Friction in Everyday Life: Walking & Knots

[Nugatory]

I am still not sure why the horizontal force that the door applies to the wedge while the door tries to close is a smaller force than the force that would be applied if the stopper did not have a sloped surface.

The horizontal force the door applies to the stopper must be exactly balanced by the friction force between stopper and floor - if the forces aren't balanced the stopper would move, and if it's doing its job it's not moving.

And how great is that force? It's a frictional force, so it's equal to the coefficient of friction times the normal (vertical!) force between stopper and floor. Now we need to know what that normal force is, and that's going to be the weight of the stopper plus the vertical force of the door on the stopper. That vertical force is what is affected by the slope of the wedge.

You may find it helpful to consider a few extreme cases. First, suppose that the stopper is very heavy - we're using a concrete block or an iron anvil as a doorstop. The doorstop will work just fine even though it's not wedge shaped at all; the normal force from its weight is sufficient to generate enough friction with the floor to hold the door. Second, consider a more normal wedge-shaped doorstop, but imagine what would happen if the door really wanted to swing; not just a powerful wind pushing on it, but a very strong person trying to shove it open. Because of the wedge shape, the harder we push the door against the stop, the greater the vertical force between door and stopper and between stopper and floor (these forces must be balanced because the stopper isn't moving) so the greater the frictional force with the floor that keeps the stopper from moving sideways and allowing the door to open. Push hard enough, and you will see the door being forced upwards as the vertical forces increase - and it's the wedge shape that allows the horizontal push to generate a vertical force.

 

[A.T.]

I am still not sure why the horizontal force that the door applies to the wedge while the door tries to close is a smaller force than the force that would be applied if the stopper did not have a sloped surface.

The horizontal force by the door isn't smaller, but the vertical force by the door allows more static friction at the base, so the wedge can resist more horizontal force by the door than a block of the same weight.

 

[fog37]

Ok, I see:

the horizontal force exerted by the door produces a normal and vertical reaction force on the slope of the wedge. This vertical normal force, which would not exist if if the wedge slope was completely vertical, contributes to the available frictional force. When we push horizontally on a slope surface, part of the push is transferred by the sloped surface downward. This would not happen if the surface was perfectly vertical like a wall...

 

[A.T.]

the horizontal force exerted by the door produces a normal and vertical reaction force on the slope

Please, forget this "force produces another force" stuff. There is a contact force by the door on the wedge which can be decomposed into components, in different ways:
normal and tangential
vertical and horzontal

This vertical normal force,

The normal force on the wedge top is not vertical
normal : perpendicular to the contact surface
vertical: perpendicular to the floor

When we push horizontally on a slope surface, part of the push is transferred by the sloped surface downward.

Again, the horizontal force not reduced by "transferring part of it downward". But it can be balanced better thanks to the vertical force by the door, which allows more horizontal friction with the floor.

 

[fog37]

Thanks A.T.

To paraphrase you, the normal (reaction force) force on the sloped surface of the wedge can be decomposed into a vertical upward component and a horizontal component directed towards the door. The vertical force component adds to the normal force that the weight of the wedge stopper produces. This increases the available friction from the floor (since the overall normal force has been increased) allowing the stopper to be more effective at keeping the door from moving.

 

[A.T.]

Thanks A.T.

To paraphrase you, the normal (reaction force) force on the sloped surface of the wedge can be decomposed into a vertical upward component and a horizontal component directed towards the door. The vertical force component adds to the normal force that the weight of the wedge stopper produces. This increases the available friction from the floor (since the overall normal force has been increased) allowing the stopper to be more effective at keeping the door from moving.

Yes, just be more clear which "normal force" you mean (stopper top vs stopper base). And don't put too much emphasis on this "action/reaction", "force produces force" stuff. It doesn't mean much and is completely irrelevant for a quantitative analysis of the forces.