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'[OT]: Physics Help - Force exerted by 1HP motor'
2002\02\25@140644 by Donovan Parks

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Hello,

I have a 1 HP motor (= 0.75kW = 553 foot lbs/sec) and need to know how much
force it can exert.  I am looking for a 'best case' estimate - that is, what
is the most force it can exert if everything where ideal.  What I am
thinking, is that if I have a load cell fixed to a wall then what will it
read.

Power = Work / time = Force * distance / time

So, Force = Power * time / distance

At this point, my physics fails me.  Certainly the time the motor acts on my
load cell will not effect the measurement and there is no distance
travelled.... help - where am I going wrong.

Regards,
Donovan Parks

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2002\02\25@142109 by Spehro Pefhany

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At 11:57 AM 2/25/02 -0800, you wrote:
>Hello,
>
>I have a 1 HP motor (= 0.75kW = 553 foot lbs/sec) and need to know how much
>force it can exert.  I am looking for a 'best case' estimate - that is, what
>is the most force it can exert if everything where ideal.

If you allow unlimited time, the force is unlimited. A tiny 1.5V motor can
produce enough force to dismember you if you have the mechanical gearing in
place and let it run long enough.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
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2002\02\25@143821 by Barry Gershenfeld

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If I can remember the few things I've picked up along the way, I think
it is correct to say that with you motor exerting a force on your load
cell, and if it is not moving, then your output is 0 HP.

HP = torque x speed

Let's see...HP doesn't directly tell you about force.  You have to know
something about how it moves...Like the way electric motors can have
lots of torque at low speed, internal combustion engines are quite
the opposite...Auto manufacturers try to get the engines turning
as fast as they can to get that torque x speed figure as high
as possible...Today's cars have tiny high-revving engines and
five forward gears...now we're at the limit of my knowledge.

Barry


{Quote hidden}

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2002\02\25@145909 by Robert Rolf

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A 1HP motor running at speed is one thing.

A STALLED motor will draw substantially more power (more like 10HP).
The stall torque will depend on the motor design. IOW, you'll have to measure it
or contact the manufacturer for the spec.

The force you see will of course depend on the length of your lever arm.

Donovan Parks wrote:
{Quote hidden}

Forgetting to account for the length of your lever arm (pully radius)

>
> Regards,
> Donovan Parks

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2002\02\25@151758 by Douglas Butler

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What you need to know is torque.  You need something like a plot of
power vs speed for your motor.  Is this a home project, or a company
project?

Sherpa Doug

> {Original Message removed}

2002\02\25@152804 by Michael Vinson

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Donovan Parks wrote:
>I have a 1 HP motor (= 0.75kW = 553 foot lbs/sec) and need to know how much
>force it can exert.  I am looking for a 'best case' estimate - that is,
>what
>is the most force it can exert if everything were ideal.  What I am
>thinking, is that if I have a load cell fixed to a wall then what will it
>read.

Power is the *rate* at which the motor delivers energy: 750W means
the motor can do 750 joules of work every second. If you like
analogies, power is to energy as speed is to position; it is the
rate at which it changes. So, if we know your motor can deliver
750W of mechanical power, we know the rate at which it can do
work. How much force is this? Unfortunately, there is not a
straightforward answer to this question: it depends on the system
your motor is doing work on. If, for example, you deliver work
to a frictionless test particle of mass m, initially at rest,
then the instantaneous force at time t is sqrt(Pm/2t), where P
is the power of the motor (750W in your case), ignoring any
nonconservative losses (friction, air resistance, mechanical
inefficiency, etc.). As for the force the motor can exert on a
wall, if you think about it for a second, you'll see it depends on
the setup. For example, if you attach a long swing-arm to the
shaft of the motor, then clearly it will deliver more force at the
end of the arm than if you use a short one.

In real life, motors are usually rated by giving the power delivered
at some particular rotation speed. You can use this number to
compute the torque the motor can produce at the given rotation
speed. Is that what you want? If so, the formula is
torque = P/w, where w (usually written as the Greek lower-case
omega) is the rotational velocity. If you have P in watts and
w in radians/second, then the torque will come out in Newton-meters
(Nm).

I'm sorry to give such a long-winded, unsatisfying answer, but
unfortunately your question is not well-posed.

Michael V

Thank you for reading my little posting.


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2002\02\25@155641 by Pic Dude

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I'm a bit confused by your setup -- are you driving a linear load (with an
"arm" on
the motor) or a rotational load?  What speed does the motor turn at?

Cheers,
-Neil.


{Original Message removed}

2002\02\25@160249 by Donovan Parks

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Hello Michael and everyone else who replied:

I obviously need to do more research into this.  Thanks for the reponses.
Special thanks to Michael for informing me on how lost I am. :)

Regards,
Donovan Parks


> Donovan Parks wrote:
> >I have a 1 HP motor (= 0.75kW = 553 foot lbs/sec) and need to know how
much
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2002\02\25@200322 by Herbert Graf

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You don't have enough info on the motor. Idealy what you need is the stall
torque, which can be determined if you know what the voltage rating of the
motor is. TTYL

> {Original Message removed}

2002\02\26@165854 by Peter L. Peres

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For a motor the solution is simple, the worst case force is produced at
stall (or at a certain rpm for a synchronous motor - aka squirrel cage and
others). The maker also gives the maximum torque of the motor (aka stall
torque but see above). Once you know this and the size of the 1st gear
(the gear on the motor axle), you can work out the maximum force and its
direction in each attachment point of the motor. You can use a graphical
method if you are lost at numbers (requires mm paper and a ruler and
pencil mostly).  Usually you over-engineer the measuring device to 100% or
more for small motors like yours, to account for mechanical abuse (dirt in
gears, people turning axle backwards etc etc).

hope this helps,

Peter

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