Calculate the Torque

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  • #1
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The Classroom door is of width 50 cm. If the Handle of the door is 20 cm from the edge and Force of 5N is applied on the handle. Calculate the torque?
I have attached what I think is the answer. But the answer on the site from where I took this question is 0.8 Nm. Here is the answer the site gives:-
Handle of the door is situated at 20 cm. Therefore line of action is 20/2 = 10 cm.
Length of the lever arm = d = 50 - 10 = 40 cm = 0.4 m
Force applied = 2N
Torque = F × d
= 2N × 0.4 m
= 0.8 Nm.
If this above answer is correct then please please tell me how this line of action and lever arm is calculated.
 

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  • #2
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Did the illustration come from the problem or did you make it? I think the issue maybe visualizing the problem.
 
  • #3
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Did the illustration come from the problem or did you make it? I think the issue maybe visualizing the problem.
I made it. I wanted to know if it is correct. I am not able to understand the question well.
 
  • #4
SteamKing
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Do you put the hinges supporting a door next to the handle, or at some other location?

This question pre-supposes you know how doors work.
 
  • #5
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Do you put the hinges supporting a door next to the handle, or at some other location?

This question pre-supposes you know how doors work.
Please explain the correct answer if possible. That diagram is what I thought the answer is,. But I am not pretty sure about what the question means. If possible could you explain with a diagram. I would be very grateful. Thank You
 
  • #6
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Please explain the correct answer if possible. That diagram is what I thought the answer is,. But I am not pretty sure about what the question means. If possible could you explain with a diagram. I would be very grateful. Thank You
I am a beginner at this. So please offer adequate help. Thank you
 
  • #7
sophiecentaur
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It isn't at all clear to me whether it's the torque to turn the handle or the torque to open the door that they want. There is no 'Force' labelled on the diagram and there should be one to make the question complete (direction and where it is actually applied (how often does one operate a door handle by pressing on the very end, for example?). If it only involves operating the latch then the radius of the door can have nothing to do with it.
Should be assigned to the 'Dodgy Question Bin" I think
 
  • #8
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It isn't at all clear to me whether it's the torque to turn the handle or the torque to open the door that they want. There is no 'Force' labelled on the diagram and there should be one to make the question complete (direction and where it is actually applied (how often does one operate a door handle by pressing on the very end, for example?). If it only involves operating the latch then the radius of the door can have nothing to do with it.
Should be assigned to the 'Dodgy Question Bin" I think
Please dont consider the diagram. There was no diagram with the question I just made it. Actually I am unable to interpret the question. It would be nice if you could help
 
  • #9
sophiecentaur
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With or without the diagram, the force in question needs to be specified (by you or your teacher) - what it does and where it is applied. Your diagram is not to blame if the info hadn't been supplied and if there was extra information in the question to confuse you, possibly.
 
  • #10
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With or without the diagram, the force in question needs to be specified (by you or your teacher) - what it does and where it is applied. Your diagram is not to blame if the info hadn't been supplied and if there was extra information in the question to confuse you, possibly.
So the question is not sufficient?
 
  • #11
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Can anyone guide where I went wrong?
 
  • #12
Nidum
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(1) Applied force has changed from 5N in problem description to 2N in the given solution .

(2) Given solution only makes sense if point of application of force on handle is offset 10 cm relative to attachment point .

Really though original question is gibberish - where did it come from ???
 
  • #13
sophiecentaur
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So the question is not sufficient?
In my opinion it is not sufficient. Does it actually talk about opening the door or turning the handle" Does it actually say where the hand force is applied? And does it actually say the axis about which it wants you to calculate the torque?
I think that constitutes a 'fail' even if the excuse is used that you should use common sense.
If I were answering that question (with my smartypants hat on) I would work it out for both axes (door hinge and handle spindle and write explicitly where I assumed the force to be applied on the handle. Now, assuming that the sums were done right then I couldn't be 'wrong' could I? :smile:
 
  • #14
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(1) Applied force has changed from 5N in problem description to 2N in the given solution .

