Conceptual question about frictional force and equilibrium

In summary, a uniform beam of length L and mass m is inclined at an angle θ to the horizontal, with its upper end connected to a wall by a rope and its lower end resting on a rough, horizontal surface. The coefficient of static friction between the beam and surface is μs. The angle θ is chosen such that the friction force is at its maximum value. A force diagram is drawn for the beam and using the condition of rotational equilibrium, an expression for the tension T in the rope is found in terms of m, g, and θ. An expression for μs is obtained, involving only the angle θ. When the ladder is lifted upward and its base is placed back on the ground slightly to the left
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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.
 
  • #8
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!
 
  • #9
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?
 

FAQ: Conceptual question about frictional force and equilibrium

1. What is frictional force?

Frictional force is a type of force that opposes the motion of an object when it comes into contact with another object or surface. It is caused by the microscopic irregularities on the surfaces of the two objects rubbing against each other.

2. How does frictional force affect equilibrium?

In an ideal situation, frictional force does not affect equilibrium as it only acts in the opposite direction of motion. However, in real-life scenarios, frictional force can cause an object to move or change its direction, which can disrupt equilibrium.

3. Can frictional force ever be beneficial?

Yes, frictional force can be beneficial in certain situations. For example, it allows us to walk on the ground without slipping and enables vehicles to have traction on the road. In some cases, frictional force can also be used to control the speed of an object, such as in brakes on a car.

4. How does the surface area affect frictional force?

The surface area of an object in contact with another surface can affect the amount of frictional force. Generally, the larger the surface area, the greater the frictional force. This is because there is a larger area for the microscopic irregularities to interact and create friction.

5. What are some strategies to reduce frictional force?

One strategy is to use lubricants, such as oil or grease, between two surfaces to reduce the contact and therefore the frictional force. Another strategy is to make the surfaces smoother, which reduces the number of microscopic irregularities and decreases the frictional force. Additionally, using wheels or rollers can also reduce frictional force by allowing the object to roll rather than slide.

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