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Homework Help: Calculating the required motor size to move a load.

  1. May 14, 2013 #1

    From what I've read in the stickies, this question may be more suited to this homework forum, although please note I'm not a student nor is this homework (it's simply for a personal project I'm doing), so please accept my apologies in advance if there are significant gaps in my knowledge that would be expected of a physics student...

    I would really appreciate it if somebody could check the following for me. There's not a lot here, however I've tried to go through every step I need in detail so things are clearer hence the size:

    1. The problem statement, all variables and given/known data

    Right, I've got a box that moves along a straight path, then once reaching the end of the path it reverses and moves back along the straight path.

    The mass of the box is 5 kg.
    The distance the box moves (from one side to the other) is 5 cm.
    The time the box must move this distance is 0.06 seconds.

    I'm trying to work out the size of the motor required to move this box within the given time constraints. Now obviously the motor is rotary and the box moves in a linear fashion, so the motor would be fixed to a crank and so would work similar to how a steam engine moves a locomotives wheel.

    2. Relevant equations

    Just a couple of pointers, I'm ignoring any inefficiencies in the motor, gearing and friction etc. So, for the object to move, I have the following:

    Distance to move object: 0.05 metres (5cm)
    Mass of object: 2 KG
    Time to move object: 0.06 second

    Force required to move object: Mass x 9.8
    Force required to move object: 2 x 9.8
    Force required to move object: 19.6 newtons

    Torque required to move object: Distance (m) x Force (newtons)
    Torque required to move object: 0.05 x 19.6
    Torque required to move object: 0.98 newton-metres

    RPM required to complete distance in time: 60 (seconds per minute) / time to move object
    RPM required to complete distance in time: 60 / 0.06
    RPM required to complete distance in time: 1000 rpm
    One RPM actually moves the object twice, so halve the RPM:
    RPM required to complete distance in time: 500 rpm

    For the motor calculations, I have the following. Unfortunately this is all I know of the motor, so I'm unsure if I even have all the information required!

    Given motor rated no-load RPM: 7000
    Given motor rated input power: 650 watts

    I don't know the motor efficiency, etc. but I know the motor the version down is 82% efficient, so I shall use that - I'm more interested in the process of these calculations rather than the accuracy as if need be I'll find a motor with full specs.

    Calculated output power: input power x efficiency
    Calculated output power: 650 x 0.82
    Calculated output power: 533 watts

    Horsepower: power / 745.69
    Horsepower: 533 / 745.69
    Horsepower: 0.71

    (The braking torque formula I got from this website: <a href="http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm">http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm</a>) [Broken]

    Braking torque (lb-ft): (5252 x HP) / rpm
    Braking torque (lb-ft): (5252 x 0.71) / 7000
    Braking torque (lb-ft): 0.53
    Braking torque (NM): lb-ft x 1.35
    Braking torque (NM): 0.53 x 1.35
    Braking torque (NM): 0.7155

    However, this is for the motor running at the given rpm, not the required RPM. So I now need to calculate the gearing required to drop the motor rpm to the required rpm:

    Gearing ratio: Motor RPM / Required RPM
    Gearing ratio: 14

    This means that the final braking torque is:

    Braking torque (NM): Braking torque (NM) x gearing ratio
    Braking torque (NM): 0.7155 x 14
    Braking torque (NM): 10.02

    So, as the required torque to move the object initially was 0.98NM, this motor is just over 10 times the required power.

    Could somebody verify the above for me please?

    Thank you.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 14, 2013 #2


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    Gold Member

    In which direction is the box moving? Horizontally or vertically? I ask because elsewhere you say

    So I don't see where g (9.8) comes into your calculation for the force?

    Your set up is similar to a piston and con rod in a car. The piston starts from rest in one position, accelerates upto some speed V in the middle then decelerates again to come to rest at the other position.

    You can use geometry to work out the peak acceleration (which won't be equal to g) and then apply f=ma.
  4. May 14, 2013 #3
    Hmmmm.... If there is no friction, then once the mechanism is up to speed, is there any more work to be done?
  5. May 14, 2013 #4
    Hello there CWatters,

    Thanks for taking the time to reply.

    My apologies, I hadn't realised the direction was needed as I assumed that the difference between the two over such a small distance would be negligible. Although to answer your question, the box is moving horizontally.

