Ladder against a wall

  • Thread starter kell
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  • #1
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Hi all, this is a problem with which I led myself in circles with for hours, because I kept forgetting to factor in a different thing each time. Long story short, I thought I'd cracked it, but my answer and the given answer are out by a square. Please see my efforts below!

Homework Statement


A uniform ladder of mass M and length L stands on rough horizontal ground and leans against a vertical wall with which it makes an angle θ. A man of mass m stands on the ladder a distance x from the bottom.

a) obtain expressions for the normal reaction and the friction force acting at the point of contact between the ladder and the ground.


Homework Equations



The normal is obviously (m+M)g.

Other than that, it's static equilibrium, and the equation I used for balancing was the moments, where τNW is the moment due to the normal FNW to the wall at the top of the ladder, and τMg and τmg are the two moments of the ladder and man:

1. τNW = τMg + τmg

The Attempt at a Solution



Pretty simple. I have the ladder to be length L, so I set the origin of turning at the base of the ladder. I want to find FNW because it's equal to the frictional force FF at the base of the ladder.

The radius component for the moment τNW is this Lcosθ. The moment is thus: FNWLcosθ.
The radius component for τMg is (L/2)sinθ, and for τmg is (x/L)sinθ.
So the two moments are: [(MgL)/2]sinθ and [(mgx)/L]sinθ

Equation 1. then becomes:

FNWLcosθ = [(MgL)/2]sinθ + [(mgx)/L]sinθ

Which evidently becomes, solving for FNW:

FNW = (M/2 + mx/L2)gtanθ

In the given answer, the mx/L2 term is simply mx/L. But the position of the man along the ladder was given as x from the bottom, so this should be written x/L, right? Which means dividing out the L on the FNW side will always create a squared term.

Any insight appreciated.

Jack

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
1,540
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The radius component for the moment τNW is this Lcosθ. The moment is thus: FNWLcosθ.
The radius component for τMg is (L/2)sinθ, and for τmg is (x/L)sinθ.
So the two moments are: [(MgL)/2]sinθ and [(mgx)/L]sinθ

The moment arm of mg should be x sinθ ,not (x/L)sinθ .The distance of the man is given from the bottom of the ladder .This distance is x not x/L .Just as the moment arm of Mg is (L/2)sinθ,likewise that of mg will be x sinθ .The torque due to mg ,τmg=mgxsinθ.

This will give you the correct answer.
 
  • #3
21,045
4,641
Why would there be an extra L? Your "man" term in the equation is not dimensionally consistent with the other two terms. Your term would be correct if he were at a fractional location x/L of the total ladder length. But that's not the case. He is at an absolute distance x up the ladder.
 
  • #4
nasu
Gold Member
3,776
433
This formula is obviously wrong:

FNW = (M/2 + mx/L2)gtanθ

The second term in the parenthesis has units of mass/length so you cannot add it to the M/2.

Now why did you get this?
You misinterpret the meaning of x.
They say "distance x from the bottom". So x is clearly a distance, in meters or whatever unit you use.

The moment of the force should be a force times a distance. For the weight of the man, the "radius component" is x*sinθ so the moment will be mgx*sinθ.
I don't really see why would you think to divide by L.

Edit. Sorry, I was writing (slowly :)) while Chestermiller posted already.
 
  • #5
6
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Yup, you're all clearly correct. Thanks so much for pointing it out. I really don't know why I was trying to divide by the L. I had established it way early on in my workings and didn't think to question it later. Losing the forest for the trees!

Cheers everyone :)
 

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