What is the angle of static equilibrium for a ladder on ice?

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The discussion focuses on determining the tension in a rope supporting a foldable step ladder weighing 150 N, positioned at a 60° angle on ice. Participants emphasize the importance of showing work, such as free body diagrams (FBD), to receive assistance in solving the problem. There are challenges in equating torques around the ladder's pivots, with users noting a lack of progress in calculations. Clarification is sought regarding the ladder's configuration and the angle's application to each half. Effective problem-solving requires a clear demonstration of the steps taken and where difficulties arise.
engineerr
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Homework Statement
A foldable step ladder weighing 150 𝑁 is made of two halves connected by a pivot on top and
by a rope in the middle. The ladder is put on ice and its angle with the surface is 60°. What is
the tension of the rope?
Relevant Equations
net torque = zero
tried equating torques around the pivots of the ladder and equating them to zero but I'm not getting anywhere with that.
 
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engineerr said:
Homework Statement: A foldable step ladder weighing 150 𝑁 is made of two halves connected by a pivot on top and
by a rope in the middle. The ladder is put on ice and its angle with the surface is 60°. What is
the tension of the rope?
Relevant Equations: net torque = zero

tried equating torques around the pivots of the ladder and equating them to zero but I'm not getting anywhere with that.
Start with some diagrams (FBD). You aren't receive help until you show an honest effort has been made.
 
engineerr said:
tried equating torques around the pivots of the ladder and equating them to zero but I'm not getting anywhere with that.
We can't help you unless you show us what you did. Just mentioning it is not enough. We have to know where you got stuck or what you did wrong.
 
Welcome, @engineerr !

Could you clarify this part?
“The ladder is put on ice and its angle with the surface is 60°.”
Each half, I assume.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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