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Motor Thrust Not equal to Force measured

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  • 27-04-2015 9:57pm
    #1
    Dubh Geannain
    Registered Users Posts: 3,393 ✭✭✭


    Hope someone can explain this one for me.

    Say I have a motor driving a door. The motor datasheet says that it is rated for 1,500 Newtons of thrust. Measuring the forced exerted by the door on a force meter yields more than double that.

    Am I making a physics 101 error or is it a little more complicated than that?


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    #2
    CatFromHue
    Moderators, Science, Health & Environment Moderators Posts: 18,142 Mod ✭✭✭✭CatFromHue


    How did you get on with this?


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    #3
    bnt
    Registered Users Posts: 12,958 ✭✭✭✭bnt


    Late to this, but - is the motor output really in Newtons (force)? I would think that Newton.Metres (torque) would be more likely. If so, since torque = force x lever arm, you would (in theory) get 1500N force at the end of a 1m lever arm, and 3000N at the end of a 0.5m lever arm.

    From out there on the moon, international politics look so petty. You want to grab a politician by the scruff of the neck and drag him a quarter of a million miles out and say, ‘Look at that, you son of a bitch’.

    — Edgar Mitchell, Apollo 14 Astronaut



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    #4
    Dubh Geannain
    Registered Users Posts: 3,393 ✭✭✭Dubh Geannain


    bnt wrote: »
    Late to this, but - is the motor output really in Newtons (force)? I would think that Newton.Metres (torque) would be more likely. If so, since torque = force x lever arm, you would (in theory) get 1500N force at the end of a 1m lever arm, and 3000N at the end of a 0.5m lever arm.

    I didn't give the exact numbers last time but here they are:

    The manual states:
    Power(W) 750
    Thrust(N) 3,500
    Gear Ratio(i) 1/28
    Force exerted by the door was measured at 9,000 N

    I would expect that the gear ratio only effects the speed and not the force driving the door.

    You're idea about leveraging is a good one and has made me think about this again.

    Would the weight of the door have an effect on the final force? The force is not being measured at the initial motion (which could be where I am erring) but rather after the door is moving at a constant speed. This made me think momentum had a big part to play in this. I was unclear of the relationship between momentum and force so had to look it up.

    I've found force = rate of change of momentum.
    11110362.gif

    I hadn't checked the speed of the door when I last looked at this so just approximating the momentum I get:
    Momentum = 1,000 kg x 0.2 m/s = 200 kg.m/s (assumed speed of 20cms per second)

    So, Force = 200/0.01 (assumed stoppage time of 1 hundredth of a second)
    Force = 20,000 Newtons

    9,000 N was actually measured but I'm making assumptions for speed and stopping time, which I believe are reasonable enough. And yes, it is a big door :D

    That appears correct to me and helps explain what I'm looking for. Cheers for the replies as I wouldn't have revisited this.


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    #5
    bnt
    Registered Users Posts: 12,958 ✭✭✭✭bnt


    I think you're over-complicating things a bit. Your original description was all about forces only, which makes it a Statics question., That is, I see no need to bring motion in to the question at all. Remember Newton's 1st law: when it comes to net forces, constant motion is equivalent to no motion, since it takes no net force to produce a constant motion. (Note the net part: weight due to gravity is a force, and it takes an opposite force to resist it.)

    OK, let's say it's a linear motor that puts out a linear force, not rotational torque: leverage would then allow you to multiply a force. For example, a simple 2:1 pulley would allow you to lift a 3000N weight with 1500N force (such a tension in a rope).

    From out there on the moon, international politics look so petty. You want to grab a politician by the scruff of the neck and drag him a quarter of a million miles out and say, ‘Look at that, you son of a bitch’.

    — Edgar Mitchell, Apollo 14 Astronaut



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    #6
    Dubh Geannain
    Registered Users Posts: 3,393 ✭✭✭Dubh Geannain


    bnt wrote: »
    I think you're over-complicating things a bit. Your original description was all about forces only, which makes it a Statics question., That is, I see no need to bring motion in to the question at all. Remember Newton's 1st law: when it comes to net forces, constant motion is equivalent to no motion, since it takes no net force to produce a constant motion. (Note the net part: weight due to gravity is a force, and it takes an opposite force to resist it.)

    OK, let's say it's a linear motor that puts out a linear force, not rotational torque: leverage would then allow you to multiply a force. For example, a simple 2:1 pulley would allow you to lift a 3000N weight with 1500N force (such a tension in a rope).

    Yep. I probably didn't give an accurate description of what I was doing in my first post. The door was moving by the time the force was measured. We used a load cell to stop the door in its tracks and measure the force exerted by the door. What we were trying to determine was what force it would hit someone or something with if it came into contact with them while moving. So motion did play a part in our measurement. We measured 9,000 N approx doing this.

    Then, looking at the drive motor datasheet it stated 3,500 N of thrust. This is what got us scratching our heads.


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    #7
    bnt
    Registered Users Posts: 12,958 ✭✭✭✭bnt


    Ah - so it's force measured for a short period during stopping, not a static force. Have a look at Impulse: another name for the change of Momentum (Δp) and has the same units (kg.m/s).

    The Wikipedia page calls it "the integral of a force, F, over the time interval, t, for which it acts" (Calculus), but it would also be simply F.t if we assume F is constant. So it can also be expressed in N.s.

    In other words: just before the door stops it has a certain momentum (p). As it hits, it dissipates that momentum (Δp) in a specific period of time. The shorter the time (t), the higher the force (F), and vice versa, as long as F.t = Δp (assuming constant F).

    Exactly why the time is shorter or longer would depend on the material: if you put a soft pad on the load cell, the impact would be softer, but the Impulse would be the same. But I see no reason why the force measured by the load cell could not be higher than the force of the motor. The motor force might be less, but if it operates for longer, then the momentum will be F.t - which is then dissipated over a shorter period of time (hitting the hard load cell) with a higher force.

    That's my take on the question, anyway.

    edit: about the question "what force it would hit someone or something with if it came into contact with them while moving" - asking about force isn't really the best way to think about the question, in my opinion, and not only for the reasons above (the force will depend on the softness). If the target is fixed, and there's no bounce, then it's going to have to absorb all the kinetic energy that the door had. If it's a person, it might not be a huge force but it could still hurt ..!

    From out there on the moon, international politics look so petty. You want to grab a politician by the scruff of the neck and drag him a quarter of a million miles out and say, ‘Look at that, you son of a bitch’.

    — Edgar Mitchell, Apollo 14 Astronaut



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    #8
    Dubh Geannain
    Registered Users Posts: 3,393 ✭✭✭Dubh Geannain


    So my final conclusion of what is happening is okay then?

    Thanks for your insights and apologies for not clarifying the problem clearly enough in my first post. It's been a while since I've looked at something like this (probably 1st year in engineering!) so am obviously quite rusty.


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