    Ah okay. So if the box is moving only horizontally then I don't count gravity as a force as I'm never having to oppose it, although if the box was moving say up and to one side I'd have gravity and the acceleration of moving to one side. Correct?

    Yes, that's a perfect description! I've dug up a few formulas (and a couple of calculators for piston acceleration - oh boy! This may take a while to work out! :D
  6. May 14, 2013 #5
    Hello BarryJ,

    Ah, so friction has to come into it somehow then? I'll have a look for some sample friction coefficients...

  7. May 14, 2013 #6
    Think about this. You could have a slot, i.e. the path, where your mass would move. At each end, put a spring. To start the system, pull your mass so it compresses one of the springs and let go. If there were no friction and everything is ideal, the mass would bounce back and forth forever. The only work needed would be to compress the spring initially. I think you will definately account for friction somewhere.
  8. May 14, 2013 #7
    Well... After putting it like that, it makes my previous statement seem a little silly! :) Haha, thank you. I shall gather an appropriate friction coefficient. :)
  9. May 14, 2013 #8
    No, the statement is not silly. It shows you are trying to learn and that is good.

    In your proposed problem, there is friction at all moving parts. The motor must be sized to overcome the friction at various parts. CWatters said your problem is sort of like a piston moving in a cylinder. In this case, friction is between the piston and the cylinder walls. If there were no friction, and you removed the cylinder head, you could spin the crank of a motor and it would spin very easily. If the cylinder head were attached, there would be work done on the gas as it is compressed and let to expand, unless there was no heat exchanged between the cylinder head and the environment.
  10. May 15, 2013 #9


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    Possibly. Don't forget it's actually an oscillating mass. It takes energy to accelerate it upto speed which you might be able to recover when it's decelerated at the other side. Using a heavy flywheel would be one way to do it. Using multiple "cylinders" out of phase might also help. That way you have one mass accelerating while another is decelerating.
  11. May 16, 2013 #10
    Revised question.

    Thanks once again for the tips on things I’ve missed such as acceleration and friction. I’ve spent the past couple of days rooting round trying to find things out. I’ve posted below a revised question that should hopefully explain things a little better along with some further calculations. I would really appreciate somebody taking a quick look over how I'm doing these calculations to make sure things are okay.


    I have an object that is moved horizontally by a given distance and is supported by a table. The object will be moved by a crank system (similar to this Crankshaft image on Wikipedia). I want to be able to source a motor that is capable of moving this object at a given speed. I’ve done a diagram (albeit, not terribly nice) that helps illustrate the problem and the components involved:


    Formulas and data:

    I have the following data for the object:
    Code (Text):
    Distance to move (cm): 8
    Mass of object (kg)  : 2.1
    Movements per second : 16
    Friction coefficient : 0.5
    I'm guessing it's obvious, but the movements per second is how many times the load is expected to move per second and the friction coefficient is between the load and the supporting table.

    It seems the only calculation that is useful from the above is the friction calculation. This was previously ignored, although the above posts makes it quite clear that friction must be taken into account otherwise the object would move forever... Quite obvious when said like that! :-)
    Code (Text):
    Normal Force (Newtons)       : (Mass of object) x 9.8
    Normal Force (Newtons)       : 2.1 x 9.8
    Normal Force (Newtons)       : 20.58
    Code (Text):
    Force for movement (Newtons) : (Normal Force) x (Friction coefficient)
    Force for movement (Newtons) : 20.58 x 0.5
    Force for movement (Newtons) : 10.29

    Previously, I took no notice of the crank only using the acceleration of the object, however after reading the above posts, it is clear that I need to calculate the acceleration of the crank instead, which would let me work out the actual force needed to move the crank (and so the object). Just in case anybody comes across this, I had quite a bit of difficulty working out the acceleration of the crank (partly due to a silly mistake and partly due to a misunderstanding of my part in what I was looking for), a thread I created helped tremendously. These calculations are correct as per the thread just mentioned, so I shall simply post the results rather than the workings:

    Code (Text):
    Crank Inputs    
    Revolutions per minute: 480
    Rod length (cm)       : 15
    Crank angle (deg)     : 0

    The RPM is derived from the desired speed of the object / 2 (one whole revolution moves the object twice).
    A Crank Angle of 0 is about the point of maximum acceleration of the crank.