(2) Given solution only makes sense if point of application of force on handle is offset 10 cm relative to attachment point .

Really though original question is gibberish - where did it come from ???
I took it from an educational site with the sole aim of practicing and this question seems to be making me more confused.
Site: http://formulas.tutorvista.com/physics/torque-formula.html
 
  • #15
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In my opinion it is not sufficient. Does it actually talk about opening the door or turning the handle" Does it actually say where the hand force is applied? And does it actually say the axis about which it wants you to calculate the torque?
I think that constitutes a 'fail' even if the excuse is used that you should use common sense.
If I were answering that question (with my smartypants hat on) I would work it out for both axes (door hinge and handle spindle and write explicitly where I assumed the force to be applied on the handle. Now, assuming that the sums were done right then I couldn't be 'wrong' could I? :smile:
Here is the site from where I got this question: http://formulas.tutorvista.com/physics/torque-formula.html :smile:
 
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  • #16
sophiecentaur
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Here is the site from where I got this question: http://formulas.tutorvista.com/physics/torque-formula.html :smile:
The question is nonsense, really. That could be something to do with the translation of the word "edge'. Looking at the answer and the "line of action" has been, somehow calculated by assuming that the door is partly open, perhaps.
It's not worth losing any sleep over it. :smile:
 
  • #17
sophiecentaur
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I just read this statement in that link you gave
"Torque is used only for rotational motion whereas moment can be consider for non rotational motions also."
That is very confusing. The torque on a torque spanner (the clue is in the name) applies to a stationary nut or bolt. I imagine that the statement applies to a particular, local, convention for the terms used.
 
  • #18
BvU
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I second Sophie. This site is so bad that it horrifies me. And the company is big, powerful and has a good reputation, that they are now ruining with enthousiasm. Weird. They should do their experiments with their internal network, not expose this nonsense to the rest of the world.

First they say "In short, torque is a moment of force" and then comes a Note in very bad english from which I must conclude that you can't use it for non-rotational motion. What ? The two are completely and utterly the same thing. I hate to write it down, but you are much better off with wikipedia explaining torque
Of course, hyperphysics is good place too (overview here).

In ## \tau = F \times d## they say: d = Perpendicular distance. Well, that is definitely not so.
In ## \tau = Fd\sin\theta## they say: "Where ##\theta## is the angle between the force applied and the axis of rotation". Well, that is definitely not so.

What they mean is $$ \vec\tau = \vec d \times \vec F$$ ##\vec d## is not a distance (a distance is a scalar -- i.e. a number) but a vector (something with a magnitude and a direction, just like a force). If you take a vector product (also known as cross product -- which google) of two vectors you get a third vector, ##\vec \tau## in this case.

I write ##\vec d \times \vec F ## and not ##\vec F \times \vec d## (the difference is a minus sign) because my convention is that ##\vec d## is the vector from the rotation axis under consideration to a point where ##\vec F## acts. (same convention as for the ##\vec r## in the Wiki lemma).

--
 
  • #19
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So the question is not sufficient?

Correct. The question is not sufficient. There are two possible cases.

My presumption is that the Force is being applied to the door and not the handle.

But they could be referring to the torque on the handle.

Consider solving both.

Torque on door:
Redraw your graphics to show the hinges on door.
Redraw your graphic to show handle location.
Make your assumptions about the Force - that is perpendicular to the door.
Then you can calculate the length of the torque arm (force to hinges) and then the torque.

Torque on handle:
Alternatively, you can identify the force as rotating the handle (not as likely)
Identify the pivot point of the handle.
Specify the location of the force on the handle
Make your assumption as to the orientation of the force (perpendicular to the handle)
The. Calculate the distance from the force to the pivot, and then calculate the torque on the handle.
 
  • #20
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Can anyone guide where I went wrong?
Consider drawing the top view of the door.
Show the hinges and the force location on the door.
 
  • #21
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The question is nonsense, really. That could be something to do with the translation of the word "edge'. Looking at the answer and the "line of action" has been, somehow calculated by assuming that the door is partly open, perhaps.
It's not worth losing any sleep over it. :smile:
Thank you. Now I am assured that this question was really not worth the time I wasted. Moreover I am just a beginner.
 
