Where to open a door the easiest?

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Opening a heavy door is easier when pushing at the handle rather than halfway between the handle and the hinges due to the principles of torque. Torque is determined by the force applied and the distance from the pivot point (hinges), meaning a smaller force at the handle can create the same torque as a larger force closer to the hinges. The discussion emphasizes that the distance to the hinges remains constant along a vertical line of the door, leading to equal force requirements at any point along that line. Understanding these concepts helps clarify the mechanics of door operation. The insights provided enhance comprehension of how force and distance interplay in practical scenarios.
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Homework Statement



Hi everyone! Here is the question:

Can you open a heavy door more easily if you push at the handle or halfway between the handle and the hinges? Why?

Homework Equations



There may be some equations but I'm more interested in a general 'word' explanation.

The Attempt at a Solution



I would think that it would be easier to open the door at the handle since it is usually placed at the center of the door, perhaps making the pressure evenly displaced. However, I'm not sure if this is correct and I'd much rather have an answer that has a solid explanation applying the laws of physics. Thanks for any help.
 
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simplyphysics said:

Homework Statement



Hi everyone! Here is the question:

Can you open a heavy door more easily if you push at the handle or halfway between the handle and the hinges? Why?

Homework Equations



There may be some equations but I'm more interested in a general 'word' explanation.

The Attempt at a Solution



I would think that it would be easier to open the door at the handle since it is usually placed at the center of the door, perhaps making the pressure evenly displaced. However, I'm not sure if this is correct and I'd much rather have an answer that has a solid explanation applying the laws of physics. Thanks for any help.

Welcome to the PF. (and nice avatar BTW)

I think the point of the question is that the handle is at the edge of the door, versus pushing half-way between the handle and the hinges. It takes a torque to rotate the door open. What is the equation for torque in terms of the force and the lever arm length? How can you use that equation to help you answer this question? :smile:
 
Thanks for your quick response! I think I understand what the question is asking now from your guidance.

The equation for torque is definitely helpful in seeing the relationship between force and distance and how it all affects torque. I see now for my question that applying a small force at the handle can produce the same torque as a larger force applied closer to the hinges; the smaller the distance, the greater the force! I have an additional question…

Imagine a vertical line drawn along the length of a door, and say the door handle is in the middle of that line. Does it take the same amount of force to open the door at any point along that line?
 
simplyphysics said:
Thanks for your quick response! I think I understand what the question is asking now from your guidance.

The equation for torque is definitely helpful in seeing the relationship between force and distance and how it all affects torque. I see now for my question that applying a small force at the handle can produce the same torque as a larger force applied closer to the hinges; the smaller the distance, the greater the force!


I have an additional question…

Imagine a vertical line drawn along the length of a door, and say the door handle is in the middle of that line. Does it take the same amount of force to open the door at any point along that line?

What are your thoughts on this? Does the distance to the hinges change anywhere on that line?
 
The distance to the hinges does not change, that is, if the distance is to the axis of the hinges. So I guess if that is the same then the force would be equal along any point on a vertical line.
 
simplyphysics said:
The distance to the hinges does not change, that is, if the distance is to the axis of the hinges. So I guess if that is the same then the force would be equal along any point on a vertical line.

Correct-a-mundo! :smile:
 
berkeman said:
Correct-a-mundo! :smile:

Thanks for your help! I may ask for more of it when I get to the electronics section…I see that is your area of expertise!
 

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