Ladder sliding down with no friction

In summary: Is there a minimum value for this contact force?There is no minimum value for the contact force. If the contact force is greater than zero, then circular motion will be maintained.
  • #1
Dustgil
42
0

Homework Statement


A uniform ladder leans against a smooth vertical wall. If the floor is also smooth, and the initial angle between the floor and the ladder is theta(0), show that the ladder, in sliding down, will lose contact with the wall when the angle between the floor and the ladder is

[tex]sin^{-1}(\frac {2} {3} sin \theta_{0})[/tex]

Homework Equations



[tex] \frac {dL} {dt}=N [/tex]

where L is the angular momentum and N is the net external torque

The Attempt at a Solution



I understand why the ladder loses contact with the wall at some point. There is a normal force on the ladder exerted by the wall that is not balanced by anything. So as the ladder falls, it sort of gains length in the x direction while at the same time being pushed away from the wall. Eventually there comes a point where the normal force exerted by the wall is zero and then negative, which will be at the angle I'm looking for.

Let's say that the angle that the ladder makes with the floor is theta. The translational motion equations are

[tex] m \ddot y=F_{n} - mg[/tex]
[tex] m \ddot x=F_{w}[/tex]I start to get a bit confused when trying to define the rotational motion. The first question is where's the best place to define the rotation at? For now I choose the point where the ladder makes contact with the ground. This is where I run into my second problem. For whatever reason I'm having trouble confidently expressing the torques exerted from both gravity and the normal force from the wall. I'm just kinda getting mixed up by it. I make a guess:

[tex]\frac {dL} {dt}= N[/tex]
[tex]I \ddot {\theta} = mgcos(\theta)(\frac {l} {2}) - F_{w}sin(\theta)(\frac {l} {2})[/tex]

Are these the correct equations? It took me a long while to get to this point so It'd be cool to be confident before I soldier on.
 
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  • #2
Dustgil said:
The first question is where's the best place to define the rotation at?
Try the corner of the wall and look at the center of mass. First understand the path it follows if both ends of the ladder are constrained to be always in contact, then figure out what must be true for the CM to deviate from this path.
 
  • #3
So the CM moves in a circle. Won't the CM stop accelerating in the x direction when it loses contact?
 
  • #4
That makes the translation motion

[tex]m \ddot {r} = r \dot { \theta }[/tex]

but I'm still unsure as to how to define the rotational motion. maybe I need to brush up on my trig..
 
  • #5
Dustgil said:
So the CM moves in a circle.
Correct. What force provides the necessary centripetal acceleration? When that force goes to zero, the CM leaves the circle. Hint: This is essentially the same problem as the small block that slides down on the outside of a hemispherical frictionless surface.
 
  • #6
For the hint it would be the contact force the block makes with the surface, so the normal force on the wall, right?
 
  • #7
Dustgil said:
For the hint it would be the contact force the block makes with the surface, so the normal force on the wall, right?
And the floor.
 
  • #8
kuruman said:
Correct. What force provides the necessary centripetal acceleration? When that force goes to zero, the CM leaves the circle. Hint: This is essentially the same problem as the small block that slides down on the outside of a hemispherical frictionless surface.
Not sure that is going to work... Not obvious to me, anyway. The particle on the sphere has no moment of inertia.
The force "providing centripetal acceleration" is not one applied force, it is the radial component of the net force. Even though the normal force from the wall becomes zero, there is still gravity and the normal force from the floor. And this net component does not go to zero, it just becomes insufficient to maintain circular motion.
 
  • #9
haruspex said:
And this net component does not go to zero, it just becomes insufficient to maintain circular motion.
I am not sure what "insufficient" means. The condition for the CM to go around in a circle is $$F_r -mg\cos \theta = - \frac{mv^2}{r} $$ where ##F_r## is the radial component of the contact force resultant. From this $$F_r = m\left(g\cos \theta - \frac{v^2}{r} \right)$$As long as ##F_r## is positive, circular motion of the CM is maintained. What is the criterion for "insufficient" other than ##F_r = 0##?
 
  • #10
kuruman said:
I am not sure what "insufficient" means. The condition for the CM to go around in a circle is $$F_r -mg\cos \theta = - \frac{mv^2}{r} $$ where ##F_r## is the radial component of the contact force resultant. From this $$F_r = m\left(g\cos \theta - \frac{v^2}{r} \right)$$As long as ##F_r## is positive, circular motion of the CM is maintained. What is the criterion for "insufficient" other than ##F_r = 0##?
You wrote that the force necessary for centripetal acceleration goes to zero. That is not Fr; it is mg cos(θ)-Fr.
 
  • #11
Yes, thank you for the correction. I made the mistake of trying to do it in my head without writing anything down - I should know better.
 
  • #12
Dustgil said:
Let's say that the angle that the ladder makes with the floor is theta. The translational motion equations are

[tex] m \ddot y=F_{n} - mg[/tex]
[tex] m \ddot x=F_{w}[/tex]

What are x and y? Are they the coordinates of the CoM? how are the related to the angle theta?
Yes, you can solve the problem by writing the equation of motion of the center of mass, and also the equation of rotation about the CoM.

One other approach can be using conservation of energy. The kinetic energy is the sum of the KE of the CoM, and the rotational energy about the CoM. As you have found, the CoM moves around a circle. That gives you a relation between θ and ##\dotθ##.
At the end, when the ladder loses contact with the wall, ##\ddot x##=0. You can write it also in terms of theta and derivatives.
 

Related to Ladder sliding down with no friction

1. What is a ladder sliding down with no friction?

A ladder sliding down with no friction refers to a scenario where a ladder is placed on a surface with no frictional force acting on it. This means there is no resistance to the ladder's motion, allowing it to slide down the surface without slowing down.

2. What factors affect the motion of a ladder sliding down with no friction?

The main factor that affects the motion of a ladder sliding down with no friction is the angle at which the ladder is placed on the surface. The steeper the angle, the faster the ladder will slide down. Additionally, the weight and length of the ladder may also impact its motion.

3. How does a ladder slide down with no friction if there is no force acting on it?

In this scenario, gravity is the only force acting on the ladder. Since there is no friction, there is no opposing force to counteract the downward force of gravity. This allows the ladder to slide down the surface without any resistance.

4. Can a ladder slide down with no friction forever?

In theory, yes. Without any friction, the ladder would continue to slide down the surface indefinitely. However, in reality, there may be other factors such as air resistance, uneven surfaces, or external forces that may eventually slow down or stop the ladder's motion.

5. How is a ladder sliding down with no friction different from one with friction?

When there is friction present, it creates a resisting force that opposes the motion of the ladder. This means that the ladder will not slide down the surface as easily and will require more force to move. It will also slow down more quickly due to the frictional force. On the other hand, with no friction, the ladder will slide down the surface with ease and will not slow down as quickly.

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