Answer: Calculate Torque: 0.8 Nm

[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 !

 

[Viraam]

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.

 

[Viraam]

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.

 

[Viraam]

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.

 

[Viraam]

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.

 

[NickAtNight]

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.

 

[Viraam]

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

 

[NickAtNight]

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.