Ladder problem - is there a mistake here ?

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The discussion revolves around a ladder problem involving a ladder propped against a wall that slides down while maintaining constant speeds at both ends. Participants are exploring the implications of the ladder's angle and the relationship between the speeds of the ends of the ladder against the wall and the floor.

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  • Mixed

Approaches and Questions Raised

  • Participants are analyzing the relationship between the lengths and speeds of the ladder's ends, questioning the implications of having equal speeds at both ends, and discussing the angular velocity of the ladder at a specific angle. Some are also considering the constraints posed by the problem's setup.

Discussion Status

The discussion includes various interpretations of the problem, with some participants suggesting that the conditions described may lead to contradictions. There is acknowledgment of differing opinions on the feasibility of the scenario presented in the problem statement.

Contextual Notes

Some participants note that the problem may contain inconsistencies, particularly regarding the speeds of the ladder's ends and the angle at which it is positioned. The exact wording of the problem has been shared, which highlights potential ambiguities in the setup.

Fermat
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I have a ladder which is propped up against a wall. The ladder slides down the wall and is constrained to move such that the speeeds of the ends of the ladder against the wall (and floor) are a constant value. i.e. [tex]\dot x = \dot y = constant[/tex]

The problem question is: What is the angular velocity of the ladder about its centre when the angle, of the ladder against the wall, is @=15 degrees ?

BUT, if the ladder length is L, then x² + y² = L² and differentiating this expression wrt time gives,[tex]2x\dot x + 2y\dot y = 2L \dot L = 0[/tex]
Therefore, [tex]x\cdot\dot x = -y\cdot\dot y[/tex]
However, if we are given [tex]\dot x =\dot y[/tex] (numerically), then this means that x = y!
But if x = y, then the ladder must be at an angle of 45 degrees, yes?
So, this means that the ladder has only one orientation, @ = 45 degrees, such that the speeds of the ends of the ladder are the same.
Is this right? Or have I made a mistake in my earlier working ?
 
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Fermat said:
I have a ladder which is propped up against a wall. The ladder slides down the wall and is constrained to move such that the speeeds of the ends of the ladder against the wall (and floor) are a constant value. i.e. [tex]\dot x = \dot y = constant[/tex]

The problem question is: What is the angular velocity of the ladder about its centre when the angle, of the ladder against the wall, is @=15 degrees ?

BUT, if the ladder length is L, then x² + y² = L² and differentiating this expression wrt time gives,[tex]2x\dot x + 2y\dot y = 2L \dot L = 0[/tex]
Therefore, [tex]x\cdot\dot x = -y\cdot\dot y[/tex]
However, if we are given [tex]\dot x =\dot y[/tex] (numerically), then this means that x = y!
But if x = y, then the ladder must be at an angle of 45 degrees, yes?
So, this means that the ladder has only one orientation, @ = 45 degrees, such that the speeds of the ends of the ladder are the same.
Is this right? Or have I made a mistake in my earlier working ?
Your conclusion is correct. The ladder could slide with one end at constant speed, but not both. I don't know how you could do the problem anyway with just the information given, unless you are supposed to find the answer in terms of the constant velocity.
 
Fermat said:
I have a ladder which is propped up against a wall. The ladder slides down the wall and is constrained to move such that the speeeds of the ends of the ladder against the wall (and floor) are a constant value. i.e. [tex]\dot x = \dot y = constant[/tex]
Perhaps you could give us the exact wording of the question. If the speed of the end against the wall is constant, the speed of the ends against the floor cannot be constant let alone the same speed.

As a function of angle [itex]\theta[/itex] of the ladder to the floor:

[tex]\dot y = \frac{d}{dt}L\sin\theta = -L\cos\theta\dot\theta = K[/tex]

means that:

[tex]\dot\theta = -K/L\cos\theta[/tex]

[tex]\dot x = \frac{d}{dt}L\cos\theta = -L\sin\theta\dot\theta = L\sin\theta(K/L\cos\theta) = K\tan\theta[/tex]

So if the speed of the end down the wall is constant, the speed along the floor is proportional to [itex]\tan\theta[/itex]

AM
 
Last edited:
The other point was:if both ends had different tangential velocities, which they would have if both ends had the same speed ( a given of the problem) but a non-45 degree inclination, then could they have an angular velocity about the ladders centre? I would have said they would only have an angular velocity about some point, the distance of which from and end would be proportional to the relative end velocites? Is that right ?
 
Here is the wording of the question as it was given to me:

What is the angular velocity of a 12-ft ladder about its center as it begins to slide off a perpendicular wall at an angle of 15 degrees to the wall, with each end having a relative speed of 8in/sec to their respective surfaces?
 
I wish I could stay for some answers, but got to go to work :(
 
Fermat said:
Here is the wording of the question as it was given to me:

What is the angular velocity of a 12-ft ladder about its center as it begins to slide off a perpendicular wall at an angle of 15 degrees to the wall, with each end having a relative speed of 8in/sec to their respective surfaces?
The short answer is that if the wall is perpendicular to the floor and the other end of the ladder is on the floor, the question poses an impossible set of facts.

AM
 
Andrew Mason said:
The short answer is that if the wall is perpendicular to the floor and the other end of the ladder is on the floor, the question poses an impossible set of facts.

AM
Many thanks.
 
Update

Just had my answer accepted :biggrin:

I told the questioner that I thought his question was wrong and that the angle should be 45 degrees, and worked out an answer based on that assumption.
He insisted that it should be 15 degrees. But he's just come back and accepted my original answer. So looks like he checked up and found out it was a typo after all!
 

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