    Crank Calculations
    Crank angle (rads)           : 0
    Crank angular vel. (rad/s)   : 50.26548246
    Crank radius                 : 4
    COS Theta                    : 1
    COS2                         : 1
    n                            : 3.75
    -rw2                         : -10106.47491
    Crank acceleration (cm/s)    : -12801.53488
    Crank acceleration (m/s)     : -128.0153488
    Crank acceleration ABS (m/s) : 128.0153488

    From the above calculations, I can work out the final force required to move the object and the crank.

    Code (Text):

    Distance to move (m) : (Distance to move (cm)) / 100
    Distance to move (m) : 8 / 100
    Distance to move (m) : 0.08
    Code (Text):
    Force required for crank (Newtons) : (Mass of object x Crank acceleration (ABS))
    Force required for crank (Newtons) : 2.1 x 128.0153488
    Force required for crank (Newtons):  268.8322325
    Code (Text):
    Total Force to move crank and object (Newtons) : (Friction force for movement + Force required for crank)
    Total Force to move crank and object (Newtons) : 10.29 + 268.8322325
    Total Force to move crank and object (Newtons) : 279.1222325
    Code (Text):
    Total Torque to move crank and object (Newtons) : (Total Force to move crank and object x Distance to move)
    Total Torque to move crank and object (Newtons) : 279.1222325 x 0.08
    Total Torque to move crank and object (Newtons) : 22.3297786

    Assuming the above is correct, in summary I need a motor that has:

    Code (Text):
    Minimum torque (Newtons) : 23
    Unfortunately, the motors I've got at hand I only know the following characteristics (I've chosen one motor at random here):

    Code (Text):

    Rated RPM (RPM)         : 12000
    Rated input power (W)   : 650
    Rated efficiency (%)    : 80

    From the above motor inputs I can work out the following:

    Code (Text):
    Output power (W) : Rated input power x efficiency
    Output power (W) : 650 x 0.8
    Output power (W) : 520
    Code (Text):
    Horsepower (HP) : Output power / 745.69
    Horsepower (HP) : 520 / 745.69
    Horsepower (HP) : 0.697
    Code (Text):
    The formula for braking torque was found at [URL="http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm"]Elec Toolbox[/URL].
    Braking Torque (Lb.ft) : (5252 x Horsepower) / Motor RPM
    Braking Torque (Lb.ft) : (5252 x 0.697) / 12000
    Braking Torque (Lb.ft) : 0.5232
    Code (Text):
    Braking Torque (Newtons) : Braking Torque (lb.ft) * 0.737
    Braking Torque (Newtons) : 0.5232 * 1.3558
    Braking Torque (Newtons) : 0.3856
    This is the braking torque of the motor running at the full RPM, however as the crank needs to rotate at 480 RPM, I can reduce the motor RPM with gearing and so increase the torque.

    Code (Text):
    Gear ratio motor to crank: Motor RPM / Crank RPM
    Gear ratio motor to crank: 7000 / 480
    Gear ratio motor to crank: 14.58
    Code (Text):
    Final braking torque of motor (Newtons) : Braking Torque * Gear ratio
    Final braking torque of motor (Newtons) : 0.3856 * 14.58
    Final braking torque of motor (Newtons) : 5.6233

    So, I can now see that the crank and the object required a torque of 22.3297786, however, the motor can only provide a torque of 5.6233, so the motor is only around 25% of the power required.
  12. May 16, 2013 #11
    You are doing a lot with torques, braking, and etc, but think about this.
    Where is the energy consumed that the motor has to overcome? It is in the friction of the mass moving back and forth. I would suggest finding out how much energy is required to move the mass 8 cm, and then continue to find the power required to move the mass at your 16 cycles per sec. Then you have the power needed from the motor. I assume you are ignoring the friction associated with the crank bearings and motor shaft.
  13. May 16, 2013 #12
    Hello there,

    Thanks for the prompt reply.

    So, am I going down the wrong route by using braking torques etc. and overcomplicating it? If so, where would these things fit in?

