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

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The discussion focuses on a physics problem involving a 75 kg man climbing a 120 N uniform ladder resting on a frictionless wall and floor, secured by a rope with a maximum tension of 400 N. Participants analyze the torque equations and forces acting on the ladder, noting the need to correctly account for angles and distances in the torque calculations. There is a suggestion to simplify the torque equation by using a single variable and to reconsider the point about which torque is calculated. The importance of correctly identifying the lever arm and the angles for torque calculations is emphasized. The conversation encourages collaborative problem-solving to clarify the approach to finding the tension and maximum climbing distance before the rope breaks.
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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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