Calculate output forces of machines

[UMath1]

Based on a force inputed in a machine, how do you calculate the force outputed? I would have thought you could do this with a simple torque balance but this does not always appear to be the case.

For example in the case of an accelerating bicycle, the force applied to the rear gear is Ft, but the force the bicycle applies on the ground is not Ft*(Rg/Rw) but slightly less than that. Rg is the radius of the rear gear and Rw is the radius of the rear wheel.

So it appears the output torque seems to lag slightly behind the input torque. Is this always the case?

 

[russ_watters]

For example in the case of an accelerating bicycle, the force applied to the rear gear is Ft, but the force the bicycle applies on the ground is not Ft*(Rg/Rw) but slightly less than that.

On what do you base that? What happens to the rest of the torque? It looks wrong to me: yes, in the real world, torque balance is used.

 

[UMath1]

It is less because if if it were equal the torque of static friction would be equal to the torque of the chain and the wheel would not angularly accelerate. In a previous thread it was established that the output force of wheel was
(F_t*R_g)/(R_w) * (M + m_f)/ (M + m_f + m_w)

Where m_f is the mass of the front wheel assembly, m_w is the mass of the rear wheel assembly, and M is the mass of the total bike+rider system.

 

[russ_watters]

Right, I remember what the issue was now. So yes, you still do a torque balance, you just may need to add a term to account for the angular acceleration, just like in some cases you might add a term to account for linear acceleration if the forces are otherwise unbalanced.

How you tackle a particular problem, though, will depend on what you want to know and how precisely you want to know it. For example, with a bike I doubt the rotational inertia is a significant factor because of how light the wheels are.

 

[UMath1]

So where does the extra torque go though? I feel like I plug numbers in equations but I don't quite understand the concept.

Why does this only happen the object is accelerating? Or is that even true?

 

[russ_watters]

So where does the extra torque go though? I feel like I plug numbers in equations but I don't quite understand the concept.

Why does this only happen the object is accelerating? Or is that even true?

The extra force is in the terms ma (linear acceleration) or i∝ (angular acceleration). That's Newton's second law.

I thought you already knew this - we beat it to death in several other threads and via pm. I really don't see what the problem is.

 

[UMath1]

Yeah but that's the difference between the input and output torques. The input torque is equal to the output torque + i∝. So there is a deficit of i∝ in the output torque. Where is this torque going?

 

[russ_watters]

Yeah but that's the difference between the input and output torques. The input torque is equal to the output torque + i∝. So there is a deficit of i∝ in the output torque. Where is this torque going?

I just answered that and your question contains it's own answer:
i∝ is the answer.

I don't understand why you keep asking variations of the same question over and over and ignoring the answers. Heck, didn't you even put together a functional equation describing the situation in a PM to me? It works fine. There is no discrepancy. What problem, specifically, do you think exists here? If you apply what you apparently know to solving a real problem, you'll get the right answer. Do you just not believe that f=ma works?

Here, solve this problem:
A 1kg block is sitting on the floor. The static friction coefficient is 0.7 and the dynamic friction coefficient is 0.5. You apply a 10N horizontal force to it. What happens?

 

[UMath1]

Initially the block has a net force of 6.86 N and it experiences an initial acceleration of 6.86 m/s^2. Once you let go the only net force is kinetic friction and it causes a net acceleration of -4.9 m/s^2.

I don't have a problem with f=ma. I just don't understand what principles are applied when you calculate the output force of a machine. Let's say for example you have a lever of 10 m with its fulcrum located at 7.5 m from the origin. An object is placed at the shorter end. Now when you apply a force downwards (Ff) over a distance (Dw) on the longer end you input energy into the system by doing work (Ff*Dw). The object then travels up a distance (Dh). Most textbooks would say in this case the plank applied a force of (Ff*Dw)/Dh on the block because work in =work out if frictional losses are neglected. There are two things I don't understand: 1) how come all the energy is transmitted solely to the block? what if there was no block, where would the energy go then? 2) as has been established before the force must be less than (Ff*Dw)/Dh in order for rotational acceleration to occur. So what causes the loss in energy?

 

[dean barry]

ignoring frictional losses, the work done (f*d) at each end is the same

 

[russ_watters]

I just don't understand what principles are applied when you calculate the output force of a machine.

