machine elements power-force formula

[anatino]

From Newtonian physics, F = ma, the net force of a system will contribute to its acceleration, and in a case of no friction or such energy losses, theoretically all forces will contribute to acceleration in that sense.

In the context of machine elements, if we ignore friction and everything... all torques added to a gear will contribute to its acceleration, described by the formula T = I*angular_acceleration

From what I understand, the relation between power, force and speed is: P = F*v
The dimensions make sense, but I've seen this formula used in such a way:
Pmotor = Fmotor * v

And I don't now how to connect this to the acceleration concept.
If Fmotor contributes to acceleration, why does P = F*v even exist, it relates the force and speed directly, but if I waited a little longer while keeping that same force the speed will increase correct?
Or maybe this formula is the force it takes to maintain the speed v, which makes sense, and so it would be equal to the friction and other forces of loss correct? Is that what it is?

If that's what it:
Does that mean it doesn't apply for ideal scenarios without friction and resistance?
Why does the power P or v even matter in such a case, wouldn't a formula such as Fmotor = F_friction be better? Cause the speed wouldn't matter in such a case since as long as Fmotor = Ffriction it will always be maintained, so I don't get why P = F * v is even used.

Perhaps I don't understand the concept or what this formula describes. Any explanation is appreciated, thank you for your time.

 

[Snickster]

And I don't now how to connect this to the acceleration concept.
If Fmotor contributes to acceleration, why does P = F*v even exist, it relates the force and speed directly, but if I waited a little longer while keeping that same force the speed will increase correct?

Power = Force x distance per unit time = F x dS/dT = F x V. Yes velocity will increase for force/torque acting on a rotating mass without any opposing force/resistance (no friction). See attached sketch/calculations where force goes only into acceleration of the mass and increase in KE of the mass. In this case there will always be a net force on the object so there will always be an acceleration and the velocity will keep increasing and thus the power input required would keep increasing although the force remains constant in accordance with Sum F = m x a.

Or maybe this formula is the force it takes to maintain the speed v, which makes sense, and so it would be equal to the friction and other forces of loss correct? Is that what it is?

Yes that is correct. Say you have a pump where the mass in the attached sketch is the mass of all rotating components (shaft, impeller, motor rotor). It takes force/torque to accelerate these rotating components up to speed. If the pump speed were self-limiting, the final constant rotational speed/angular velocity "w" of the rotating components will be such that when the force/torque input by the motor on the shaft/rotor end will be equal to the resistance force/torque on the shaft/impeller end by the flowing fluid and friction of rotation of the shaft. At some speed these forces will balance and there will be a zero net force on the rotating elements so there will be no more acceleration of the rotating elements once it reaches that constant velocity since Sum F = 0 = m x a, then "a" acceleration of the rotating masses must be zero (constant speed is reached) of the rotating elements (shaft, impeller, motor rotor with windings).

However, a motor is designed to run a constant speed depending on the number of magnetic poles so the motor design determines the final rotational speed. In this case the motor will reach this speed regardless and attempt to maintain this speed as long as it has the rated/available horsepower to do it. Then the opposing force of fluid flow and friction on the impeller end will equal the force of the rotating magnetic field on the motor rotor end under steady state conditions. This will be the force F in the P = FxV = TxW equation. At constant speed no more force is required to accelerate the mass of the rotating elements, so the only force the motor has to overcome opposing force of flow of fluid and friction.

Note that if motor has a design rated speed say 3600 RPM the shaft will always turn a little slower than the rated speed. This is called "Slip" for induction type motors. The rated speed of the motor is the speed of the rotating magnetic field. The more opposing load on the shaft the lower the speed of the shaft/rotor relative to the design magnetic field rotation speed such that the shaft may turn at 3400 RPM if the magnetic field rotates at 3600 RPM. This is because if the shaft turns at the same speed of the magnetic field there would be no magnetic force on the motor rotor windings since the motor winding will be rotating at the same speed of the magnetic field and the magnetic field would not have a velocity relative to the electrons in the winding to create the electro-magnetic force. Therefore the higher the external load on the shaft the slower it will turn and the faster relative velocity the rotating magnetic field will cut across the rotor windings creating a greater force the more the shaft/rotor slows down under greater load. In this way the motor output force on the rotor end always matches the external resistant force and at that point velocity of rotation is constant since F = m x a = 0, no net force on rotating components when motor force balances external force on rotating components and P = FxV=TxW.

If that's what it:
Does that mean it doesn't apply for ideal scenarios without friction and resistance?

It is looking at constant velocity of rotation of motor at equilibrium where force of motor on rotor/shaft equals opposing external force on the shaft such as for a pump the force of fluid on the impeller plus friction force of shaft rotation.

Why does the power P or v even matter in such a case, wouldn't a formula such as Fmotor = F_friction be better? Cause the speed wouldn't matter in such a case since as long as Fmotor = Ffriction it will always be maintained, so I don't get why P = F * v is even used.

See above explanation and attached calculation.

Perhaps I don't understand the concept or what this formula describes. Any explanation is appreciated, thank you for your time.

 

[3DDave]

Simply - if the power is absorbed some place, like pumping water or in lifting an elevator, then when the power required to do that is applied there is a steady state and no acceleration, even if there is no loss due to friction.

why P = F * v is even used: If you want the elevator to lift twice the weight or go twice as fast with some given weight, then there will have to be twice the power. Also, this is why counterweights are used - to run counter to the lifted weight and reduce the amount of power required.

Certainly, in a vacuum, as long as the propellant lasts, a rocket motor will continue to accelerate the mass of the rocket and remaining propellant. On Earth a major obstruction is the presence of water or air that gets displaced by motion, so even those devices, such as cars, which are not intended to move the surrounding fluid (air) still will, at some speed, have displacing the air absorb all the power output of the motor.

 

[rb1957]

frictionless is only a simplification to solving the problem.

without friction Power_in = Power_out
with friction Power_in = Power_out + P_friction (you lose some of the input power to overcome friction)

Why does P = Fv even exist ? because power is a quality of it's own. Motors are designed to produce power (by consuming energy) and this power allows useful work to be done.

consider Force is a quality, Work/Energy is different, Power is different too; Power is Work per unit time.

 

[Compositepro]

I recall having somewhat similar thoughts as a new graduate. F equal ma is fundamentally used to determine when the equations of statics applies. When nothing is accellerating, a is zero. Thus, F is zero. Which means all the forces must be exactly balanced. This concept seems trivial at first. But in Physics it is the logical first step in making the distinction of when to use statics or dynamics to solve a problem.

 

[GregLocock]

The connection between power and acceleration is the change in Kinetic energy per unit time.

 

[MintJulep]

Who loads their (equal) pile of rocks onto their (identical) truck fastest?



When finished, they've both done the same amount of work, but one has done that work is less time.

That's power.

 

[rb1957]

maybe the weedy (nerdy?) guy would drive his truck closer to the pile of rocks ?

but yes I get the point !

better, he might say to the other guy "hey, load my rocks first and I'll give you $10"

 

[3DDave]

I've seen many cases where big-muscle guy cannot keep up with the string bean.