Calculate force to open a door with a gas spring

[Camiloae]

Homework Statement

How much force (F) in Newtons needed to open a 138 cm long (L) 50 kg door (W) at a contact point 98 cm. (A) away from the hinge ata 13 degree (C) angle?

Relevant Equations

I need the ecuation to calculate F

 

[hmmm27]

Is that the entire question ?
Is it a trapdoor, ie: does it open upwards ?

 

[haruspex]

kg is a mass, not a weight. Better to label it M, not W.
If the diagram is looking down from above the door then the force needed depends on friction. If there is no friction, even the tiniest force will open the door, it just might take a while.
If it’s a horizontal view, the door opening upwards, then the weight of the door matters, but it will be 45g, i.e. ##45kg*9.8m/s^2=440N##.

 

[bob012345]

First, figure out what condition is necessary to open the door then compute what value of force F meets that condition. A hint is to think about moments.

 

[rudransh verma]

kg is a mass, not a weight. Better to label it M, not W.
If the diagram is looking down from above the door then the force needed depends on friction. If there is no friction, even the tiniest force will open the door, it just might take a while.
If it’s a horizontal view, the door opening upwards, then the weight of the door matters, but it will be 45g, i.e. ##45kg*9.8m/s^2=440N##.

If it’s a horizontal view then how a=g. It’s not freefall motion.

 

[haruspex]

If it’s a horizontal view then how a=g. It’s not freefall motion.

If it’s a horizontal view ('elevation') then the door is opening upwards, like a trapdoor, as @hmmm27 suggested. But I doubt this is what is intended.

 

[rudransh verma]

If it’s a horizontal view ('elevation') then the door is opening upwards, like a trapdoor,

Then how a=g? a should be equal to acceleration of the door in upward direction!

 

[haruspex]

Then how a=g? a should be equal to acceleration of the door in upward direction!

I don't know where you are getting acceleration a from. What I wrote in post #3 is that if m is the mass then the weight is mg.

 

[rudransh verma]

If it’s a horizontal view, the door opening upwards, then the weight of the door matters, but it will be 45g, i.e. 45kg∗9.8m/s2=440N.

I don’t understand this. So I have to apply a force equal to its weight to raise the door?

 

[haruspex]

I don’t understand this. So I have to apply a force equal to its weight to raise the door?

Of course you have to apply a force to lift a massive door. Whether it is equal to the weight depends on the geometry. If the applied force were vertical and at the side furthest from the hinge, it would only need to be half the weight. In the diagram, the force is applied at an angle, and only about 70% of the way from the hinge.

 

[Camiloae]

I think my consultation needs extra information. The door is in horizontal position. The question is aimed at finding the force, in Newtons, for a custom made gas spring arm, like those used in SUV's and hatchback cars so that it is easier to lift the trapdoor. The F vector stands for the force that will be excerted by the gas-spring, at an angle of 13 degrees in order to achieve Static Equilibrium. The force (F) needs to compensate the weight of the door at the pivot point so that the sum of forces is cero and the door can be opened with little effort. The weight of the door at the pivot point of the arm is aprox 33 kilograms, (the point of excertion is not on the end, otherwise it will be 25 kg) and if I divide 33kg by sine of the angle, sin(13)= 0.224951, and then multiply it by 9.8 (gravity pull), the result for F will be aprox 1466 Newtons, which is the force needed for there to be Static Equilibrium when the door is in closed position. I'm asking this question because even though I have done some research, I want to be sure the answer is close to 1466 Newtons because if I order the gas-spring with too many Newtons the door will not close, and too little Newtons the door will be too heavy to open. I just need someone to confirm if my calculations are correct or if I need to correct something. Thanks!

 

[Camiloae]

Is that the entire question ?
Is it a trapdoor, ie: does it open upwards ?

Its, a trapdoor, please look at explanation above

 

[bob012345]

Ok so this is not really a homework problem for a class. The 13 degrees is the angle the gas spring is constrained to operate at during the initial opening. Your math seems correct but it seems to me that you can speak to the company you plan to order it from and they may be able to confirm the size and operating characteristics of the gas spring you need.