As far as I can tell, you already know the relevant/basic principles.

Lets say for example you have a lever of 10 m with its fulcrum located at 7.5 m from the origin. An object is placed at the shorter end. Now when you apply a force downwards (Ff) over a distance (Dw) on the longer end you input energy into the system by doing work (Ff*Dw). The object then travels up a distance (Dh). Most textbooks would say in this case the plank applied a force of (Ff*Dw)/Dh on the block because work in =work out if frictional losses are neglected. There are two things I don't understand: 1) how come all the energy is transmitted solely to the block?

Where else would it go?

what if there was no block, where would the energy go then?

What energy? If you describe the situation more specifically (and attach equations or better yet, numbers), the answer should be obvious.

2) as has been established before the force must be less than (Ff*Dw)/Dh in order for rotational acceleration to occur. So what causes the loss in energy?

What loss in energy? As far as I can tell, you fully accounted for all of it. Again: write out the equation for what you describe and it should all be there.

It seems you have some vague discomfort here, caused in part by your vague treatment of the subject. If you be more specific with your descriptions and your math, you will find that you understand the issues fine and the math works fine.

 

[UMath1]

Work in= (Ff*Dw)
Work out = "a little less than (Ff*Dw)/Dh" * Dh

Therefore work in does not equal work out.

If there were no block and I pushed the long end of the plank down a distance Dw and held it in place as it touched the ground. Where would the energy from from my work (Ff*Dw) go?

While I vaguely understand how to calculate the output force of a lever, I feel completely lost on calculating the force output of a bicycle wheel. Given Fc is the force of the chain on the Rear gear, Rg is the radius of the rear gear, and Rw is the radius of the rear wheel, why is the force output close to or a little less than (Fc*Rg)/Rw. What would would be work in and work out in this case?

 

[russ_watters]

Work in= (Ff*Dw)
Work out = "a little less than (Ff*Dw)/Dh" * Dh

Therefore work in does not equal work out.

C'mon; you already know that's not the only work being done. What are we discussing here?

If there were no block and I pushed the long end of the plank down a distance Dw and held it in place as it touched the ground. Where would the energy from from my work (Ff*Dw) go?

How could you have pushed the long end of the plank down if there is no block on the short end? What was the plank doing before you started pushing on i?

 

[UMath1]

Is the difference in work the change in energy of the plank?

Sorry I mean the short end of the plank.

 

[russ_watters]

Is the difference in work the change in energy of the plank?

Sorry I mean the short end of the plank.

You need to be clearer because now I don't know what your scenario looks like. And again, your lack of clarity is almost certainly what is tripping you up.

 

[UMath1]

I mean in the case where there is a block on the short end. When you push down on the long end, why doesn't some of the energy inputed go to the plank, why does all of it go to the block in the form of kinetic energy.

2) How do we calculate the forces for the bike wheel? What is work in and work out in this case? The force of the chain on the rear gear doesn't act over a distance so it doesn't to work right? So how exactly do we calculate the output force?

 

[russ_watters]

I mean in the case where there is a block on the short end. When you push down on the long end, why doesn't some of the energy inputed go to the plank, why does all of it go to the block in the form of kinetic energy.

How real-world do you want it to be? Based on previous constraints, I'd say that when you push down on a lever to lift an object, applying a constant force, energy goes to:
1. Increase in potential energy of the object.
2. Increase in kinetic energy of the object.
3. Increase in kinetic energy of the plank.
4. Decrease in potential of the plank (it starts with the long end in the air).

2) How do we calculate the forces for the bike wheel?...

So how exactly do we calculate the output force?

You already constructed an equation for this. It works fine.

What is work in and work out in this case?

You have the equation for the forces. What is the definition of work? So you tell me: how do you use the equation for the forces to get the work(s)?

The force of the chain on the rear gear doesn't act over a distance so it doesn't to work right?

Is the chain/gear moving (rotating)? If it is moving it acts over a distance.

 

[UMath1]

Is it torque of chain*theta = force of wheel on ground*distance traveled?

The equation for output force is derived from the concept of work in=work out. Thats why I want to understand this.

 

[russ_watters]

Correct, since you convert from radius to circumference by multiplying by 2pi.

 

[jack action]

So where does the extra torque go though? I feel like I plug numbers in equations but I don't quite understand the concept.

