What is the torque needed to open a 2000kg door?

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SUMMARY

The torque required to open a 2000kg door measuring 3m wide and 4m high, assuming negligible friction, is calculated using the formula torque = force x distance from the hinge. The force acting through the center of gravity (C.O.G) is 2000kg multiplied by 9.81m/s² (gravity), resulting in a torque of 2000 x 1.5 x 9.81. To achieve a specific angular velocity, such as opening the door in ten seconds, one must apply the principles of angular motion, including torque, moment of inertia, and angular acceleration.

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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
 
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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?
 
want door to open in ten seconds.
 
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.
 

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