# Work in tightening a nut, given max torque and friction torque?

• Kermit_the_Phrog
In summary: If you are trying to tighten a nut, then the final equation would be Wmaxforce = Tw θ.If you are trying to tighten a nut, then the final equation would be Wmaxforce = Tw θ.
Kermit_the_Phrog
Homework Statement
As a nut is tightened through an angle θ , the friction between the nut and the threads on the bolt increases. This causes an increasing frictional torque, Tf = γθ as illustrated. The maximum forces, F, you can apply to the wrench gives a maximum torque Tw. What is the total work you do in tightening the nut?

A. Twθ
B. Tw (Tf/y)
C. (Tw^2)/(2y)
D. (Tw^2)/y
Relevant Equations
W = T θ
NOTE: THIS IS THE GRAPH PROVIDED

At first I approached this problem attempting to solve for the total work done, using the formula

Wtotal = Wmaxforce - Wfriction

I then subbed into the formula, representing the values of work as their torque value times theta, which gave me a longer algebraic solution that was not one of the multiple choice options listed.

ie,

Wtotal = Twθ - Tfθ
= Twθ - (Yθ)θ
= θ (Tw- yθ)

(this is very clearly wrong)

However, now that I've thought about it, the question is phrased in a specific way. It never really asks for the total work, only for the work that YOU do while tightening the nut :

So would it just be

Wmaxforce = T θ
= Tw θ

and thus option A?

Or is the real solution closer to my first approach?

NOTE: i am not really given the final answer so unfortunately I have no idea which final multiple choice answer is correct ¯\_(ツ)_/¯

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You need to do an integral because the torque is not constant but depends on the angle. The upper limit of the integral is the final angle which you have to find.

kuruman said:
You need to do an integral because the torque is not constant.

Oh okay, so I'd find the area under the graph of the frictional torque and use that?

Kermit_the_Phrog said:
Oh okay, so I'd find the area under the graph of the frictional torque and use that?
Yes, that's another way to integrate.

kuruman said:
Yes, that's another way to integrate.

Okay so I took my formula from before and use the integral of the friction torque this time, it gave me this:

Wtotal = Wmaxforce - Wfriction
= Tw θ - (1/2)yθ ^2
= θ (Tw - (1/2)yθ)

which unfortunately doesn't seem to match any of the answers either - did I do something wrong, or am i just missing something?

Kermit_the_Phrog said:
Wtotal = Wmaxforce - Wfriction
This is incorrect. The maximum torque is a limit and it can be reached only when the nut can no longer be turned. Until that point the torque is less than its maximum value. Forget your formula and go back to finding the area under the curve.

Also, since the nut has zero initial and final angular speeds, the change in kinetic energy is zero. Therefore, by the work-energy theorem, the net work is also zero. Since you and friction are the only agents that do work in this case, the work done by you is the negative of the work done by friction.

kuruman said:
Yes, that's another way to integrate.

okay, thank you for the help.

I think I have an idea though!

If we take the expression I just made, and find its derivative, can we find the value of theta when force is a maximum? (I.E., derive with respect to theta, and solve for the value of theta when W' is zero, then use this value in the original expression to find the value of W?)

I.E.,

W= Tw θ - (1/2)yθ ^2

so

W' = Tw - yθ

Set W' equal to zero

θ = Tw/y

Sub into the first equation

W = Tw (Tw/y) - (1/2)y (Tw/y)^2
= (Tw^2)/y - (Tw^2)/(2y)
= (Tw^2)/(2y)

However, upon reading your newest response, I see that the above solution is also likely wrong, so I'll keep working i guess.

kuruman said:
This is incorrect. The maximum torque is a limit and it can be reached only when the nut can no longer be turned. Until that point the torque is less than its maximum value. Forget your formula and go back to finding the area under the curve.

Also, since the nut has zero initial and final angular speeds, the change in kinetic energy is zero. Therefore, by the work-energy theorem, the net work is also zero. Since you and friction are the only agents that do work in this case, the work done by you is the negative of the work done by friction.

Okay, so if the total work is zero, shouldn't the final equation be

Wmaxforce = Wfriction
Twθ = (1/2)y(θ^2)
Tw = (1/2)yθ and at this point, I'm mostly lost, as I'm unaware of what to solve for any more, seeing that the "Wtotal" term no longer exists (its value is zero). (I might be more confused now that I was to begin with)

Kermit_the_Phrog said:
Wtotal = Wmaxforce - Wfriction
That is for the net work done on the nut. Since it finishes with no angular velocity, this is zero. And you don’t mean Wmaxforce, just Wforce.
Kermit_the_Phrog said:
Twθ =
It has already been pointed out to you that the work done by the applied torque is not equal to the maximum torque multiplied by the final angle.

haruspex said:
That is for the net work done on the nut. Since it finishes with no angular velocity, this is zero. And you don’t mean Wmaxforce, just Wforce.

It has already been pointed out to you that the work done by the applied torque is not equal to the maximum torque multiplied by the final angle.

Ah, I apologise, I am still learning, and this problem has really got me stumped for some reason. Thanks for bearing with me.

Anyway, in that case, would the work done by the applied torque be it's anti derivative?

Kermit_the_Phrog said:
would the work done by the applied torque be it's anti derivative?
Yes.

haruspex said:
Yes.

Alright, then I need to find an expression for the maximum torque's anti derivative. However, the question doesn't give me an expression for Tw in the way it does for Tf - Can you point me in the right direction please?

haruspex said:
Yes.
Okay, so I gave it a try any way and set Tw = Fθ, Meaning that its antiderivative would be

1/2F(θ)^2

however, using this in my other calculations didnt get me anywhere, as it did this:

Wforce = Wfriction
1/2F(θ)^2 = (1/2)y(θ^2)
so F =y

Which hit a dead end pretty quickly, since this expression didn't lead me anywhere in my calculations.

