m=20 kg and has a length L=1.5 m, and the angle that each side of the ladder makes with the vertical is alpha=20 degrees. A person of mass M=119 kg is standing at a distance x=0.026 m from the center of the ladder.

What is the tension on the string?

2. Relevant equations
Fy=0 (both sides)
Fx=0 (both sides)
Torques=0

3. The attempt at a solution

Ok, so for the left side:
Fy= MpG + MlG=Fn1 + Fay=0

Where MpG is Fg of the person, and Mlg Fg of the ladder. Fn1 is the normal force on the end and Fay the articulation force at the top in the y direction.

Fx=0=Fax=Ts
Where Ts is tension string and Fax articulation force in x direction.

for the right side:
Fy=Mlg + Fay=Fn2

Same goes, but Fn2 normal force on this side of the ladder

Fx: Ts=Fax

Torques. Eh this is where I get confused with the angles and all.

Net Torque is 0.
So....I calculated the torque the person exerts on the ladder relative to the top point is 30.32 (this was the first question).
So I have 0=30.32 + 2Ml(A/2)G - Fn1A - Fn2A - 2(A/2)Ts
Where A is the ladder length
So I have
0=30.32 + A(MlG-Fn1-Fn2-2Ts)
So
A(Fn1 + Fn2 + 2Ts)= 30.03 + A(Mlg)

Now I'm supposed to solve for Ts.

Pretty sure I've gone wrong in a few places, not sure though. And if not I just hit a roadblock, I seem to have too many unknowns as Fn1 doesn't equal Fn2 (I believe).

Last edited by a moderator: May 3, 2017