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Here we go again! Newtons and Watts

  1. Mar 2, 2010 #1

    So I thought I knew the laws of physics and electronics, but I've recently stumbled on to a puzzling bit of "Math vs Pratical application". I'm messing up the forces and equations in my strained little head, so I'll let you guys have a go.

    I'll go into practical applications of this and RC-helicopters if needed later. This should cut to the core of my problem:

    I know a Newton is the force required to continuously accelerate a frictionless 1kg brick by 1 m/s^2 (horizontal motion). My question is how many watts would I need to do the same thing?

    Say I have a motor with 85% efficiency mounted to a coil of wire tied to my frictionless 1 kg brick. How many watts do I give the motor to acheive that same continuous horizontal acceleration?

    I've really tried to find solutions in this forum and the rest of the net, but all I come up with is watts needed lift stuff. I've found that Watt = Joule/s = Nm/s, but this doesn't really help me.

    I would appreciate any help!
    Thank you!
  2. jcsd
  3. Mar 2, 2010 #2


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    Staff: Mentor

    Work is force multiplied by distance: W = F * D

    Power is work per unit time: P = W/delta_t

    So the power required depends on how far and how fast you push the block or whatever.

    Does that help?
  4. Mar 2, 2010 #3
    Wow, fast reply! thanks!

    Still have some issues to understand, though. My brick is not actually moving so distance is kind of irrelevant. Its got the 1m/s^2 acceleration force (1 N), but what if a spring is holding it in place? It's not moving anywhere, but the force is still there and my motor is working hard maintaining this force.

    What am I missing?
  5. Mar 2, 2010 #4


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    How can the motor be moving and the object not?
  6. Mar 2, 2010 #5
    No, the motor is not moving, but it is connected to a power supply. This applies a force on its output shaft trying to pull on the object even if it is not able to actually move it.
  7. Mar 2, 2010 #6


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    Energy is being consumed by the motor but no mechanical work is being done on the brick since it is not moving.

  8. Mar 2, 2010 #7
    The object is not moving, but the force of one Newton is there and it is consuming power. How much?

    I'm sure you allready know this, but to be sure we and anyone else reading this are on the same page I'll write it anyway. Inside the motor is a copper coil facing a magnet. These are unresponsive until an electric current is passed through the coil. Based on the direction of the current, it will push the coil (and the output shaft of the motor) clock wise or counter clock wise. This "push" or force will be there even if nothing is actually moving as long as the current is running through the coil.

  9. Mar 2, 2010 #8
    i guess you have the average power in mind!? The power the motor must provide at any time is

    [tex]P = \frac{\mathrm dW}{\mathrm dt} \qquad \mbox{with} \qquad W = \int \mathrm dr ~ F = \int \mathrm dt ~ v(t) \, F[/tex]​

    (i treat the problem as one dimensional) so

    [tex]P = \frac{\mathrm dW}{\mathrm dt} = \int \limits_{t_0}^t \mathrm dt^\prime ~ a \, F = a F (t - t_0) \qquad \mbox{with} \qquad a = \mathrm{const} [/tex]​

    the average power must be

    [tex]\left<P\right> = \frac{1}{T} \int \limits_{t_0}^{t_0+T} \mathrm dt^\prime ~ P(t^\prime)[/tex]​

    for the simple case of [tex]t_0=0[/tex] this yield

    [tex]\left<P\right> = \frac{1}{2} a F T[/tex]​

    so you can prove it with the law of conservation of energy. Assumed that the mass is initially at rest and we stop our treatment at [tex]t=T[/tex], then the energy in your system must be

    [tex]E=m a s \qquad \mbox{with} \qquad s = \frac{1}{2} a T^2[/tex]​

    so the average power is

    [tex]\left<P\right> = \frac{\Delta W}{\Delta t} = \frac{E}{T} = mas \cdot \sqrt{\frac{a}{2s}}[/tex]​

    [tex]T[/tex] applied to the equation of the average power i derived above yields

    [tex]\left<P\right> = \frac{1}{2} a F T = \frac{1}{2} a F \sqrt{\frac{2s}{a}} = \frac{1}{2} a^2 m \sqrt{\frac{2s}{a}}[/tex]​

    That's an indication that the power of the motor must change with time! To be exact

    [tex]P(t) = aFt[/tex]​

    That are my thoughts!
    Last edited: Mar 2, 2010
  10. Mar 2, 2010 #9


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    Since the motor is not rotating, only holding the brick stationary, the amount of power used to hold the brick stationary will be determined based on the locked rotor current of the motor.

    The catch is you are presuming the motor can only put out enough torque that is equivalent to 1 N of force on the brick. The locked rotor current depends on the specifics of the motor so I don't know of any analytical way to get the "equivalent power" for 1 N of force on the brick. It would depend on the motor specification.

  11. Mar 2, 2010 #10
    Hi, stewartcs!

    The motor output force does NOT depend on the motor specification. I can with high accuracy (using dedicated electric equipment) control how many amps and volts (or if you will watts) should be applied to the motor. The more I supply the motor, the more the motor will respond with a powerful force on the output shaft.

