# Physical meaning of a Metre-Second? [or (Kg · s) or (N · s) ]

1. May 17, 2013

### CF.Gauss

We all understand what metres PER second (m/s) physically means... but,
What I was wondering was what does a 'metre-second' actually mean? Or does it have a physical meaning at all!

As an example:

If we take the units for Dynamic viscosity we have:

Kilogram per metre second ( Kg / m.s )

On a side note, in relation to my above example:
The SI physical unit of dynamic viscosity is the pascal-second (Pa·s), which is identical to
Kg. m-1.s-1 = Kg / (m s)

Another example, of a similar type, would be the standard unit of momentum, which is:
Kilogram-metre per second (kg · m/s or kg · m · s-1 ) which in SI units is equal to Newton-second (N · s)

In this example we have both a Kilogram-metre and, in SI, a Newton-second!
I can easily conceptualise the idea of, say, 1 Metre every 1 Second (m/s) but Im finding it hard to conceptualise this concept!

How exactly could one have, in the physical world, a metre-second (m · s), kilogram-metre (Kg · m) or Newton-second (N · s)?

2. May 17, 2013

### jbriggs444

If you could snort a white powdery substance from the surface of a mirrored conveyer belt moving at a fixed speed using a stationary rolled up piece of currency with a diameter measured in meters and had a certain number of seconds to do so, the "meter second" could be used as a relevant measure.

3. May 17, 2013

### rock.freak667

:rofl: :rofl:

Never used metre-second but for something like the N*s, it would be the force acting over a period of time such that the Impulse = Force*time

For something like kg-m, this could be something like a mass acting a distance. For example, if you have a shaft with an unbalance in it (center of rotation does not coincide to where the mass acts), you quantify the unbalance as a mass-eccentricity.

Mass eccentricity me = mass*distance, usually given in gram-cm or some unit like that.

4. May 17, 2013

### CF.Gauss

would the force acting over a period of time not be Newtons per second ( N/s )?

Last edited: May 17, 2013
5. May 17, 2013

### rock.freak667

That would give you the rate of change of force. A typical Force-time graph looks like this:

The area under the graph gives the impulse which is the same as the change in momentum.

6. May 18, 2013

### lightarrow

Very good!
I thought of a tunnel's price (tunnel lenght multiplied time of realization) but you beat me...