Man, i don't think this problem is supposed to be this hard, I feel like I am missing something obvious

The work done by torque ##\tau=y\theta## is $$W=\int_0^{\theta_{max}}\tau~d\theta.$$Find ##\theta_{max}## in terms of ##T_w##, then integrate. It's that simple.

kuruman said:
The work done by torque ##\tau=y\\theta## is $$W=\int_0^{\theta_{max}}\tau~d\theta.$$Find ##\theta_{max}## in terms of ##T_w##, then integrate. It's that simple.

Okay, I am with you on the integration part, but how do I relate Tw and θmax?

If the general expression for the torque is ##T_f=y\theta##, what is the specific expression for the torque when ##\theta=\theta_{max}##?

Kermit_the_Phrog
kuruman said:
If the general expression for the torque is ##T_f=y\theta##, what is the specific expression for the torque when ##\theta=\theta_{max}##?

In that case,

Tf = y θ

so

Tfmax = yθmax ?

which rearranges to give

θmax = Tfmax/y

If I am correct, then I should integrate the first expression now:

W = (1/2)y(θmax)^2

I then sub in the equation for θmax:

W = 1/2y(Tfmax/y)^2
=(Tfmax)^2/2y

Which is very close to answer C EXCEPT for the fact that it's in terms of Tfmax and NOT Tw. I'm guessing that these terms are equal in this case? If they are, that would make the proof work perfectly, but is this assumption unwarranted? How do I prove that Tfmax = Tw?

oh and many thanks for helping me all this way, I really appreciate it !

[Though I'm feeling pretty foolish now, seeing that the answer was that simple... :( ]

From the problem statement
Kermit_the_Phrog said:
The maximum forces, F, you can apply to the wrench gives a maximum torque Tw.
Tfmax is your invention; it is not given by the problem.

kuruman said:
From the problem statement

Tfmax is your invention; it is not given by the problem.
Hang on, if net work is zero, does net torque have to be zero too?

If that's true, then

0 = Tw - Tfmax

and therefore,

Tw = Tfmax , which works with my solution

Is this correct?

kuruman said:
From the problem statement

Tfmax is your invention; it is not given by the problem.

And I know that, but I can't simply take the expression

Tf = yθ

and replace Tf with Tw when I sub in θmax: the two terms are not necessarily the same, right?

Thus, I made a term for the torque of friction when theta is at its maximum, titled "Tfmax" as a place holder

kuruman said:
From the problem statement

Tfmax is your invention; it is not given by the problem.
Or is it simpler than this? we just assume that the made up term "Tfmax" is equal to the term "Tw" because the question states:

"The maximum forces, F, you can apply to the wrench gives a maximum torque Tw"

Thus, since Tfmax occurs when the force of friction is a maximum, it would give a "maximum torque Tw"?

Is it really that simple?

Kermit_the_Phrog said:
And I know that, but I can't simply take the expression

Tf = yθ

and replace Tf with Tw when I sub in θmax: the two terms are not necessarily the same, right?

Thus, I made a term for the torque of friction when theta is at its maximum, titled "Tfmax" as a place holder
##T_f## is a placeholder for any torque less than the maximum and ##T_w## is a placeholder for the maximum torque that you can apply to the nut. There is only one maximum value but many torque values less than that maximum. The problem called the placeholder for the maximum ##T_w## but you chose to rename it Tfmax. That's all.
$$Torque = \begin{cases} 0, & \theta =0 \\ y\theta, & 0 \lt \theta \lt \theta_{max} \\ T_w, & \theta = \theta_{max} \\ \end{cases}$$

kuruman said:
That's all.

Man, I really over thought this, huh.

Kermit_the_Phrog said:
Or is it simpler than this? we just assume that the made up term "Tfmax" is equal to the term "Tw" because the question states:

"The maximum forces, F, you can apply to the wrench gives a maximum torque Tw"

Thus, since Tfmax occurs when the force of friction is a maximum, it would give a "maximum torque Tw"?

Is it really that simple?
Yesssss! It's that simple.

kuruman said:
Yesssss! It's that simple.

Thanks again by the way - I really do appreciate the help. Enjoy the rest of your day, and you're the best.

kuruman

## 1. What is the purpose of tightening a nut?

The purpose of tightening a nut is to securely fasten two objects together. This is important in many applications, such as construction, automotive, and machinery, to ensure the safety and stability of the objects.

## 2. What is max torque and how is it determined?

Max torque, also known as maximum torque, is the amount of force that can be applied to tighten a nut without causing damage. It is determined by the material and size of the nut, as well as the material and strength of the bolt or screw it is being tightened onto.

## 3. What is friction torque and how does it affect the tightening process?

Friction torque refers to the amount of resistance or friction between the nut and the bolt or screw. It is caused by the threads and surfaces of the nut and bolt rubbing against each other. Friction torque can affect the tightening process by requiring more force to be applied to achieve the desired torque.

## 4. How do you calculate the amount of torque needed to tighten a nut?

The amount of torque needed to tighten a nut is calculated by considering the max torque and friction torque. The max torque is usually specified by the manufacturer and the friction torque can be estimated based on the materials and conditions. The total torque needed is the sum of these two values.

## 5. What are the consequences of over-tightening a nut?

Over-tightening a nut can lead to several consequences. It can cause damage to the threads of the nut or bolt, making it difficult to remove in the future. It can also cause the nut to break or strip, which can compromise the strength and stability of the objects being fastened. Additionally, over-tightening can cause unnecessary stress and potential failure of the surrounding components.

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