    Saunderson: I'm not as versed in the math as you clearly are, but I don't see why the supplied power would need to be dependent on time. One amount of watts will give one amount of Newtons pulling the brick, no matter when or for how long it is done. Maybe you can explain in more detail what you mean?
    Last edited: Mar 2, 2010
  12. Mar 2, 2010 #11
    The connection between the force exerted on an object and the power which is dissipated is

    [tex]P = \frac{\mathrm dW}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} \Bigl( \int \mathrm dt ~ \frac{\mathrm d\vec r(t)}{\mathrm dt} ~ F(\vec r,t) \Bigr) = \vec v \cdot \vec F[/tex]​

    Maybe it is helpful for you to imagine the power as a kind of energy current. So, the faster the object is, the more energy must flow on the object.

    You can treat the force [tex]F=ma[/tex] with [tex]a=\mbox{const}[/tex] as to be conservative. It's easy to see that

    [tex]\Delta T = \frac{1}{2} m \Bigl( v_2^2-v_1^2 \Bigr)[/tex]​

    rises for higher velocities (presumed [tex]\Delta v = v_2 - v_1 = \mbox{const}[/tex]), so the energy current (Power) must rise too.

    I hope i could help!?
    Last edited: Mar 2, 2010
  13. Mar 2, 2010 #12


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    Just to make sure that you understand something -- what is the efficiency of the motor when it has a locked shaft? How much useful work is it doing? (Remeber, W = F*D)
  14. Mar 2, 2010 #13
    Hi, berkeman!

    The motors I use is normally specified at 85% efficiency under normal load. This is not at locked shaft where it probably will be a bit lower. It will also heat up due to the higher current being pulled, but a proper motor controller can continuously monitor this and reduce this current to protect the motor.

    It seems that "useful work" is a relative term. I would say that keeping a spring tight is quit useful in some applications even if the spring is not going anywhere.

    Maybe another way of approaching the problem is in order:

    This is a slightly different scenario, but poses the same problem:
    An small RC-helicopter (electric) is able to hover (float in the air without moving) consuming about 60 Watts of power from the buildt in battery. I have one of these in my living room :-)
    Much of this power is lost as heat in the electric circuits and controllers and not to forget the low efficiency of the propeller. All losses can be known, but since I don't have exact specifications for this scenario we can assume that a total of 3W (5% of input power) end up as pure lift. How much does the helicopter weigh? Or rather, how many newtons of lift is being produced with these 3 Watts?

    Remember that the helicopter does not move (up/down or sideways), but is continuously fighting gravity of 9.81 m/s^2 midair.

    Why do I feel this should be easy, but is still so difficult for me to calculate.
  15. Mar 2, 2010 #14
    Another possibility to convince you: how much is the acceleration of an object if the energy flow (Power) on it is constant? (as one dimensional treatment)

    [tex]P = v(t) F(t) = m v(t) a(t) = m v(t) \dot v(t) = \frac{m}{2} \frac{\mathrm d}{\mathrm dt} \Bigl( v^2(t) \Bigr)[/tex]​

    integration of both sides yields

    [tex]\frac{2P}{m} \cdot t = v^2(t) \qquad \Leftrightarrow \qquad v(t) = \sqrt{\frac{2P}{m} ~ t} [/tex]​

    so the acceleration of the object is

    [tex]a(t) = \frac{P}{m} \Bigl(\frac{2P}{m} ~ t \Bigr)^{-\frac{1}{2}}[/tex]​

    Significant: acceleration is not constant! In addition the acceleration diminishes with time.

    you can prove the validity of [tex]v(t), a(t)[/tex] by applying them to the first equation.

    (for simplicity i omit to set the initial conditions and the special theory of relativity)

    very crazy if you look at the singularity at the origin :)
    Last edited: Mar 2, 2010
  16. Mar 2, 2010 #15
    Hi, Saunderson!

    I'm sure your equations are correct, but I'm more practical than theoretical, and have difficulties accepting the results.

    My view is something like this:
    If you are a waiter and holding a tray of food in one hand, you are continuously working against gravity by applying a force to counteract it. This is burning watts and calories in your arm and although it may feel like the tray gets havier as time passes its still the same forces acting on the tray (gravity pulling down and your arm keeping it up).

    This is a very interesting subject and I really appreciate all your answers. I'm really eager to get to the bottom of this:-)
  17. Mar 2, 2010 #16


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    The simple answer is power = force x speed or power = torque x angular_velocity at any moment in time. In order for your motor to accelerate the brick using a continuous 1 newton of force and the corresponding constant acceleration, the power at any moment increases linearly with speed (until the motor reaches it's maximum power output). The previous posts work out the details.
  18. Mar 2, 2010 #17


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    Staff: Mentor

    Nope. Try a lot lower. Like zero % efficiency.

    No useful work is done if the object isn't moving.

    And on your helicopter example, air is most defintely being accelerated and moved. That is where the engine power is being put to use.
  19. Mar 2, 2010 #18
    Physically you do no work because you are physically a rigid body! At least at the macroscopic level.
  20. Mar 2, 2010 #19


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    There's a similar thread somewhere her on the PF that's been active for a couple of days. Something about pusing against a wall and the concept of work. I haven't read it, but from the title it sounds like it might help you out.

    EDIT -- link: https://www.physicsforums.com/showthread.php?t=382155&highlight=wall

  21. Mar 2, 2010 #20
    Hi, Jeff!

    Thanks! So Power = force * speed. OK, but my speed is zero m/s. Nothing is moving, but the spring is kept tight and my motor is consuming watts. Clearly there is power being consumed, but it doesn't look like these equations cover the cases I'm explaining. The power that is usually transferred into speed and acceleration is now only seen as a tight spring and thermal build up on the motor. I think that somehow this transfer need to be taken into account.
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