Here is one reference with a calculator to help with the design; This company offers guidance for your application.

https://www.guden.com/news/article/...uoIogLzThzBvcXLkIj2gj6_lxikldvORoCdnsQAvD_BwE

 

[bob012345]

As an aside, could you design this trap door with a balanced counterweight instead?

 

[Camiloae]

Ok so this is not really a homework problem for a class. The 13 degrees is the angle the gas spring is constrained to operate at during the initial opening. Your math seems correct but it seems to me that you can speak to the company you plan to order it from and they may be able to confirm the size and operating characteristics of the gas spring you need.

Here is one reference with a calculator to help with the design; This company offers guidance for your application.

https://www.guden.com/news/article/...uoIogLzThzBvcXLkIj2gj6_lxikldvORoCdnsQAvD_BwE

Hi there, thanks for your response, I'll take a look at your recommendation. I had to do all the math on my own since the guys down here in Colombia I am ordering the gas-spring from won't provide any support at all, which is a big bummer. Regarding your counterweight suggestion, the way the trapdoor is designes doesn't allow space for a counterwight system. The proyect requires a gas spring. Thanks again!

 

[bob012345]

Hi there, thanks for your response, I'll take a look at your recommendation. I had to do all the math on my own since the guys down here in Colombia I am ordering the gas-spring from won't provide any support at all, which is a big bummer. Regarding your counterweight suggestion, the way the trapdoor is designes doesn't allow space for a counterwight system. The proyect requires a gas spring. Thanks again!

I am just curious but could you show a simple sketch showing the restrictions? Also, what safety margin are designing into the system?

 

[Camiloae]

Hi, this is the sketch I made to make my calculations. I am going to order the arms with a little less force so that the door stays closed when left alone since there is no latching system. If I go over in force than it won't stay closed in equilibrium and I'd would rather have to exert a little bit more force when opening. I guess that is the "safety margin" you are wondering about.

 

[haruspex]

Hi, this is the sketch I made to make my calculations. I am going to order the arms with a little less force so that the door stays closed when left alone since there is no latching system. If I go over in force than it won't stay closed in equilibrium and I'd would rather have to exert a little bit more force when opening. I guess that is the "safety margin" you are wondering about.

I cannot understand your new diagram. The arm (black line) seems to be L shaped, but the two ends of the unequal arms would be adjacent when the door is closed, which is not possible.
Where is the arm hinged at the floor end?
If it is not a straight line, what shape is it and where on it is the spring?

 

[bob012345]

This is one kind of design;

Or perhaps the gas springs you will buy look like this?;

 

[Camiloae]

I cannot understand your new diagram. The arm (black line) seems to be L shaped, but the two ends of the unequal arms would be adjacent when the door is closed, which is not possible.
Where is the arm hinged at the floor end?
If it is not a straight line, what shape is it and where on it is the spring?

It looks L shaped because you are actually seeing both arm positions depicted simultaneously in the same drawing. The upright line represents the arm in open door position (110cm.) and the more slanted line represents the arm in closed door position. The data for closed position is the one I make my calculations with in order to find the static equilibrium. The arm will be hinged at the floor end against the lateral wall of the entrance (similar to your photo) but the arm door pivot will be farther from the hinge than the wall pivot, as in the new image:

 

[haruspex]

It looks L shaped because you are actually seeing both arm positions depicted simultaneously in the same drawing. The upright line represents the arm in open door position (110cm.) and the more slanted line represents the arm in closed door position. The data for closed position is the one I make my calculations with in order to find the static equilibrium. The arm will be hinged at the floor end against the lateral wall of the entrance (similar to your photo) but the arm door pivot will be farther from the hinge than the wall pivot, as in the new image:

Ah, ok.
So the equation you need is the torque balance in the closed position.
The torque from the weight of the door is its weight (mg) multiplied by the distance of its mass centre from the hinge (138cm/2).
The torque from the spring is the force it exerts multiplied by the perpendicular distance between its line of action and the hinge (98cm sin(13°)).

