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The question is:

A door is hinged at one end and is free to rotate about a vertical axis. Does its weight cause any torque about this axis? Give reason for your answer.

I think that the answer would be that the weight does produce a torque, since the position vector of each particle and its corresponding weight give the same sense of torque and hence the respective multiplied magnitudes are added up.

If my answer is correct, then there must be a counter-moment which prevents a door from rotating downwards and coming in contact with the floor. Which force produces this torque? I think that the answer might be in the fact that the the axis chosen and the screws attached to the door are distinct. The forces due to these screws must provide the counter-moment.

Are the answers true or need some tinkering/improvement?

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All your thinking is very good and correct. BUT you are missing a point in the question :)

Yes, the door's weight (which pulls from the center og gravity - no need to think about each particle of the door) creates a torque. The hinges then gives a counter torque. True.

BUT let's read the question:

A door is hinged at one end and is free to rotate about a vertical axis. Does its weight cause any torque about this axis?

Does the weight try to turn the door about a vertical axis? Nope! It tries to turn it about a horizontal axis. See the picture below.

To the comment:

There are several forces on the door. All of those cause torques about some axis. But when you choose an axis to look at, a torque will always be zero, no matter the size of the force, in precisely two cases:

  • When the force acts on the axis and
  • when it acts parallel to the axis.

It's all about what axis you look at. About a horizontal axis (that does not go through the center of gravity), you are right, there is a torque from the weight, which the hinges have to hold back. This causes stress in the hinges. But about any vertical axis, the weight is unable to make anything turn.

enter image description here

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  • $\begingroup$ r is the perpendicular distance (horizontal) of COG from the given axis and the force of gravity acts vertically downwards. So the torque must be non-zero. How can it be zero? A mathematical demonstration please. $\endgroup$ – user117913 Dec 14 '14 at 17:03
  • $\begingroup$ It's not about the size of anything. It's about the direction. Look at a door in your home. Make sure what axis it is turning about. Then ask yourself, which way is the weight pulling. Is it pulling in a direction that will turn the door about that axis? Nope. $\endgroup$ – Steeven Dec 14 '14 at 17:05
  • $\begingroup$ @user117913 Clarifying picture is added. There is a torque caused by the weight. That is true. But that torque is not turning the door around the vertical axis as the question asks for. It is turning the door about a horizontal axis. $\endgroup$ – Steeven Dec 14 '14 at 17:16

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