  • #22
66
2
I second Sophie. This site is so bad that it horrifies me. And the company is big, powerful and has a good reputation, that they are now ruining with enthousiasm. Weird. They should do their experiments with their internal network, not expose this nonsense to the rest of the world.

First they say "In short, torque is a moment of force" and then comes a Note in very bad english from which I must conclude that you can't use it for non-rotational motion. What ? The two are completely and utterly the same thing. I hate to write it down, but you are much better off with wikipedia explaining torque
Of course, hyperphysics is good place too (overview here).

In ## \tau = F \times d## they say: d = Perpendicular distance. Well, that is definitely not so.
In ## \tau = Fd\sin\theta## they say: "Where ##\theta## is the angle between the force applied and the axis of rotation". Well, that is definitely not so.

What they mean is $$ \vec\tau = \vec d \times \vec F$$ ##\vec d## is not a distance (a distance is a scalar -- i.e. a number) but a vector (something with a magnitude and a direction, just like a force). If you take a vector product (also known as cross product -- which google) of two vectors you get a third vector, ##\vec \tau## in this case.

I write ##\vec d \times \vec F ## and not ##\vec F \times \vec d## (the difference is a minus sign) because my convention is that ##\vec d## is the vector from the rotation axis under consideration to a point where ##\vec F## acts. (same convention as for the ##\vec r## in the Wiki lemma).

--
Ohh my! Thanks a lot for acquainting me about this. I am really grateful for the detailed solution you provided but I am just 14 and not aware about the trigonometric ratios and vector algebra yet. Well this question I posted I just for extra knowledge as I don't like to limit myself to the borders of the textbook. The syllabus we follow has only few topics for physics like Linear Motion, Equations of motion, Laws of Motion, Gravitation, Sound, Work and Energy. But I really. whole heatedly, appreciate your effort. Thanks a lot.
 
  • #23
66
2
Correct. The question is not sufficient. There are two possible cases.

My presumption is that the Force is being applied to the door and not the handle.

But they could be referring to the torque on the handle.

Consider solving both.

Torque on door:
Redraw your graphics to show the hinges on door.
Redraw your graphic to show handle location.
Make your assumptions about the Force - that is perpendicular to the door.
Then you can calculate the length of the torque arm (force to hinges) and then the torque.

Torque on handle:
Alternatively, you can identify the force as rotating the handle (not as likely)
Identify the pivot point of the handle.
Specify the location of the force on the handle
Make your assumption as to the orientation of the force (perpendicular to the handle)
The. Calculate the distance from the force to the pivot, and then calculate the torque on the handle.
Actually I am a beginner at this so your highly professional language is perplexing. As I mentioned above: I am just 14 and not aware about the trigonometric ratios and vector algebra yet. Well this question I posted I just for extra knowledge as I don't like to limit myself to the borders of the textbook. The syllabus we follow has only few topics for physics like Linear Motion, Equations of motion, Laws of Motion, Gravitation, Sound, Work and Energy. Thanks a lot for your response.
 
  • #24
66
2
Consider drawing the top view of the door.
Show the hinges and the force location on the door.
If possible would you guide me at the basic level. Without any complex Trigonometric Rations, Vector Algebra, Logarithms etc.
For me the ## \tau = F \times d ## equation is comprehensible
If possible do help using this formula only or else I better forget this question. After all you all have stated that the question is meagre.
 
  • #25
sophiecentaur
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In τ=F×d \tau = F \times d they say: d = Perpendicular distance. Well, that is definitely not so.
In τ=Fdsinθ \tau = Fd\sin\theta they say: "Where θ\theta is the angle between the force applied and the axis of rotation". Well, that is definitely not so.
That's a bit harsh. Millions (literally) of people use these elementary definitions, which work very well in most practical situations. You may find it hard to believe that there is life outside the world of Strict Mathematics, once that's where you live. 2D is more than enough for many people to cope with.
 
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