## Calculation of force

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### Calculation of force

Can the brilliant minds help me please ?

I need a calculation to measure the effort required to move an object.

The scenario

There is a mass on steel rollers weighing 250kg suspended at height on parallel steel tracks.

Assume no friction.

The height between the mass and handler is 4metres

The mass will be moved via a wire or chain

The distance of travel is 8 metres

Assume walking speed is 0.5m/s

I need the effort to start moving the mass, the effort to maintain movement and the effort to stop the mass.

Many thanks

I need a calculation to measure the effort required to move an object.

The scenario

There is a mass on steel rollers weighing 250kg suspended at height on parallel steel tracks.

Assume no friction.

The height between the mass and handler is 4metres

The mass will be moved via a wire or chain

The distance of travel is 8 metres

Assume walking speed is 0.5m/s

I need the effort to start moving the mass, the effort to maintain movement and the effort to stop the mass.

Many thanks

Blessed are the Cheese makers

Better to be an hour early than 1 minute late

Better to be an hour early than 1 minute late

### Calculation of force

I've been retired a long time now and so my brain doesn't work as well as it did so I might be missing a trick here, but....

The weight of the mass acts vertically down, any force required to move the mass requires a horizontal component. As there is no friction there is nothing to oppose the force required to move it, so NO force is required.

If there WAS friction, then the frictional force depends on the weight, F = uW, where u is the coefficient of friction between the material of the mass and track, though there would be rolling resistance too.

Sorry, but I need to go and lie down a bit now

The weight of the mass acts vertically down, any force required to move the mass requires a horizontal component. As there is no friction there is nothing to oppose the force required to move it, so NO force is required.

If there WAS friction, then the frictional force depends on the weight, F = uW, where u is the coefficient of friction between the material of the mass and track, though there would be rolling resistance too.

Sorry, but I need to go and lie down a bit now

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### Calculation of force

Effectively I will be pulling or pushing the weight which is above so some effort is required

This is the figure I need please

This is the figure I need please

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### Calculation of force

Fixed speed requires only the force to overcome friction (rollers and air resistance). Force required to move it from stationary is dependant on acceleration?

It’s the inertia you need to calculate IIRC? That defines the force required from what I remember.

Been a while since I did this too..

It’s the inertia you need to calculate IIRC? That defines the force required from what I remember.

Been a while since I did this too..

Phil Howard

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### Calculation of force

Yes it's the inertia force I need

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### Calculation of force

Force required is dependant on the friction (which there will be some), and the acceleration required, and time taken. Walking speed is the constant speed, and zero acceleration is obviously impossible - so how long is available to move the 8m? Constant speed says 16 seconds, but that's not going to happen (that's how initial calculations for BIW cells are always wrong as they forget the acceleration of robots and cylinders...and then complain when we can't achieve cycle time).

F = mass * acceleration + friction (of which there will be some).

Friction can be roughly calculated if you know how long it takes to come to a standstill itself from a fixed speed? If, for example, when moving at 0.5m/s and let go, it stops itself in 4m, then that should be the same as accelerating it to 0.5m/s in 4m so you can add that back in. That does mean, of course, the force to stop is less than the force to get it moving.

F = mass * acceleration + friction (of which there will be some).

Friction can be roughly calculated if you know how long it takes to come to a standstill itself from a fixed speed? If, for example, when moving at 0.5m/s and let go, it stops itself in 4m, then that should be the same as accelerating it to 0.5m/s in 4m so you can add that back in. That does mean, of course, the force to stop is less than the force to get it moving.

Phil Howard

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### Calculation of force

As Phil says, Force equals mass times acceleration.

Mass equals weight, due to gravity (I think!)

Acceleration is from zero to 0.5 m/s, you need to specify the distance and the time to get up to speed. Same goes for slowing it down at the other end.

If I remember right, V squared plus U squared equals 2FS

If you assume you can get up to speed in 1 metre (S) then 0 plus 0.5 squared equals 2 times acceleration times S, i.e. 0.125 metres per second squared.

Therefore force equals 250 times 0.125 with the answer in kg/m/s squared. Someone then needs to convert this to Newtons?

I hope this is along the right lines - it's years since I used my brain.

Mass equals weight, due to gravity (I think!)

Acceleration is from zero to 0.5 m/s, you need to specify the distance and the time to get up to speed. Same goes for slowing it down at the other end.

If I remember right, V squared plus U squared equals 2FS

If you assume you can get up to speed in 1 metre (S) then 0 plus 0.5 squared equals 2 times acceleration times S, i.e. 0.125 metres per second squared.

Therefore force equals 250 times 0.125 with the answer in kg/m/s squared. Someone then needs to convert this to Newtons?

I hope this is along the right lines - it's years since I used my brain.

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### Calculation of force

Overall, if you assume there is no friction, then once it is moving it won't need any force to keep it moving, so the only energy used is in accelerating and decelerating.

