Solving Tension on a Ladder with Frictionless Surface

[jromega3]

Homework Statement

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.

https://tycho-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys121/autumn08/homework/07b/ladder_STAT/ladder.gif

What is the tension on the string?

Homework Equations

Fy=0 (both sides)
Fx=0 (both sides)
Torques=0

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).

 

[anathema]

Thank you for your post. It seems like you have made some good progress in your attempt to solve this problem. However, there are a few areas where I can offer some clarification and guidance.

Firstly, in your attempt to solve for the tension on the string, you have correctly identified that the net torque on the ladder must be zero. However, the equation you have written is not quite correct. The torque exerted by the person on the ladder should be negative, as it is trying to rotate the ladder in the opposite direction to the other forces. So the correct equation should be:

0 = -30.32 + 2Ml(A/2)G - Fn1A - Fn2A - 2(A/2)Ts

Secondly, you are correct that there are two unknown normal forces (Fn1 and Fn2) in this problem. However, there is a way to eliminate one of these unknowns by considering the forces in the x-direction. Since the ladder is not moving in the x-direction, the sum of the forces in that direction must also be zero. This means that:

Fax = Ts

So now you have two equations and two unknowns (Ts and Fn2) and you can solve for both of them.

Lastly, it is important to note that the angle alpha is not the same as the angle A in your equations. The angle alpha is the angle that each side of the ladder makes with the vertical, while the angle A is the length of the ladder. So be careful when using these angles in your equations.

I hope this helps to clarify some things and guide you towards the correct solution. Good luck!

 

[thomate1]

I would like to offer some suggestions for solving this problem. First, it is important to clearly define all of the variables and forces involved. In this case, the ladder and person have different masses, so it may be helpful to label them as Ml and Mp, respectively. Also, the normal forces on each side of the ladder should be labeled as Fn1 and Fn2, as you have correctly done.

Next, it may be helpful to draw a free body diagram for each side of the ladder and the person. This will help visualize all of the forces acting on each component and make it easier to set up the equations.

When setting up the equations, it is important to consider all of the forces acting on each component. For example, on the left side of the ladder, there is the weight of the person and the ladder, the normal force from the ground, and the tension in the string. On the right side, there is the weight of the ladder and the normal force from the ground.

In terms of the torques, it may be helpful to choose a pivot point and consider the torques acting on each component relative to that point. This will help in setting up the equations for torque equilibrium.

Lastly, it is important to remember that the tension in the string is the same on both sides of the ladder, as the string is assumed to be massless and frictionless. This means that Ts on the left side is equal to Ts on the right side.

I hope these suggestions are helpful in solving this problem. Remember to carefully consider all of the forces and torques acting on each component and to use the equations of equilibrium to solve for the unknowns.