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steve_5959
Feb6-11, 06:30 AM
My door is 2000kg, it is 3m wide, 4m high and is on a hinge. for the moment i am saying that friction is negligable. How much torque is required to open my door. I have already said that the torque needed to open the door is acting through the C.O.G so it is 2000(1.5)(9.81), i don't know if that is correct though, please help :) If I can find this then I can find out the angular velocity and acceleration

russ_watters
Feb6-11, 08:10 AM
If friction is negligible and the door hinges are perfectly aligned, then any amount of torque will open the door.

How fast do you want the door to open?

steve_5959
Feb6-11, 09:18 AM
want door to open in ten seconds.

edgepflow
Feb6-11, 10:02 AM
You could apply the angular equal of Newton's laws for linear motion:

F = ma
torque = moment of inertia X angular acceleration
angular acceleration = change in angle / change in time.

Let me know if you need any help with this.