What makes a door move, the torque or the force or both?

[A.T.]

i think everyone knew what i meant, this is not english forums LOL

You keep asking people to confirm if your formulations are correct. What's the point if you don't want to improve them.

 

[alkaspeltzar]

You keep asking people to confirm if your formulations are correct. What's the point if you don't want to improve them.

Who said I wasn't improving, no need to be rude.

 

[alkaspeltzar]

(in fact a full description requires both), it is much more simple to analyze the torque. From only the torque we can determine the full motion and thereby determine the unknown hinge force. But since the hinge force is unknown in advance it can only be used in this case as an "after the fact" explanation.

Agreed, we need both force and torque for full description and understanding of the motion of the door. But I see now from yourself and others that in rotational motion, it is easier to use/focus on the torque since the forces underway are not clear and/or are changing, at the hinge and such. thks!

 

[sophiecentaur]

Even when there is only one force there's still torque

Yes - I could have put that better. What I meant was that , if you want to apply a certain torque, you can't use just one force - unless you know what it is you're trying to apply the torque to - i.e. where you are pushing relative to a fulcrum or the CM. Hence my point about a screwdriver vs a wrench.

i think everyone knew what i meant, this is not english forums LOL

Sometimes it's necessary to put the Transatlantic members right about the use of some words.

 

[snorkack]

So given one leaf door, there are both force and torque on the door, as it turns and translates. Is that correct?

With revolving door, there is only rotation therefore no net force

Precisely.
Of course, you need gross forces to get a torque, but you can have zero net force if you have a pair of opposite and equal forces.
So in case of revolving door, since there is no translation, you know hinge forces must needs be equal and opposite to the force exerted on door. In case of one leaf door, since there is translation, the hinge forces are different (and smaller).

There are two plausible pivot points to consider. Door hinges, or door centre of mass. If hinges do not hold, and you push the door at exactly the centre of mass, there would be no torque, and the door would be pushed out of doorframe without rotating.

 

[nasu]

Sometimes it's necessary to put the Transatlantic members right about the use of some words.

But the Transatlantic-ness is a frame dependent quality. A simple frame translation and you become the Transatlantic neighbour.

 

[sophiecentaur]

But the Transatlantic-ness is a frame dependent quality. A simple frame translation and you become the Transatlantic neighbour.

It's always 'the other guy' who needs educating.

 

[A.T.]

So given one leaf door, there are both force and torque on the door, as it turns and translates. Is that correct?

The correct statement is:
Given a one leaf door, there are both net force and net torque on the door, as its angular velocity changes and its center of mass accelerates.

With revolving door, there is only rotation therefore no net force

The correct statement is:
With a revolving door, only the angular velocity changes therefore there is no net force

 

[alkaspeltzar]

The correct statement is:
Given a one leaf door, there are both net force and net torque on the door, as its angular velocity changes and its center of mass accelerates. The correct statement is:
With a revolving door, only the angular velocity changes therefore there is no net force

A.T.,

I see the clarification. Yes you are correct. It is always net torque or net force that induces accelerations.

You were right, I was being lazy, it easier to explain 1 on 1 when you can work and draw it out.

Thanks for hearing me out and like other confirming my thoughts.

 

[PeterO]

Don't forget that the door, via the hinges, is connected to the wall, and through those hinges the door can be subject to a reaction force of what ever size and direction is needed for the door to move the way it does (it opens or closes?), when you apply the force you do apply, in the direction you applied it, and at the point(s) of contact you chose - so analysis may be far more difficult than you imagine.
before you touched the door, the hinges were already applying an upward force so it didn't fall to the floor, and the upper hinge was perhaps applying an inward force so the door didn't rotate - and perhaps the lower hinge was applying an outward force to prevent the same. If you removed all but one screw from the top hinge, that last screw might "pull out" showing that just one screw was not capable of providing the required force.

 

[cmb]

Torque is indeed defined in terms of force, but we can have torque even when there is no net force. An example might be spinning a wheel on a shaft by placing my hands on opposite sides of the wheel and pushing with one hand, pulling with other. The center of mass of the wheel stays put so we know that there's no net force on the wheel, but it starts to spin telling us that there is a net torque on the wheel.
Even when there is only one force there's still torque (in general - we can always choose a point about which to calculate the torque in such a way that it comes out zero). If there is only one force involved then the net force is necessarily non-zero and the center of mass will accelerate along with any rotation caused by the torque

Yes, a pure torque resulting in pure angular acceleration with no net translation of the CoM of the body to which the forces are applied is always generated by a balance of (at least two) forces, sum total in anyone axis being zero.

Some very complicated answers here that I think miss the question about the door, because it is not a pure angular acceleration. I think the point you were asking is whether the vector sum of forces from a) the door handle, and b) the hinge, are zero?

Answer; not exactly but close. (answer to you title is 'both')

There are two forces on the door when you push/pull the handle; 1) that force, and 2) the force from the hinge. The magnitude and vector direction of the two forces are not quite precisely equal while the vector direction is not quite opposite.

The reason is that the door not only receives a torque sending it into a spin but also receives a translating acceleration as its CoM begins to accelerate. The force on the handle is always very slightly greater than the force on the hinge if the door has angular acceleration greater than the torque-friction is inducing at the hinge.

Take the case of a slammed door; at the instant the door slams shut, there is a force from the retaining edge of the door which is equal and opposite to the forces from the hinge that result in torque. However, the door also has translational inertia too, perpendicular to the aperture, and there has to be an EXTRA impulse from the door edge and hinge which are both in the same direction to decelerate the translational motion.