There is also the question of what happens as the door opens. Both torques and the spring force will change. There could be a risk that the spring will exert too much torque, producing a hazard as the door opens too fast, or making it difficult to reclose. But I imagine the gas pressure will drop too much for that to be a worry.

 

[Camiloae]

Ah, ok.
So the equation you need is the torque balance in the closed position.
The torque from the weight of the door is its weight (mg) multiplied by the distance of its mass centre from the hinge (138cm/2).
The torque from the spring is the force it exerts multiplied by the perpendicular distance between its line of action and the hinge (98cm sin(13°)).

There is also the question of what happens as the door opens. Both torques will change. There is a risk that the spring will exert too much torque, producing a hazard as the door opens too fast, or making it difficult to reclose. To analyse that we need the location where the spring is mounted at the floor end. There will be an ideal position. I'll try to calculate it. How deep is the joist on which it will be mounted?

I appreciate your help a lot. At the floor end the arm will be attached to a lateral wall.

I just revised my design and made sure all the input data is right. Please disregard previous diagrams, enclosed you will find the final design with some adjustments.

Regarding the pivot position of the arm against the wall, the horizontal distance from the door hinge is 42 cm and the vertical distance from the closed door is 15 cm. The arm will be connected to the door at a distance of 98 cm from door hinge and at a vertical distance from the door of 3 cm. Tis diagram is made on a scale of 1:10 on milimmiter paper.The angle is actually twelve (12) degrees. The gross door weight is 60 kg. That will mean 30 kg if lifting perpendicularly from the edge (138 cm from hinge). Since the force excerted by the arm is not going to be from the edge but 98 cm. from the hinge, my calculations result in aprox 39 kg. at that point. Now, what I think I should do is take the 39 Kg, divide it by sin(12) and multiply by 9.8. The result is 1838 Newtons. Since it there are going to be two (2) arms, and I want to be on the safeside (the door staying closed even if these means a little bit heavier to pull up) I am planning on ordering the arms with 850 Newtons each.

Note: The arm dimensions (57cm closed and 110 cm. open) are a fixed variable which I can't change. The arm fixing points positions are the best I could come up with regarding the fixed arm dimensions but making sure the door opened at least 75 degrees. I can play around with the fixing positions as long as they are not in the area with diagonal lines (concrete) and as long as the two pivot points are 57 cm apart and at least at a 12 degree angle (a smaller angle will mean too much force required).

Basically all variables are kind of fixed except the Newtons for the arms which is what the manufacturer asks me. It's a small shop down here in my country and they won't do calculations for me.

I don't have the force curve for the arms but I guess as long as the arms have up to 10% less force than the one required for static equilibrium I should be ok. I am even be considering the arms ordering with 20% less force (1500 Newtons for both arms) to be on the safe side. The door won't have a latch or locking system and this is not a possibility because of design requirements.

I just want to be somewhat sure that I am al least close to "ballpark" I don't either want the door to not close when left alone or to be too heavy for lifting!

 

[haruspex]

The angle is actually twelve (12) degrees. The gross door weight is 60 kg. That will mean 30 kg if lifting perpendicularly from the edge (138 cm from hinge). Since the force excerted by the arm is not going to be from the edge but 98 cm. from the hinge, my calculations result in aprox 39 kg. at that point. Now, what I think I should do is take the 39 Kg, divide it by sin(12) and multiply by 9.8. The result is 1838 Newtons.

So that's 60x138x9.8/(2x98sin(12))? I get 1991N, but that's a minor matter.
The only other thing I would check concerns my guess about how the force exerted changes as the arms extend. If it falls off a lot, fine; otherwise might need to ensure the torque is not excessive in the open position.

 

[Camiloae]

So that's 60x138x9.8/(2x98sin(12))? I get 1991N, but that's a minor matter.
The only other thing I would check concerns my guess about how the force exerted changes as the arms extend. If it falls off a lot, fine; otherwise might need to ensure the torque is not excessive in the open position.

1991N is correct, I was a bit sloppy with the rule of thirds calculating the mass of the door at 98cm.

I'll check with the manufacturer what the fall off is to be on the safe side.

Problem solved, thanks so much!