Conceptual question about frictional force and equilibrium

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A uniform beam inclined at an angle θ is analyzed for equilibrium, where its upper end is secured by a rope and the lower end rests on a rough surface with static friction coefficient μs. The discussion emphasizes that the angle θ is at a critical point where the friction force is maximized, meaning any further increase in angle would cause the beam to slip. When the beam's base is shifted slightly to the left, the required frictional force to maintain equilibrium increases; if it does not increase sufficiently, the beam will slip rather than fall. The length of the beam does not affect this qualitative analysis, and the focus remains on the relationship between angle, friction, and equilibrium. Understanding these dynamics is crucial for solving related problems in physics.
lichenguy
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A uniform beam of length L and mass m is inclined at an angle θ to the horizontal. Its upper end
is connected to a wall by a rope, and its lower end rests on a rough, horizontal
surface. The coefficient of static friction between the beam and surface is μs.
Assume that the angle θ is such that the friction force is at its maximum value.
a) Draw a force diagram for the beam.
b) Using the condition of rotational equilibrium, find an expression for the tension T in the rope in terms of m, g and θ.
c) obtain an expression for μs, involving only the angle θ.

e) What happens if the ladder is lifted upward and its base is placed back on the ground slightly to the left of its former position? Explain.

I did a, b, and c, but I'm not sure on e. The graph shows μ in terms of the angle θ.

For e i wrote, so far, that FN would decrease because FT would now have a component in the same direction as FN.
If I'm understanding this right, μs is changing to maintain the equilibrium?

Also, I'm not sure what is meant by "Assume that the angle θ is such that the friction force is at its maximum value.".
 

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lichenguy said:
FT would now have a component in the same direction as FN.
True, but note the word "slightly". If you go through the algebra you will find that for a small displacement x the change in FT is of the order x2, so I think you are supposed to ignore that.
lichenguy said:
Assume that the angle θ is such that the friction force is at its maximum value.".
As θ changes, the necessary frictional force changes. They are saying that the position of the ladder is at the extreme angle for the given frictional coefficient. Any further (which way?) and the ladder would slip.
 
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haruspex said:
True, but note the word "slightly". If you go through the algebra you will find that for a small displacement x the change in FT is of the order x2, so I think you are supposed to ignore that.

As θ changes, the necessary frictional force changes. They are saying that the position of the ladder is at the extreme angle for the given frictional coefficient. Any further (which way?) and the ladder would slip.
Ok, ty, i get the maximum frictional force thing now.

On the e question, pulling it back will decrease the angle, but am guessing it won't change the angle in the same way as before? Before we moved the top, not the bottom.
 
haruspex said:
True, but note the word "slightly". If you go through the algebra you will find that for a small displacement x the change in FT is of the order x2, so I think you are supposed to ignore that.

As θ changes, the necessary frictional force changes. They are saying that the position of the ladder is at the extreme angle for the given frictional coefficient. Any further (which way?) and the ladder would slip.
Does it depend on the length of the ladder? Because i found something, but I'm not sure if it's right.
 
lichenguy said:
Does it depend on the length of the ladder? Because i found something, but I'm not sure if it's right.
It's a qualitative question. No, the length of the ladder does not matter.
If the base is shifted to the left, what will happen to the frictional force that is needed to maintain equilibrium? Will it increase, decrease or stay the same?
 
haruspex said:
It's a qualitative question. No, the length of the ladder does not matter.
If the base is shifted to the left, what will happen to the frictional force that is needed to maintain equilibrium? Will it increase, decrease or stay the same?
There will be need for an increased frictional force to stay up. If the frictional force doesn't increase it will fall over, i guess.

I tried to find out if the rate of change of μ that we found earlier was enough to maintain equilibrium while moving the base back. I guess that's not what was asked.
 
lichenguy said:
If the frictional force doesn't increase it will fall over,
Yes, but it is a bit inaccurate to say it will fall over. The base might hit the wall first. Just say the ladder will slip, and give your reason.
 
haruspex said:
Yes, but it is a bit inaccurate to say it will fall over. The base might hit the wall first. Just say the ladder will slip, and give your reason.
Alright :D i didn't think of it hitting the wall, hehe.
Thanks lad!
 
lichenguy said:
There will be need for an increased frictional force to stay up. If the frictional force doesn't increase it will fall over, i guess.

Assuming the surface is uniform, what affects the available friction force? Does that increase or decrease if the base is shifted to the left?
 
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