# Calculate output forces of machines

1. Sep 22, 2015

### 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?

2. Sep 22, 2015

### Staff: Mentor

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.

3. Sep 22, 2015

### 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
(Ft*Rg)/(Rw) * (M + mf)/ (M + mf + mw)

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

Last edited: Sep 22, 2015
4. Sep 22, 2015

### Staff: Mentor

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.

5. Sep 23, 2015

### 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?

6. Sep 23, 2015

### Staff: Mentor

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.

7. Sep 23, 2015

### UMath1

Yeah but thats 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?

8. Sep 23, 2015

### Staff: Mentor

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?

9. Oct 2, 2015

### 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. 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? 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?

10. Oct 3, 2015

### dean barry

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

11. Oct 4, 2015

### Staff: Mentor

As far as I can tell, you already know the relevant/basic principles.
Where else would it go?
What energy? If you describe the situation more specifically (and attach equations or better yet, numbers), the answer should be obvious.
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.

12. Oct 4, 2015

### 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?

13. Oct 4, 2015

### Staff: Mentor

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

14. Oct 4, 2015

### UMath1

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

Sorry I mean the short end of the plank.

15. Oct 4, 2015

### Staff: Mentor

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.

16. Oct 6, 2015

### 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?

17. Oct 6, 2015

### Staff: Mentor

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).
You already constructed an equation for this. It works fine.
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)?
Is the chain/gear moving (rotating)? If it is moving it acts over a distance.

18. Oct 7, 2015

### 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.

19. Oct 7, 2015

### Staff: Mentor

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

20. Oct 7, 2015

### jack action

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 $F_{in} = F_{out}$.

21. Oct 8, 2015

### UMath1

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

22. Oct 8, 2015

### 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:

23. Oct 11, 2015

### 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?

24. Oct 12, 2015

### 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:
$$I = I_r + I_f\left(\frac{R_r}{R_f}\right)^2 + m R_r^2$$