Torsion spring force required to open a door

[GreaseMonkey83]

I have a heavy door that I am adding torsion springs too in order to make it easier for a human to open.

-The door weighs 460lbs.
-It is hinged on one side by two heavy duty hinges and uses a thrust bearing in each hinge with a friction coef. of .008
-the effective radius of the thrust bearing is 31/32"
-The door handle is 56" from the hinge
-The torsion spring will apply its force 0.75" from the hinge.

Id like to beable to pull with only 5lbf normal to the door at the door handle.
I should beable to open the door 90* in 4 seconds

Do i have enough information to estimate the amount of force needed from the torsion spring?

If so, where do i start?? thanks in advance for your help.

 

[berkeman]

I have a heavy door that I am adding torsion springs too in order to make it easier for a human to open.

-The door weighs 460lbs.
-It is hinged on one side by two heavy duty hinges and uses a thrust bearing in each hinge with a friction coef. of .008
-The door handle is 56" from the hinge
-The torsion spring will apply its force 0.75" from the hinge.

Id like to beable to pull with only 5lbf normal to the door at the door handle.
I should beable to open the door 90* in 4 seconds

Do i have enough information to estimate the amount of force needed from the torsion spring?

If so, where do i start?? thanks in advance for your help.

Welcome to the PF.

I think the other thing you need is some idea of the effective radius of the thrust bearings in the hinges. That is what you will use to calculate the frictional torque that is opposing the opening torque. The smaller the effective radius of the thrust bearings (where the frictional force & torque are generated), the lower the force required to overcome that friction.

 

[GreaseMonkey83]

Ah yes, good point sir. It has been added.

 

[berkeman]

Ah yes, good point sir. It has been added.

Perfect! Now you just need to add up the torques and sum them to zero.

Torque(spring) + Torque(handle) + Torque(friction) = 0.

Do you know how to calculate the torques from the force X distance of each? Use the right-hand rule to get the directions of the torques right (+ is up and - is down). Are you familiar how to do that? And be sure to use consistent units -- maybe convert everything into meters, kilograms, etc. (MKS).

 

[GreaseMonkey83]

Yes... that seems a lot easier than what I was thinking. I thought the moment of inertia of the door would have to be factored in somehow. Thanks!

 

[Mech_Engineer]

Moment of inertia needs to be used to determine your angular acceleration for the net accelerating force. You say you want to open the door 90 degrees in 4 seconds, if you assume you're accelerating the door for the entire 4 seconds you can figure out how much torque (and therefore force) will be needed to achieve that acceleration and final travel distance in the time specified. That force will need to be added to your friction to determine the force needed to open it in the time required.

It may be the force to overcome inertia is much larger than the frictional force if you have good bearings...

See here: http://en.m.wikipedia.org/wiki/Angular_acceleration

Keep in mind also, this added spring force will "fight" the user when closing said door, assuming it's a one-way sort of mechanism.