# B Newtons and Newton meters for pushing a weight

1. Jun 29, 2017

### neil h

Hi. I’m trying to build something and having very little basic physics am having to try and learn as I go along. Today I spoke to a company that makes damping units and I was asked how much force in Newton meters would be applied to the unit at the top speed. It’s a small cart weighing 70Kg and the top speed is .36 meters per seconds. All the wheels are bearing mounted and there is no motor or anything attached it’s just pushed by hand. I felt there must be too little friction for it to be relevant so as I remember it’s force times acceleration. So let’s say someone gives it a shove and pushes it to its max speed in one second that’d be 70 x .33 = 23.1 N I guess, is this right? And how do I get from there to Nm?

I’m pretty confused about what the basic equations are and any help or just being pointed in the right direction to learn about this would really help me out?

Thanks, N.

2. Jun 29, 2017

### A.T.

Force is measured in Newtons. Torque and work are measured in Newton meters. You should ask back what they mean.

3. Jun 29, 2017

### neil h

Well I'm presuming they are asking as the rotary damper would be attached to the axle by gears or a belt

4. Jun 29, 2017

### sophiecentaur

That could be a bit misleading. Work is measured in Joules, which involves a force times a distance but the metres are different metres from the ones used to calculate torque. If you EVER read about Newton metres, they are talking about Torque and not Work. So the Newton metres that the OP was being quoted would have to have referred to torque (from a rotary damper)
Force is Mass times acceleration. The mass is 70kg and the acceleration is 0.36 m/s2 so the Force needed would be 25.2N (applied for 1s).
But how is that relevant? What do you need the damper to do? Is it supposed to act as a speed limiter? How were you planning to use the Damper? If we can have a few more details then we can work out what spec the damper needs to have. Tell us more about your "something".

5. Jun 29, 2017

### neil h

No not a speed limiter it's just to be used to smooth out the motion as it's hand pushed, it's a variable continual rotary damper with a x10 adjustment on the damping force so the question is which viscosity of fluid to have in the pot, I need to get it in the right ball park then it'll be a suck it and see process as it's ultimately a tactile and not a mechanical problem.
I guess I started with the N as I hadn't got a clue how to figure out the Nm and was hoping that would of got me halfway there.

6. Jun 29, 2017

### neil h

All the examples of calculating Nm I can find use the distance from the fulcrum x force. Newtons x meters = Nm, but how do I get from someone pushing this cart to the torque on the axle?

7. Jun 29, 2017

### sophiecentaur

Have you actually drawn this out? If so, could we see a rough layout? If not then you really need to. It strikes me that, as the applied force is linear, you need a damped spring connecting the humans and the cart , similar to the linear suspension system on the front of many cars - just a lot lighter weight system with a much longer throw. To use a rotary damper, you would need a rack or belt and it still needs a compressible shaft with a spring to return it to an equilibrium position.

8. Jun 29, 2017

### sophiecentaur

Do you have a link for the dampers you are considering? I am very confused about your real requirement and how you intend to solve the problem.

9. Jun 29, 2017

### neil h

Ok, deep breath, this is what I did, tell me if I'm mad. Using 50 Kg's on the cart as the load I have 16.5 N as a max and 3.65 N at a more normal speed. So to calculate the torque in Nm on the axle I first split the force in half for my max and min as there is two axles supporting the weight giving me 8.25 N and 1.825 N respectively. Then I took the radius of the wheel ( 0.03 m ) and multiplied by that to get the Nm, 2.475 Nm and 0.055 Nm. Then finally I divided the radius of the wheel by the radius of the axle to get something called the IMA which with my 15mm axle is 4 and multiplied both by that to get the Nm coming of the centre of the axle where the damper would be connected via a gear or a belt. Giving me a range of 0.22 Nm to 0.99 Nm.

10. Jun 29, 2017

### sophiecentaur

They say a picture speaks a thousand words. If it's a workable system then you should be able to draw it out. A link to the dampers you are considering would help, too. Do they actually do what you want?

11. Jun 29, 2017

### neil h

I appreciate your interest in the design of the whole assembly and how it would work but it's a bit much to go into in detail here, don't take me as rude i know your looking to help but I have to call back and sort out ordering one of these things first thing tomorrow and the design is a whole other can of worms. I really need to get my head around this maths and get some sleep.

12. Jun 30, 2017

### A.T.

Force is related to acceleration of mass, not to speed.

13. Jun 30, 2017

### neil h

Yep, that's in the first post. Dose it sound like I'm doing the Nm calculations right? I've used Mass x Acceleration to get a maximum and average Newton value. Then I divided the total in half to calculate for each axle and multiplied N by the radius of the wheels to get the Nm. Lastly I divided the radius of the wheel by the radius of the axle, treating it like a first class lever, and multiplied my Nm values to arrive at the torque on the axle.

Dose it sound like I've done the maths right?

14. Jun 30, 2017

### A.T.

Where does the value for acceleration come from?

15. Jun 30, 2017

### Nidum

Is this something like :

You have a cart where the wheels can each move up and down independently relative to the cart .
The vertical movement of each wheel is controlled by a spring and a damper .
The damper is of the rotary type and is connected to the wheel axles(s) by a lever linkage .

?

16. Jun 30, 2017

### neil h

Just a comment on if any part of it looks sound or not would really help.

17. Jun 30, 2017

### neil h

The value for acceleration comes form the expected accelerations of the cart.

18. Jun 30, 2017

### neil h

I see, sorry I should of been clearer. It's just a normal cart on a horizontal plane with 4 wheel, two axles and someone pushing it. There are no springs or anything like that just two fixed axles rolling in sets of bearings.

19. Jun 30, 2017

### Nidum

ok . The only other application of a damper that I can think of is between the push bar and the cart itself . Arranged so as to possibly reduce jolting of the cart contents during initial start away and as the cart hits random bumps in the pathway ?

20. Jun 30, 2017

### A.T.

Expected accelerations by what? You seem to use it as the braking acceleration by dampers attached to the axles, when you let the cart go. This has little to do with the accelerations when pushing the cart.