Answer: Calculate Torque: 0.8 Nm

[Viraam]

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.

 

[UMath1]

Did the illustration come from the problem or did you make it? I think the issue maybe visualizing the problem.

 

[Viraam]

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.

 

[SteamKing]

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.

 

[Viraam]

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

 

[Viraam]

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

 

[sophiecentaur]

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

 

[Viraam]

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 don't 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

 

[sophiecentaur]

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.

 

[Viraam]

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?

 

[Viraam]

Can anyone guide where I went wrong?

 

[Nidum]

(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 ?

 

[sophiecentaur]

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?

 

[Viraam]

(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

 

[Viraam]

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?

Here is the site from where I got this question: http://formulas.tutorvista.com/physics/torque-formula.html

 

[sophiecentaur]

Here is the site from where I got this question: http://formulas.tutorvista.com/physics/torque-formula.html

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.

 

[sophiecentaur]

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.

 

[BvU]

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).

--

 

[NickAtNight]

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.

 

[NickAtNight]

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.

 

[Viraam]

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.

Thank you. Now I am assured that this question was really not worth the time I wasted. Moreover I am just a beginner.

 

[Viraam]

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.

 

[Viraam]

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.

 

[Viraam]

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.

 

[sophiecentaur]

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.

 

[NickAtNight]

Sure.

If possible would you guide me at the basic level.
For me the $\tau = F \times d$ equation is comprehensible.

The answer in your picture was correct for the problem you assumed.

You assumed the 5N force was applied at the end of a 0.2m (20cm) handle. You then correctly calculated the answer for that problem at 1 N-m. (5N times 0.2 m)

So Aplus on the math and applying the Torque equation.

However, that was probably not their intended question. So you would not get the books answer.

You stated the books answer. So we will make some guesses as to what they intended by that answer.

You want to open that door! yes? so grab the handle of the door and pull on it ! let's say you are pulling on the handle with a force of 5N.

Now the door will open by pivoting on its hinges. So the pivot length is from the hinges to your hand (see picture). So how long is that?

Where is your hand? Did you grab the edge of the door? the door is 50cm long so your d would be 50cm if you did.

Did you grab the edge of the handle? The handle is 20 cm long, so you are only 30 cm away from the hinges. (50cm- 20 cm)

Did you grab the middle of the handle? The answer you gave from the book assumes you did because they took 20cm and split it in half. 20cm/2= 10 cm. In this case, then you are 40 cm away from the hinges.

A better problem statement by the company would clear up this misinformation !

Once you have defined the problem you are going to solve, you already know how to do the math. It is the force times the distance.

You gave us two different forces, 5N and 2N, so staying with the 5 N times 0.4 m gives an torque of 2.0 N-m. If the force was only 2N, then the answer is 0.8 N-m.

PROBLEM SOLVING TECHNIQUE.
when faced with a badly written problem (usually on a test), you want to write down how you read the problem. Draw a simple graphic. write down your equations. Apply the equations. And then check that your units cancel properly. If allowed, ask the teacher to clarify the problem.

Given that you solved an acceptable interpretation of the written problem correctly, I would have given you full credit if you came in and reviewed the test with me.

See attached graphic

 

[NickAtNight]

If possible would you guide me at the basic level.

Torque is a term that is often used when tightening bolts with a wrench (see picture). The typical English units of measurement is usually foot-lbs. so if the manufacturer wants you to tighten the bolt to 30 ft-lbs and you have a 1 foot long wrench, you know you need to apply 30 lbs of force at the end of the wrench.

But perhaps you are only 14 and can only apply 10 lbs of force. What do you do. Why call your mom (or dad) of course. But they are not home. So you calculate how long a wrench you need. 30 ft-lbs/10 lbs = 3 feet.

Well, you do not have a longer wrench, but there is a nice 5 foot section of pipe that will fit over the wrench. So you put it over the wrench, select a distance 3 foot away and pull.

Now you have to be careful. I made the pipe longer than you need. if you grab it 5 foot away, you have to reduce the force you apply.

If you grab it at the end, 5 ft away, and apply 10 ft of force then you have just applied 50 ft-lbs ! You might have snapped off your bolt ! The manufacturer asked for 30 ft-lbs for a reason.

When you tighten bolts, you are actually stretching them. If you stretch a metal too much, it will snap.

At the plant, if the tightness of the bolt is critical to the sealing of the flange, we switch from using torque to measuring how tight the bolt is to (with a very fancy machine, measuring how much we have actually stretched the bolt.

Got it?

 

[NickAtNight]

If possible would you guide me at the basic level Without any complex Trigonometric ...

trig? You do not need any stink in' trig. You just need a ruler and a protractor ! And you probably learned to use both of them if first grade. right?

The next level of complication that gets thrown at you is that the force will be applied at an angle to the door.

We use trig to solve those problems. But a ruler, some gridded graph paper and protractor will do just as well. See attached picture.

