Calculate the Torque

  • Thread starter Viraam
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  • #26
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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
 

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  • #27
108
34
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?
 

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  • #28
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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
 

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  • #29
66
2
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} ##
 
  • #30
66
2
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 dont 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.
 
  • #31
66
2
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} ##
 
  • #32
108
34
## 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.
 
  • #33
108
34
. 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.
 
  • #34
108
34
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
 
  • #35
108
34
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 !
 
Last edited:
  • #36
66
2
Please let them know they have a math errors in their problem !



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 !
You are right. No classroom door can be so small.
 
  • #37
66
2
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
Thanks for the tip.
 
  • #38
66
2
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.
I agree that one force is perpendicular but how is the other force parallel.
Now the force is 3.5N and thec lever arm is 0.3m so torque is ##3.5N \times 0.3m=1.05Nm##
Am I right.
 
  • #39
66
2
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.
I do understand that when force is applied parallel then there is no perpendicular distance so it's 0 and the troque is also zero.
 
  • #40
108
34
I do understand that when force is applied parallel then there is no perpendicular distance so it's 0 and the troque is also zero.
Correct.

When we split a force into 2, we have to calculate the torque from both forces.

But because of how we split the forces, one of the Torques is zero. So we ignore that Force.
 
  • #41
66
2
Correct.

When we split a force into 2, we have to calculate the torque from both forces.

But because of how we split the forces, one of the Torques is zero. So we ignore that Force.
How is one force parallel when we split a right angle equally. Thanks for helping
 

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  • #42
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How is one force parallel when we split a right angle equally. Thanks for helping

Nice picture. Suggested upgrades:
5N Force line
1) Put an arrow at the end of the 5N force. This shows what direction it is going in.
2) Did you use a compass to get the angle precise?
3) Did you make the length a nice length so that you can use a ruler and graph paper to solve the problem easily? If you have a ruler with inches, then let 1" equal 5N (for example). Then this line should be 5" long. This size easily fits on a single sheet of paper. The larger you make the problem, the more accurate the ruler method is for getting the right answer. If you have a metric ruler, perhaps you let 4 cm = 1 N. So then the line would be 5*4 = 20 cm long. Got it?

Vertical Force line
2) Shorten the vertical line for the second force. It may not be any taller than the 5N Force line.
3) Now if you measure the length of the line, you have your answer directly with no maths...
4) Use a different color for this line. So it is easily visually different from the original line.
5) Put an arrow on the end of it so that we know what direction it is going.

Horizontal Force line
5) You didn't draw your horizontal force line.
6) Same rules apply as for the vertical force line.
7) Angle between the Vertical and Horizontal lines must be 90 degrees. This makes the lines 'orthoganal'.
8) Draw this line in the same color as the vertical line. They are related. They are the two resulting lines when you split the 5N Force line.
9) What is the length of this line going to be? Same rule applies as did for the vertical line. It must be the same horizontal length as the 5N force line.
Note: Typically we would start at the 0 point for the 5N line and go left (in this case) until we reach the vertical line.
10) You can measure the length of this line, apply your scale factor (1"=5N or 4 cm = 5N) and read the Force straight of your graph with no additional maths.

Note: You do not have to draw the lines 'vertical' and 'horizontal'. You can draw them at any angle you wish. The only rule is that they start at the start of the 5N force, they must be perpendicular (orthogonal) to each other, and they must end at the end of the 5N force.
 

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