This then adds summatively to one force negatively and one force being positively (in some datum direction) giving two forces. When a door slams, the force is greater on the closing edge than the hinge, the delta between the two being the force which decelerates the door motion in the direction perpendicular to the opening.

If you watch a door with a loose hinge being slammed, you may notice the hinge jumps 'outwards' momentarily, this being the reversal of the force vector from one direction when being pushed closed then at the moment of the 'slam' through zero and then the force other direction.

A revolving door is quite different as there is no translational momentum. The forces on the central pivot of a revolving door are always equal and opposite at all times as its CoM never moves, unlike a domestic door.

 

[alkaspeltzar]

Yes, a pure torque resulting in pure angular acceleration with no net translation of the CoM of the body to which the forces are applied is always generated by a balance of (at least two) forces, sum total in anyone axis being zero.

Some very complicated answers here that I think miss the question about the door, because it is not a pure angular acceleration. I think the point you were asking is whether the vector sum of forces from a) the door handle, and b) the hinge, are zero?

Answer; not exactly but close. (answer to you title is 'both')

There are two forces on the door when you push/pull the handle; 1) that force, and 2) the force from the hinge. The magnitude and vector direction of the two forces are not quite precisely equal while the vector direction is not quite opposite.

The reason is that the door not only receives a torque sending it into a spin but also receives a translating acceleration as its CoM begins to accelerate. The force on the handle is always very slightly greater than the force on the hinge if the door has angular acceleration greater than the torque-friction is inducing at the hinge.

Take the case of a slammed door; at the instant the door slams shut, there is a force from the retaining edge of the door which is equal and opposite to the forces from the hinge that result in torque. However, the door also has translational inertia too, perpendicular to the aperture, and there has to be an EXTRA impulse from the door edge and hinge which are both in the same direction to decelerate the translational motion.

This then adds summatively to one force negatively and one force being positively (in some datum direction) giving two forces. When a door slams, the force is greater on the closing edge than the hinge, the delta between the two being the force which decelerates the door motion in the direction perpendicular to the opening.

If you watch a door with a loose hinge being slammed, you may notice the hinge jumps 'outwards' momentarily, this being the reversal of the force vector from one direction when being pushed closed then at the moment of the 'slam' through zero and then the force other direction.

A revolving door is quite different as there is no translational momentum. The forces on the central pivot of a revolving door are always equal and opposite at all times as its CoM never moves, unlike a domestic door.

I found the example I posted in my college physics book. I see what you are saying.

There has to be force and torque on a door which hangs on hinges. Without net force, center of mass wouldn't move. Without torque, it wouldn't rotate but would move forwards.

I kinda think of it like Dale mentioned. The torque created set the door in rotation and the net forces must follow from the difference of applied force minus hinge force so that the c of m moves and satisfies f=ma.

Thanks for the insight!

 

[jbriggs444]

Without torque, it wouldn't rotate but would move forwards.

You need to be a little careful about this. You can have rotation (i.e. rotational acceleration of a body) without any net torque on that body. The catch is that a force whose line of action passes through the selected reference axis has zero torque. There is no rule that says you have to choose the center of mass of the door as your reference axis. You can put your reference axis at the position of the door knob if you like.

If net torque (about a chosen axis) is zero then the rate of change of angular momentum (about that axis) is zero. But angular momentum is not the same thing as rotation.

 

[sophiecentaur]

You need to be a little careful about this.

Agreed but also you need to avoid introducing more and more 'examples' and ways of looking at it. If a force is applied anywhere but through the (fixed) hinge of a door then there is a torque. A torque, only can be applied ` (with a 'reaction wheel', perhaps) and that will have the effect of rotating (accelerating) the door around the hinge. There will be a force through the hinge (centripetal), after acceleration has started.

 

[jbriggs444]

Agreed but also you need to avoid introducing more and more 'examples' and ways of looking at it. If a force is applied anywhere but through the (fixed) hinge of a door then there is a torque. A torque, only can be applied ` (with a 'reaction wheel', perhaps) and that will have the effect of rotating (accelerating) the door around the hinge. There will be a force through the hinge (centripetal), after acceleration has started.

If there is a net force, the torque depends on the choice of reference axis. That choice is free.

 

[sreerajt]

May be a relevant read:
Leo Rausér's answer to Why exactly is it scientifically impossible to lift someone up by the neck? How strong do you have to be to do so? - Quora

 

[sophiecentaur]

If there is a net force, the torque depends on the choice of reference axis. That choice is free.

I may have just confused myself. If you have a reaction wheel system, attached to a free object, that cannot apply any net force - or you would have a reaction less drive à la Eric Laithwaite. Ah - but attaching a hinge provides another reaction torque in addition to that, due to the MI of the door.

 

[jbriggs444]

I may have just confused myself. If you have a reaction wheel system, attached to a free object, that cannot apply any net force - or you would have a reaction less drive à la Eric Laithwaite. Ah - but attaching a hinge provides another reaction torque in addition to that, due to the MI of the door.

The scenario I have in mind is a door with the hinge pins pulled out, just barely stuck in the jamb (*). You pull on the knob. The door comes out of the doorway, rotating as it does so. But if your choice of reference axis is at the knob then there was no torque and no change in angular momentum. Nonetheless, the door is unambiguously changing its orientation -- it is rotating without a torque having been applied.

The counter-balancing angular momentum is, of course, accounted for by the linear motion of the door's center of mass.

(*) Assume that you've also trimmed away the wood in the door frame so that the hinge-ward 1/3 of the door width is free to push into the jamb without impediment.