How Does Tension Affect a Climbing Man on a Frictionless Ladder?

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Homework Help Overview

The problem involves a 75 kg man climbing a uniform ladder that is resting against a frictionless wall and floor, with a rope providing tension at the base. The discussion focuses on calculating the tension in the rope at a specific point and determining the maximum distance the man can climb before the rope breaks.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of torque equations and the forces acting on the ladder, questioning the treatment of normal forces and angles. There are suggestions to clarify the point about which torque is being calculated and to ensure consistency in the variables used.

Discussion Status

Some participants have offered insights into the torque equation and the variables involved, while others are seeking clarification on the approach taken by the original poster. The conversation is ongoing, with multiple interpretations being explored regarding the setup and calculations.

Contextual Notes

The problem is constrained by the conditions of a frictionless environment and the maximum tension limit of the rope, which are central to the discussion but not fully resolved.

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Homework Statement



A 75 kg man climbs up a 120 N uniform ladder of length 4 m. The ladder rests on a frictionless wall and a frictionless floor, however the lower end of the ladder is fastened to the wall by a horizontal rope which can withstand a maximum tension of 400 N.

Find
a) The tension in this rope when the man is one third up the ladder.
b) The maximum distance (d) that the man can climb before the rope breaks.

Given:
Fg man: 735 N
Fg ladder: 120 N
l (length) : 4 m

Homework Equations


∑ τorque
∑ Fx
∑ Fy


The Attempt at a Solution


I drew all the forces acting on the ladder and came up with a torque equation but I'm having a hard time dealing with the normal forces and possibly some angles.

∑ τorque = -COS 37 Fg man (d) - COS 37 Fg ladder (l/2) + SIN 53 T (l/3) + SIN 53 n1 (l)
∑ τorque = 586.9 N d -191.7 N + 1.06 T + 3.19 n1 = 0

∑ Fx = -n1 + T = 0
∑ Fy = n2 = Fg man + Fg ladder

Much help with be appreciated. :D
Please provide some detail and way to solve the problem.
Thank you.
 
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welcome to pf!

hi firefly.ember! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
firefly.ember said:
∑ τorque = -COS 37 Fg man (d) - COS 37 Fg ladder (l/2) + SIN 53 T (l/3) + SIN 53 n1 (l)
∑ τorque = 586.9 N d -191.7 N + 1.06 T + 3.19 n1 = 0

how did you get both T and N1 in the torque equation? :confused:
 
T=n1 so your Torque equation only has one variable. I think your lever arm for T is incorrect, it is .15 m from the ground so you have to account for the distance from where the ladder rests on the ground. Check the direction of the torque caused by T and the angle between the forces caused by the ladder and man perpendicular to the ladder, I think it is Cos 53 rather than cos 37. What do you think?
 
You are finding torque about what point?

Looking at individual terms in your expression for torque, it looks like you used a different point for each term.

I would find it about the point at which the ladder contacts the wall.
 
Last edited:
I'm summing the moments where the ladder touches the ground.
 
RTW69 said:
I'm summing the moments where the ladder touches the ground.
RTW69,

That's fine. I was aiming my comments at OP (firefly.ember). Your comments appear to be on target.

Let's see if firefly.ember replies before we do much more.
 

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