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
> >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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