You little sister is standing in the way. So you cannot push directly on the door with your 5N force. You have to stand off to the side and push it so that you do not get cooties. Let's say that the angle is 30 degrees from plum. What happens to your 5 N force?

Well, draw draw the door and the. Draw a 5" long line at a 30 degree angle to the vertical line to the door. We are going to say 5N is a 5" line.
Now draw a straight line over to underneath where you should have been pushing from. and then a straight line up to the door. measure both lines.

The first line should be 1/2 of 5" ! One 2.5". That would be 2.5 N
The second. Line should be 4.33". That would be 4.33 N

So of your 5N push, only 4.33 N goes towards opening the door (is that still enough?) and 2.5 N is wasted effort.

Got it.

Now try some different angles. Try. 0 degrees. 30 degrees. 45 degrees. 60 degrees and 90 degrees.

What happens to the force used to push open the door as the angles change?

Now why did I choose those angles, because they are special. Do you see the easily memorized relationship below?

0 degrees. Sqrt 0/ sqrt 4 = 0/2 = 0
30 degrees. Sqrt 1/ sqrt 4 = 1/2 = 0.5
45 degrees. Sqrt 2 / sqrt 4 = sqrt 2/ 2 = 0.707
60 degrees. Sqrt 3 / sqrt 4 = sqrt 3/ 2 = 0.866
90 degrees. Sqrt 4 / sqrt 4 = 2/ 2 = 1

 

[Viraam]

Sure.The answer in your picture was correct for the problem you assumed.

You assumed the 5N force was applied at the end of a 0.2m (20cm) handle. You then correctly calculated the answer for that problem at 1 N-m. (5N times 0.2 m)

So Aplus on the math and applying the Torque equation.

However, that was probably not their intended question. So you would not get the books answer.

You stated the books answer. So we will make some guesses as to what they intended by that answer.

You want to open that door! yes? so grab the handle of the door and pull on it ! let's say you are pulling on the handle with a force of 5N.

Now the door will open by pivoting on its hinges. So the pivot length is from the hinges to your hand (see picture). So how long is that?

Where is your hand? Did you grab the edge of the door? the door is 50cm long so your d would be 50cm if you did.

Did you grab the edge of the handle? The handle is 20 cm long, so you are only 30 cm away from the hinges. (50cm- 20 cm)

Did you grab the middle of the handle? The answer you gave from the book assumes you did because they took 20cm and split it in half. 20cm/2= 10 cm. In this case, then you are 40 cm away from the hinges.

A better problem statement by the company would clear up this misinformation !

Once you have defined the problem you are going to solve, you already know how to do the math. It is the force times the distance.

You gave us two different forces, 5N and 2N, so staying with the 5 N times 0.4 m gives an torque of 2.0 N-m. If the force was only 2N, then the answer is 0.8 N-m.

PROBLEM SOLVING TECHNIQUE.
when faced with a badly written problem (usually on a test), you want to write down how you read the problem. Draw a simple graphic. write down your equations. Apply the equations. And then check that your units cancel properly. If allowed, ask the teacher to clarify the problem.

Given that you solved an acceptable interpretation of the written problem correctly, I would have given you full credit if you came in and reviewed the test with me.

See attached graphic

Thank You! I am actually able to understand through your wonderful explanation. You seem to be proficient at this. Even I had the doubt how does the force convert into 2N from 5N. Here is the link from where I got the question: http://formulas.tutorvista.com/physics/torque-formula.html#torque-problems
Now can you tell me if my solution is right, if I apply the force of 5N at the edge of the handle.
$F = 5N \\ d = 50 cm - 20 cm = 30 \ cm = 0.3 m \\ \tau = F \times d \\ \Rightarrow \tau = 5N \times 30 cm \\ \Rightarrow \tau = 5N \times 0.3 m = \fbox{1.5 Nm}$

 

[Viraam]

Now why did I choose those angles, because they are special. Do you see the easily memorized relationship below?

0 degrees. Sqrt 0/ sqrt 4 = 0/2 = 0
30 degrees. Sqrt 1/ sqrt 4 = 1/2 = 0.5
45 degrees. Sqrt 2 / sqrt 4 = sqrt 2/ 2 = 0.707
60 degrees. Sqrt 3 / sqrt 4 = sqrt 3/ 2 = 0.866
90 degrees. Sqrt 4 / sqrt 4 = 2/ 2 = 1

Thanks again. I have understood the example you gave with the help of a diagram(The 5N at 30$^{\circ}$ example). The only thing that I don't seem to understand well is the relationship you have mentioned about the angles. It would be great if you could throw some more light on that. I also wanted to ask: If I do happen to apply a force of 5N on the door at a 45 degree angle then the force that is actually applied on the door is $3.5N$. Is this right.

 

[Viraam]

Torque is a term that is often used when tightening bolts with a wrench (see picture). The typical English units of measurement is usually foot-lbs. so if the manufacturer wants you to tighten the bolt to 30 ft-lbs and you have a 1 foot long wrench, you know you need to apply 30 lbs of force at the end of the wrench.

