Force of the wall against the ladder is from static friction?

  • #1
annamal
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A ladder is leaning against a frictionless wall and the floor. The force F of the wall against the ladder is what is opposing static friction f at the floor. I don't understand how the force F causes the ladder to slide to the left on the floor unless opposed by friction force f. See picture below.

Screen Shot 2022-04-04 at 3.55.49 PM.png
 
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  • #2
The force ##\vec{F}## applies to the ladder as a whole so it is applied to all parts of the ladder, including the base. Think of the ladder as having wheels at the base that can be locked. If the wheels are unlocked, pulling the top of the ladder away from the wall will cause the whole ladder to roll to the left. If the force of static friction ##\vec{f}## (e.g. by locking the wheels) is sufficient to counteract this force ##\vec{F}## at the base, the base of the ladder will not move.

AM
 
  • #3
Andrew Mason said:
The force ##\vec{F}## applies to the ladder as a whole so it is applied to all parts of the ladder, including the base. Think of the ladder as having wheels at the base that can be locked. If the wheels are unlocked, pulling the top of the ladder away from the wall will cause the whole ladder to roll to the left. If the force of static friction ##\vec{f}## (e.g. by locking the wheels) is sufficient to counteract this force ##\vec{F}## at the base, the base of the ladder will not move.

AM
In order to roll to the left, you would have to put a force at the center of mass (middle) of the ladder. The force F is being applied to the top of the ladder
 
  • #4
annamal said:
In order to roll to the left, you would have to put a force at the center of mass (middle) of the ladder. The force F is being applied to the top of the ladder
No, this is not true. The center of mass will accelerate to the left no matter where the force is applied. As long as it has a component to the left. The equation of motion for a body which is not point-like is $$ \vec{F}_{net}=M\vec{a}_{CM}$$.
If there is a net torque then there is also rotation around the CM but this does not change the translation of the CM. This is one of the reason the CM is ann useful concept.
 
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  • #5
annamal said:
I don't understand how the force F causes the ladder to slide to the left on the floor unless opposed by friction force f.
Think of it this way.
1. The frictionless wall can only exert force F away from its surface and perpendicular to it. Therefore, F is directed to the left.
2. The floor can exert force of static friction fs parallel to its surface in either direction.
3. The ladder is in equilibrium which means that the net force in both the vertical and horizontal direction must be zero.
4. The only horizontal force, other than F, is fs.

Question
In what direction must fs be so that the sum of the two horizontal forces is zero?
(a) To the left.
(b) To the right.
(c) No direction, the force is zero.
(d) Not enough information is given.

Your choice but make sure that items 1-4 are satisfied. Then consider that fs has an upper limit that it cannot exceed.
 
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  • #6
annamal said:
A ladder is leaning against a frictionless wall and the floor. The force F of the wall against the ladder is what is opposing static friction f at the floor. I don't understand how the force F causes the ladder to slide to the left on the floor unless opposed by friction force f. See picture below.
It seems important to me to keep the natural order in which cause and effect happens in this case.
The weight of the ladder is the only driving force in this mechanism; that is the cause of any other force and movement that may appear.

The nature of the mechanism imposes two main restrictions to any movement of our ladder:
* Top end can only be moved vertically.
* Bottom end can only be moved horizontally.

No friction force restricts any vertical movement of the top end.
There is certain available amount of legs-floor grip or friction force that restricts (up to certain point) any horizontal movement of the bottom end.

That legs-floor grip is the only thing stopping the center of mass of the ladder from relocating itself to a position of lower potential energy (floor level in this case).

As discussed in your hinged door thread, summation of moments is your friend in understanding and solving this problem.
 
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  • #7
annamal said:
I don't understand how the force F causes the ladder to slide to the left on the floor unless opposed by friction force f.
The problem simply states that F opposes f. This does not imply that the force F causes the ladder to move to the left when unopposed by f. Perhaps you are thinking that the ladder should move in the direction of F because objects seem to move in the direction in which a force is applied, but this is not so. There are lots of situations where the direction of motion is not the same as the direction of an applied force, or even the net force! Newton's 2nd Law tells us that the acceleration and the net force are in the same direction, it tells us nothing about the direction of motion.
 
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