Newtons and Newton meters for pushing a weight

[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.

 

[A.T.]

I was asked how much force in Newton meters...

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

 

[neil h]

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

 

[sophiecentaur]

Torque and work are measured in Newton meters.

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)

as I remember it’s force times acceleration

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".

 

[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.

 

[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?

 

[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.

 

[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.

 

[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.

 

[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?

 

[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.

 

[A.T.]

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.

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

 

[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?

 

[A.T.]

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.

Where does the value for acceleration come from?

 

[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 .

?

 

[neil h]

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

 

[neil h]

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

 

[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.

 

[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 ?

 

[A.T.]

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

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.

 

[neil h]

There is no push bar, I've described it as a cart being pushed to give a simplified idea of the maths I'm trying to do, trying to get the torque value off the two axles if it's loaded with 50 Kg and pushed by a person to the values I specified. It's not actually a cart, it's a part mounted inside a larger machine on rails and the reasons for the minimum and maximum acceleration values it is pushed are rather complicated. I can not publish the drawings for this here and even if I could it would take a great deal of explanation to clarify how all of it works together and why.

It's a set of 4 wheels fixed on 2 axles mounted with several sets of deep groove bearing in a small steel frame carrying 50 Kg, there is very low friction and we measured .073 m/s as the minimum acceleration applied and .33 m/s as the maximum. To arrive at 3.65N and 16.5N, this is as far as I could get with any confidence but need to know the Nm torque on the axles to chose a range for the damper.

 

[Nidum]

I'll have one more go . You are talking about a retarding damper ? ie one that stops the 'cart' smoothly at the end of it's travel ?

In which case calculations could involve forces , accelerations and kinetic energy .

Industrial dampers are sometimes rated by the energy they can absorb in a certain time or within a certain contraction distance .

 

[A.T.]

we measured .073 m/s as the minimum force applied and .33 m/s as the maximum.

These are velocities, not forces.

To arrive at 3.65N and 16.5N,

How?

Nm torque on the axles to chose a range for the damper.

What is the damper supposed to achieve quantitatively?

 

[sophiecentaur]

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

If he is using a damper then it could relate to speed too. A rotating vane type would provide a speed limit but the OP is so unspecific of the type and where it would actually be used that it's hard to comment usefully.
Actually, if the requirement is to achieve nearer constant speed then a flywheel would add effective mass and 'resist' small speed changes.

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.

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 ?

Where do you propose to put a damper so that will have the effect you want? It may all be in your head but it hasn't appeared on this thread yet. Without a diagram it is impossible to advance this.

 

[neil h]

This is a continual rotary damper dashpot attached to the axle by a timing belt or gears, the effect of the damping will be continual drag during motion. It dose not limit speed, though it would depending on the setting and it is not used as a break, it's just to apply a very small amount of resistance when pushing the part.

 

[neil h]

The dashpots are rated by the viscosity of the filling and I need a rough idea of the range of Nm force applied to the pot and radians per second to continue the discussion with the supplier about which dashpot to order, according to my maths it's the lightest viscosity one they make, if my maths is right.

 

[neil h]

But I'm not so confident on how to get from the meters per second acceleration and the weight of the cart to the Nm torque on the axle where it would be fitted.

 

[A.T.]

it's just to apply a very small amount of resistance when pushing the part.

Their resistance depends on the speed, not on the push force. You cannot control the maximal acceleration with them. You can only control the maximal speed for a given push force.

 

[neil h]

When the guy was trying to help me out on the phone, which was good of him, I've learned over the last year or so that industrial suppliers do not have much time for folk calling in without the correct specifications and drawings. Anyhow he started asking questions and he did ask me the radius of the wheels but we were stopped with the force applied. It's a bit of an odd one having someone pushing, motors are easy to look up but a push is a vague amount so I said I'd go away and work it all out and get back to him.

The part will do the job if I can put it in the right range and I can always put a gear ration on the transition to the damper to fine tune it but I need to be able to communicate what ball park I'm playing in.

 

[neil h]

That's the logic I called up with. I don't really understand why he wanted the Nm force. He asked for the speed in radions and the max Nm applied. Anyhow I have to call him back now so I'll see how I did I guess.