    Well, I know that I need to apply more than 10.29 newtons of force to get the object to move due to the static friction, so if I applied 11 newtons:

    Code (Text):
    Energy (J) = Force x Distance
    Energy (J) = 11 x 0.08
    Energy (J) = 0.88
    I know that the total force needed to move both the crank (running at the required RPM) and object is 279.1222, which gives an energy of:

    Code (Text):
    Energy (J) = Force x Distance
    Energy (J) = 279.1222 x 0.08
    Energy (J) = 22.329776
    So that is the energy consumed per second, however I'm not 100% sure where to go from here...
  14. May 16, 2013 #13
    Actually, I've been searching about and I wonder would I be better getting a moderator to move this thread to the "Mechanical Engineering" forum?
  15. May 16, 2013 #14
    I don't see where you got the 279 number. Your force of 10.29 is correct. The energy to go back and forth one time will be 10.29 X 0.08 X 2 = 1.64 J. At 16 cycles per sec you will use 1.64 X 16 = 26.34 joules/sec or 26.34 watts if you motor is 100% efficient.
  16. May 16, 2013 #15

    The 279 number is in the "Crank & Object Force Calculations" section, but here it is below:

    Code (Text):
    Total Force to move crank and object (Newtons) : (Friction force for movement + Force required for crank)
    Total Force to move crank and object (Newtons) : 10.29 + 268.8322325
    Total Force to move crank and object (Newtons) : 279.1222325
    Is what I've actually got in the revised post (#10) wrong?
  17. May 16, 2013 #16
    Continuing, 26 watts is about 26/760 = 0.034 hp. at 80% efficiency, you would need a .043 hp motor. Now you will have to figure out the rpm of the motor and the desired gear ratio to reduce the speed.
  18. May 16, 2013 #17
    I don't see why the crank force would be so high??
  19. May 16, 2013 #18
    Well, the crank has a maximum acceleration of around 128.015 metres per second and the weight is 2.1 KG, which gives a force of approx. 268 newtons as per the first post by CWatters:

  20. May 16, 2013 #19
    The equation for d(t) = .04sin(2Xpi/(1/16))t = .04sin 32pit
    v(t) = 4.02cos(2Xpi/(1/16))t
    a(t) = -404sin(2Xpi/(1/16))t

    This is just for the moving 2.1 kg mass. The maximum acceleration will be 404m/sec^2
    However, the average acceleration is 0.

    If you have a motor driving a crank and flywheel, the inertia of the flywheel will average out these what seem to be high forces.

    Bottom line, the only consumer of energy is friction. If you use a small hp motor, yes, it might take a time to get up to speed but there is really little work done by the motor.
  21. May 16, 2013 #20
    Hello Barry,

    Thanks very much for that. I have to admit though, I don't understand what's happening with the maths there, it seems you're calculating distance, velocity and acceleration. Nevertheless, I'm a little more interested in the writing there than the maths...

    My apologies, there isn't a flywheel present - I'm sorry if I gave you that impression...

    This makes perfect sense after your earlier posts - obviously initially I need to overcome static friction, then to keep the velocity constant I need to overcome the kinetic friction... However...

    I calculated the force needed to move the crank a given time through a given distance as follows:

    Code (Text):
    Distance to move (m) : (Distance to move (cm)) / 100
    Distance to move (m) : 8 / 100
    Distance to move (m) : 0.08
    Code (Text):
    Force required for crank (Newtons) : (Mass of object x Crank acceleration (ABS))
    Force required for crank (Newtons) : 2.1 x 128.0153488
    Force required for crank (Newtons):  268.8322325
    Code (Text):
    Total Force to move crank and object (Newtons) : (Friction force for movement + Force required for crank)
    Total Force to move crank and object (Newtons) : 10.29 + 268.8322325
    Total Force to move crank and object (Newtons) : 279.1222325
    The object will have a velocity of 0 whenever it changes direction and so the acceleration of the object must be made each time the crank reverses direction, so surely this force is always present so to bring the object up to speed - I'm struggling to see how we can skip over it? If so, I'm quite confused why CWatters said I needed to calculate this.

  22. May 16, 2013 #21
    When the mass changes direction, you are correct in that the velocity will be 0. At this time, the acceleration will be the greatest. Any motor will have some sort of flywheel even if it is just the rotor of the motor. The crank will also act as a flywheel. Even if there were no specific flywheel, there would be one. The example I gave with the mass moving at a sinusoidal rate is not exact but close. The connecting rod between the flywheel, or crank, and the mass will cause a slight deviation. In any case, you do not need a large hp motor.
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