Ladder question (Static equilibrium)

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An 8.0 m ladder weighing 200 N rests against a smooth wall at a 50.0° angle, and the problem involves determining how far an 800 N person can climb before the ladder slips. The discussion highlights the challenge of setting up torque equations, particularly when using the base of the ladder as the pivot point. It emphasizes the importance of considering both friction and the normal force in the calculations, contrary to the initial assumption that they could be ignored. Participants stress the need for equilibrium of forces and torques to solve the problem accurately. Understanding these forces is crucial for determining the maximum climbing height before slipping occurs.
bjgawp
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Here's my problem:
An 8.0 m, 200 N ladder rests against a smooth vertical wall. The coefficient of static friction between the ladder and the ground is 0.60, and the ladder makes a 50.0° angle with the ground. How far up the ladder can an 800 N person climb before the ladder begins to slip?

The problem is that I cannot set up the equation for the torques.
Tnet = 0
If we use the bottom of the ladder as our pivot point, we won't have to worry about friction and the normal force - leaving the 200 N ladder, the 800 N person, and the force provided from the wall (I may be wrong). But, after looking through my notes, I am lost on setting up the equation for the torques. Any help would be appreciated.

Thanks in advance!
 
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Have you taken the fact that the forces need to be in equilibrium into account?
 
bjgawp said:
...If we use the bottom of the ladder as our pivot point, we won't have to worry about friction and the normal force...

It's quite the opposite - you should worry about friction and the normal force.
 
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