OK, I'll give it a try.

Imagine you are pushing with a force $F_{in}$ on a block with a mass $m$ that is pushing onto something else with a force $F_{out}$.

Now, let's look at this on an energy point of view. Energy must balanced out, it cannot be created nor destroy. So the energy in will be equal to sum of the energy out and the kinetic energy change of the block. In math form, we write:
$$dE_{in} = dE_{out} + dE_m$$
$$F_{in}dx = F_{out}dx + mvdv$$
$$F_{in}dx = F_{out}dx + m\frac{dx}{dt}dv$$
$$F_{in}dx = F_{out}dx + m\frac{dv}{dt}dx$$
$$F_{in}dx = F_{out}dx + madx$$
$$F_{in} = F_{out} + ma$$
You can see now that the $ma$ term is in fact necessary to account for the change in kinetic energy of the block. If the block doesn't accelerate or if it has no mass, then there is no change in kinetic energy and $F_{in} = F_{out}$.

 

[UMath1]

I think I understand it better now..but how did you get mvdv?

 

[jack action]

Actually, the $mvdv$ comes from $ma$ and that's how we get to the definition of kinetic energy $E = \frac{1}{2}mv^2$:
$$\begin{align*} dE &= Fdx \\ dE &= madx \\ dE &= m\frac{dv}{dt}dx \\ dE &= m \frac{dx}{dt}dv \\ dE &= mvdv \\ \int_{E = E_0}^{E_f} dE &= \int_{v = v_0}^{v_f} mvdv \\ E_f - E_0 &= \frac{1}{2}m(v_f^2 - v_0^2) \end{align*}$$
Or, if $E_0 = 0$ and $v_0 = 0$:
$$E_f = \frac{1}{2}mv_f^2$$
But since the concept of energy is often easier to visualize, I went backward to explain why $ma$ has to exist and where the torque goes.

So how did Newton came up with his second law of motion? Answer: Observation. The basis for all science.

Simply put, motion was first quantified. If an object was going twice as fast, therefore it had twice the «motion» and if its mass was twice as big, then there was twice the quantity of «motion». Motion was called momentum ($p$) and, in math form, the previous statement reads $p=mv$. It's an arbitrary definition made by humans based on observation, so it is true by definition.

Then Newton observed that the force needed to vary the quantity of «motion» was proportional to that variation over a time period $t$. In math form, $F = \frac{p_f-p_0}{t}$ or in a more precise differential format: $F = \frac{dp}{dt}$ which in turn lead to $F = \frac{d(mv)}{dt} = m\frac{dv}{dt} = ma$.

Here's a nice text that explains the context in which that observation was made:

Newton Clarifies the Concept of Force
Michael Fowler

Initial Confusion…
Actually, it took him years. That may seem surprising now, since we are so used to the force of gravity, and ordinary “push” forces, not to mention electrical forces, and so on. In Newton’s time, though, it was still widely believed that if something was moving, there had to be a force acting (the old Aristotelian view) and, if you couldn’t see an external force acting, there must be one inside the body. This was perhaps partially a confusion of force and momentum—if a body at rest was struck and it began to move, the idea was that the force that struck it was now in the body. After all, if the body in motion now struck you, you would feel the force! And in fact Newton himself believed this at first. In his first foray into mechanics, in 1665, he wrote an essay “On Violent Motion” in which he accepted the view that bodies are kept in motion by a force inside them. (Never at Rest, by Richard S. Westfall, page 144. These notes are largely based on that book.) A little later, after reading Galileo and Descartes, he espoused the concept of inertia, as stated in his First Law (but essentially discovered by Galileo and stated very clearly by Descartes.) Maybe the difficulty here is notational more than conceptual—if the word “force” is used to describe an impulse, which in modern terminology would be force×time, then this “force” has the same dimensions as momentum, and such an (impulsive) “force” on a body delivers an equivalent amount of momentum to it. If the body were initially at rest, this momentum could be labeled (confusingly!) as the “force” of the motion, which would be delivered as an impulse if the body hit a wall. If this is the correct interpretation of usage at that time, perhaps the viewpoint wasn’t really Aristotelian, just the language. But I’m not an expert on this—the interested reader can find references in Westfall’s book.

