Static Equilibrium - When will this guy bail off his ladder?

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Homework Help Overview

The problem involves a window cleaner's ladder positioned against a frictionless wall, examining the maximum distance the cleaner can climb before the ladder slips. The scenario includes considerations of static friction and the angle of the ladder.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the coefficient of static friction and the angle of the ladder on the likelihood of slipping. There are attempts to verify the calculated distance of 9.41 m and whether it seems reasonable given the conditions.

Discussion Status

Multiple participants have arrived at the same calculated distance of 9.41 m, expressing uncertainty about its reasonableness due to the wet floor and the friction involved. Some participants are seeking validation of their calculations and exploring the assumptions made in the problem.

Contextual Notes

Participants are considering the effects of the wet floor and the angle of the ladder on the static friction, questioning whether the given coefficient of static friction is sufficient to prevent slipping.

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Homework Statement


A window cleaner of mass 95 kg places a 22kg ladder against a frictionless wall at a angle 65 degrees with the horizontal. The ladder is 10 m long and rests on a wet floor with a coefficient of static friction equal to .40. What is the max length that the window cleaner can climb before the ladder slips?


Homework Equations



sum F(x)=F(fr)-F(w)=0
sum F(y)=F(n)-mg=0
sum torque=F(w) * l (sin theta) - mg * 1/2 cos theta=0

(Using the point of the ladder on the ground as the pivot point)

The Attempt at a Solution




Using this equation: FW (10) sin(65) - 22 (9.8) (5) cos(65) - 95 (9.8)(x)cos(65) = 0

I then used Fr-Fw=0
Fn(Us) = FW
(1146.6*.40) = 458.64

I then input this into the equation

FW=22 [(9.8) (5) cos(65) - 95 (9.8)(x)cos(65)]/ (10) sin(65)

x = 9.41 m for the distance of the guy up the ladder. I've checked and re-checked and keep getting this answer, but intuitively it seems like that's too high since it's a wet floor that the ladder's on. Is the coefficient of static friction of 0.40 high enough to make this reasonable? Thanks for anyone who is willing to take a look!

-Lauren
 
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Is the coefficient of static friction of 0.40 high enough to make this reasonable?
Not necessarily. It is the angle of ladder that prevents the slipping. Walking on the slippery floor we take smaller steps.
 
Hmm..ok so being one who doesn't often hang out on ladders, I'm trying to figure out whether 65 degrees is enough to prevent slipping. Is it possible that 9.41 m would be a reasonable answer? I came out with the same answer after trying it multiple times, so I can't see where I could be going wrong. Thanks for anyone's opinion!
 
Last edited:
What is the answer?
 
Hi rl.bhat, the answer I came up with x=9.41 m on the 10 m ladder. What do you think?
Thanks!
-Lauren
 
I too got the same answer. Is it wrong?
 
yay! That's good news. I'm submitting it for an assignment so I'm glad to hear it!
I'll let you know whether it's right when I get it back :) Thanks for helping!
 

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