Work done equals force times distance moved, again it is an assumption that you can get up to speed in 1 metre, so you only need to use force for the first and last metre.

I don't know what the modern terms are for work done (horse power per minute? Kilowatt-hours?)

Going to stop now, brain hurting....

Work done equals force times distance moved, again it is an assumption that you can get up to speed in 1 metre, so you only need to use force for the first and last metre.

I don't know what the modern terms are for work done (horse power per minute? Kilowatt-hours?)

Going to stop now, brain hurting....

Philip Needham

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### Calculation of force

Multiply kg by 9.81 (gravity) to get to Newton's IIRC a Newton has very approximately the same mass as an apple! (1/4 lb)

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### Calculation of force

Does gravity have any bearing on this when it's a horizontal move?

Phil Howard

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### Calculation of force

Gentlemen I admire your knowledge but I'm in way over my head unless I start to study

Bottom line is that an ergo study needs a figure to calculate whether to keep this load movement manual or motorised .

Therefore I am out of plain logic contesting the vendors figure of 3kg to move 250kg from rest.

Any takers , as from what I have here so far , gives me a figure of 31.25kg which is close to what I calculated it as , but would prefer some educated back up before I open my mouth and suggest they are wrong

Thank you

Bottom line is that an ergo study needs a figure to calculate whether to keep this load movement manual or motorised .

Therefore I am out of plain logic contesting the vendors figure of 3kg to move 250kg from rest.

Any takers , as from what I have here so far , gives me a figure of 31.25kg which is close to what I calculated it as , but would prefer some educated back up before I open my mouth and suggest they are wrong

Thank you

Blessed are the Cheese makers

Better to be an hour early than 1 minute late

Better to be an hour early than 1 minute late

### Calculation of force

Maybe your decision manual/motorized is more dependent on the frequency at which it will be required?

I would think that a heavy duty garage door opener would probably do the trick, although the chain drive would almost certainly need lengthening.

I would think that a heavy duty garage door opener would probably do the trick, although the chain drive would almost certainly need lengthening.

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### Calculation of force

for my tuppence, steel rollers on steel rails should be near zero friction and so getting it moving might be less issue than stopping at end of travel, if you assume walking speed and 250kg that is one hell of a load of inertia to stop in a short distance/time.

inertia is just mass x speed, you need to get rid of the kinetic energy at end of track which is 1/2 mass x velcity squared (mass in kg, velocity in m/s)

this is the energy disapated when one car hits another, hence the need for 'impact absorbing bumpers' they reduce the speed from 30mph to zero in a s long a time as possible , and the V squared means four times the damage when car doing 60mph not 30mph

so a VERY large cushion could be required!!!

inertia is just mass x speed, you need to get rid of the kinetic energy at end of track which is 1/2 mass x velcity squared (mass in kg, velocity in m/s)

this is the energy disapated when one car hits another, hence the need for 'impact absorbing bumpers' they reduce the speed from 30mph to zero in a s long a time as possible , and the V squared means four times the damage when car doing 60mph not 30mph

so a VERY large cushion could be required!!!

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### Calculation of force

Gents I respect these calculations .

Can a figure be generated please

Can a figure be generated please

Blessed are the Cheese makers

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Better to be an hour early than 1 minute late

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### Calculation of force

OK!

Given the two assumptions of no friction and getting up to walking speed in 1 metre, using the 4 laws of motion (which I had to look up, failing memory!):

taking 1 metre to reach 0.5m/s will take 4 seconds.

If v squared equals 2as then a (acceleration) equals 0.125 metres per sec. squared

Force = ma = 250 x 0.125 equals 31.25 kgm/sec squared

Instead of 31.25kg, a push of around 50kg should be possible by a healthy young bloke, working backwards, gives an acceleration of 0.2m/s squared.

If v squared equals 2as and the terminal velocity is still 0.5m/s then the distance to get up to this speed is now only 0.625 metres.

At the other end you'll need 50kg force starting 0.625 metres before the end of the track to bring it to a halt.

Seems to agree with your calculation, so we might have both made the same mistake!

Does this help?

Given the two assumptions of no friction and getting up to walking speed in 1 metre, using the 4 laws of motion (which I had to look up, failing memory!):

taking 1 metre to reach 0.5m/s will take 4 seconds.

If v squared equals 2as then a (acceleration) equals 0.125 metres per sec. squared

Force = ma = 250 x 0.125 equals 31.25 kgm/sec squared

Instead of 31.25kg, a push of around 50kg should be possible by a healthy young bloke, working backwards, gives an acceleration of 0.2m/s squared.

If v squared equals 2as and the terminal velocity is still 0.5m/s then the distance to get up to this speed is now only 0.625 metres.

At the other end you'll need 50kg force starting 0.625 metres before the end of the track to bring it to a halt.

Seems to agree with your calculation, so we might have both made the same mistake!

Does this help?

Philip Needham

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