Once he accepted the concept of inertia, that a body will continue in steady motion if no forces are acting on it, it was natural to restrict the definition of force to that which caused change in motion: “Force is the pressure or crowding of one body upon another” and he concluded that there must be a direct relation between the magnitude of the external impact and the change in motion that occurs. Westfall (page 146) quotes Newton: “So much force as is required to destroy any quantity of motion in a body so much is required to generate it, and so much as is required to generate it so much is also required to destroy it.” He also wrote a little later: “Tis known by the light of nature, that equal forces shall effect an equal change in equal bodies.” The major advance here is the concept of force as an external agent acting on a passive body, and causing a proportionate change in the body’s motion—in other words, making a clear distinction between force and momentum.

- How Did Newton Get to the Second Law?​

​

 

[UMath1]

Thanks that clears up most of my confusion! I just want to confirm I understand it correctly for a bicycle wheel undergoing acceleration.

dEin = Torque of Chain dΘ + Force of Friction dx. dOut = Force output=force of friction * R dΘ. So delta KE= (Torque of Chain dΘ + Force of Friction dx) - (Force output=force of friction * R dΘ), correct?

Also, my teacher said differentials aren't fractions, so would steps 3 and 4 in your simplification still work if they aren't fractions?

 

[jack action]

For rotation, it is the same, except replacing force with torque, mass with inertia and velocity & acceleration with their rotational equivalent.

$$dE_{in} = dE_{out} + dE_m$$
$$T_{in}d\theta = T_{out}d\theta + I\omega d\omega$$
$$T_{in}d\theta = T_{out}d\theta + I\frac{d\theta}{dt}d\omega$$
$$T_{in}d\theta = T_{out}d\theta + I\frac{d\omega}{dt}d\theta$$
$$T_{in}d\theta = T_{out}d\theta + I\alpha d\theta$$
$$T_{in} = T_{out} + I\alpha$$

In the case of a bicycle wheel that is actually moving a bicycle, the $I$ component will carry the equivalent inertia of the bicycle of mass $m$ going at speed $v$, plus the inertia of the rear rotating wheel $I_r$, plus the inertia of the front rotating wheel $I_f$, both with radius $R_r$ and $R_f$. To find out $I$, the following equation must be true:

$$\frac{1}{2}I\omega_r^2 = \frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_f\omega_f^2 + \frac{1}{2}mv^2$$
$$\frac{1}{2}I\omega_r^2 = \frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_f\left(\frac{v}{R_f}\right)^2 + \frac{1}{2}m \left(R_r\omega_r\right)^2$$
$$\frac{1}{2}I\omega_r^2 = \frac{1}{2}I_r\omega_r^2 + \frac{1}{2}I_f\left(\frac{R_r}{R_f}\right)^2\omega_r^2 + \frac{1}{2}m R_r^2\omega_r^2$$
Simplifying:
$$I = I_r + I_f\left(\frac{R_r}{R_f}\right)^2 + m R_r^2$$

For more info on how to calculate equivalence between mass and rotational inertia, see this page.

 

[UMath1]

How is there a torque out though? The force the wheel applies to the ground has no moment arm.

 

[jack action]

What about the tire radius?

 

[UMath1]

The tire is considered a part of the wheel though, right? The tire+wheel imparts a force parallel to the ground and then static friction imparts an equal but opposite force back to the wheel in the forward direction. So energy out would be F dx, where F is the force the wheel puts on the ground. But then energy comes back in thriugh static friction, but this energy is divided between the wheel and the whole bike. So energy in would be F Mwheel/Mbike dx?

 

[jack action]

The tire is considered a part of the wheel though, right? The tire+wheel imparts a force parallel to the ground and then static friction imparts an equal but opposite force back to the wheel in the forward direction. So energy out would be F dx, where F is the force the wheel puts on the ground.

Yes.

But then energy comes back in thriugh static friction, but this energy is divided between the wheel and the whole bike. So energy in would be F Mwheel/Mbike dx?

No. Where the energy goes is no related to the force per say. And there is no «equal and opposite» concept with energy, as with forces. Energy is transferred from one point to another.

Energy comes from the cyclist, more precisely from the cyclist burning calories. This energy is mostly transformed into kinetic energy to the bicycle. Some is lost in heat, whether from the heat loss that the cyclist have during effort or through the friction between the tire and ground or the chain and sprocket, etc. If the cyclist goes uphill, some will be transformed into potential energy.