But perhaps you are only 14 and can only apply 10 lbs of force. What do you do. Why call your mom (or dad) of course. But they are not home. So you calculate how long a wrench you need. 30 ft-lbs/10 lbs = 3 feet.

Well, you do not have a longer wrench, but there is a nice 5 foot section of pipe that will fit over the wrench. So you put it over the wrench, select a distance 3 foot away and pull.

Now you have to be careful. I made the pipe longer than you need. if you grab it 5 foot away, you have to reduce the force you apply.

If you grab it at the end, 5 ft away, and apply 10 ft of force then you have just applied 50 ft-lbs ! You might have snapped off your bolt ! The manufacturer asked for 30 ft-lbs for a reason.

When you tighten bolts, you are actually stretching them. If you stretch a metal too much, it will snap.

At the plant, if the tightness of the bolt is critical to the sealing of the flange, we switch from using torque to measuring how tight the bolt is to (with a very fancy machine, measuring how much we have actually stretched the bolt.

Got it?

Awesome Explanation. Just understood each word of it. Now, if I do want to apply a Force at the end of 5 foot pipe then the force must be reduced. So, is my calculation for finding the reduced force correct:
$\tau = F \times d \\ 30 = F \times 5 \\ F = \frac{30}{5} = 6 \text{ lbs}$

 

[NickAtNight]

$F = 5N \\ d = 50 cm - 20 cm = 30 \ cm = 0.3 m \\ \tau = F \times d \\ \Rightarrow \tau = 5N \times 30 cm \\ \Rightarrow \tau = 5N \times 0.3 m = \fbox{1.5 Nm}$

Correct. For the problem specified.

 

[NickAtNight]

. I also wanted to ask: If I do happen to apply a force of 5N on the door at a 45 degree angle then the force that is actually applied on the door is $3.5N$. Is this right.

Close. The math is correct. (5N * 2^.5/5 = 3.5 N

The Force on the door is 5N at 45 degrees

You are splitting that force into two forces!

Since the angle is 45 degrees, the two new forces are the same size.

One Force is perpendicular to the door.
The other Force is parallel to the door.

Each Force has a distance from the door hinges.

For the Force perpendicular, the lever are Is the width from the hinges.

For the Force parallel to the door, the distance is zero. So the parallel force does not apply any torque.

 

[NickAtNight]

So, is my calculation for finding the reduced force correct:
$\tau = F \times d \\ 30 = F \times 5 \\ F = \frac{30}{5} = 6 \text{ lbs}$

Correct.

Pointer: always write down the units.
Do a unit analysis and make sure you end up with the right result.

Test problems like to mix and match units to catch you sleeping. Like the sample problem using cm when you are going to use m.

In this case. F is in N-m
The distance is in m
So when you divide the F by the d
The meters cancel and you are left with N

$\frac {N-m} {m} = N$

If you distance was in cm, then you would have to multiply by 100 cm/m so that you can cancel out the cm and the m to gave only N left

$\frac {N-m} {cm} = N-m/cm$

$\frac {N-m} {cm} \frac {100 cm} {m}= N$

Unit analysis is an important skill to learn. It allows you to find silly math errors caused by using the wrong units in an equation. It also allows your error checker/ reviewer to quickly check your results

 

[NickAtNight]

Please let them know they have a math errors in their problem !

Here is the link from where I got the question: http://formulas.tutorvista.com/physics/torque-formula.html#torque-problems
...
Solved Examples
Question 1: The width of a door is 40 cm. If it is opened by applying a force of 2 N at its edge (away from the hinges), calculate the torque produced which causes the door to open?
Solution:

Force applied = F = 2 N
Length of lever arm = d = 40 cm = 0.40 m (since distance between axis of rotation and line of action of force is 40 cm)
Torque = F × d
= 0.40 × 20
= 8 Nm.

Question 2: 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?
Solution:

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.

Question 1 geez ! Let's fix it.

Torque = F × d
= 2 N x 0.4 m
= 0.8 N-m.

Fixes
1) The force is 2N not 20 N !
2) You put the numbers into the equation in the same order. They flipped them and put in distance force instead of Force Distance
3) the answer is 0.8 N-m not 8 N-m

Question 2 - aw man ! Question is bad.

1) why did they change the door width?
They should have left it the same for the concept they showing - moving in the Force from the doors edge

2) the fancy door handle should have wen simpler. No dividing by two. Just position the handle x distance in and be done with it. I would do a fancy handle in the third exercise, not the second.

3) they changed the force from 2 N to 5 N from the first problem, but then revert to 2N in the problem!

Let's leave the Force at 2N
Leave the width at 40 cm
Put the handle at 10 cm from the doors edge
So handle to hinges is door width less handle distance = 40 - 10 = 30 cm

So force times distance is 2 N * 0.3 m is 0.6 N-m

POINT: Oh, and are they kidding about the 'classroom' door width. Is this a classroom for midgets? How wide should a classroom door be... 36" is probably standard these days. At 2.4 cm to the inch these doors need to be 92 cm wide !