For kinetic energy, you have to look at the velocity changes of every component (both linear and in rotation);
For potential energy, you have to look at the height variation of the components;
For heat energy, you have to look at temperature differences;
And for work energy, you have to look at the forces and the displacement of these forces.

But there are more forms of energy that may have to be considered.

In your problem, we start at the motion of the sprocket, so we know how much work (Tdθ) comes out of the cyclist. There is no need to know how much heat was lost through the cyclist transpiration. The job is to determine how the machine that is the bicycle will transform this energy input. By experience, we know the heat loss from the tire-ground friction force is negligible, so we don't look at it. The cyclist rides on a plane surface, so there is none going to potential energy. All that is left is the speed variation (kinetic energy) and work that the cyclist does on his surrounding, which is mostly found with the aerodynamic force and a little bit from rolling resistance (not to be confused with friction). These forces will be related to a displacement dx of the bicycle.

 

[UMath1]

I know there is no equal and opposite energy concept. But the force the wheel imparts to the ground is equal and opposite to the force of static friction, which neglecting rolling resistance and air resistance is the net force on the bike. So Fsf dx is energy in, where Fsf is static friction. But this energy is delivered to the entirety of the bike, so I was thinking that the wheel receives Mwheel/Mbike of this energy.

So energy in the wheel would be Tindθ + Some fraction of Fsf dx ( i think Fsf Mwheel/Mbike dx). Energy out would be F dx, where F is the force the wheel imparts on the ground? And the difference between the two would be Madx?

 

[jack action]

For the sum of forces for the entire bicycle (neglecting rolling and air resistance):
$$F_{friction} = ma$$
For the sum of moments for the rear wheel:
$$T_{in} = T_{out} + I\alpha$$
$$T_{in} = F_{friction}r_w + I\alpha$$
Mixing the two:
$$T_{in} = mar_w + I\alpha$$
Multiply both side by $d\theta$:
$$T_{in}d\theta = mar_wd\theta + I\alpha d\theta$$
$$T_{in}d\theta = madx + I\alpha d\theta$$
$$E_{in} = E_{accelerating \hspace{4 pt} the\hspace{4 pt} bike} + E_{accelerating\hspace{4 pt} the\hspace{4 pt} wheel \hspace{4 pt} in\hspace{4 pt} rotation}$$
Knowing $a = \alpha r_w$, you could rewrite it also this way:
$$T_{in}d\theta = m(\alpha r_w)r_wd\theta + I\alpha d\theta$$
$$T_{in}d\theta = m\alpha r_w^2d\theta + I\alpha d\theta$$
So the portion of the energy necessary to rotate the wheel is:
$$\frac{I\alpha d\theta}{m\alpha r_w^2d\theta + I\alpha d\theta}$$
$$\frac{I}{mr_w^2 + I}$$
And not $\frac{m_w}{m}$ like you mentioned.

If I understand you correctly, you think that somehow there is a difference between the friction force and the force the tire imparts on the ground. There is not, they are two equal and opposite forces.

The wheel torque creates the friction force. The reaction force of that friction force is an equal and opposite force acting at the wheel center. The distance between the two is the wheel radius and this couple formed is equal and opposite to the input torque. That force is the one propelling the entire bike. This combination of wheel forces moves the bike AND rotates the wheel at the same time. So you cannot separate the two (One force that pushes the bike and another one that rotates the wheel). It's the design of the machine that make this possible. What you actually have is not two different forces acting on the same displacement, but the same force creating two different displacements: one linear and one in rotation. So the way you looked at it was $E_{total} = (F_1 + F_2)dx$, but you should look at it as $E_{total} = F(dx + rd\theta)$.

 

[UMath1]

Ok. I know that there is no difference between the friction force and the force the tire imparts to the ground and they equal and opposite. What I was saying is that Eout is not entirely Eout, because the wheel is part of the bike. I know that Iadθ is the energy that goes into rotating the wheel giving it rotational kinetic energy, but the wheel also gains translational kinetic energy from a portion of mαr^2dθ. I thought this portion was Mw/Mbike.

 

[UMath1]

Does that make any sense?

 

[jack action]

Yes.