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Starting force of a falling object on a compound lever

  1. Feb 25, 2017 #1

    I am building a bicycle with suspension on it. The rear wheel is connected to the bike via an axle on a swing arm (a long lever), connected to a linkage lever which is connected to a shock, which has a spring on it. Ultimately I'm trying to calculate the force applied to the shock under certain cases.

    Generally, I think I understand how to calculate the force on the shock given some force at the axle on the end of the swing arm. I use the total torque and force equations and N forces.

    What I'm not sure I understand is the vertical upward force (Newtons) applied to the axle when the bike is sitting at rest and when it is dropped from some height, which is probably funny cause it's much easier to calculate than the forces through the swingarm and linkage.

    I just want to be sure I'm not missing something.

    For this discussion, we can assume the rear axle bears 60% of the total weight of the bike and rider.

    Bike and rider weight = 100kg

    When the bike and rider are just sitting there on the ground, not moving:
    Vertical upward force on the bike as a whole is 100 * 9.8 = 980N.
    Upward force on the rear axle is 980 x .6 = 588N.

    At the point the bike and rider touch the ground after falling 1m:
    Up force = 980 + (980 * 1) = 1960N * .6 = 1176N. This is the value I use as the vertical up force being entered into my torque and force calculations through the swing arm and linkage.

    At the point the bike and rider touch the ground after falling 2m:
    Up force = 980 + (980 * 2) = 2940N * .6 = 1764N.

    At the point the bike and rider touch the ground after falling .5m:
    Up force = 980 + (980 * .5) = 1470N * .6 = 882N.

    Am I doing this right? Specifically I not completely sure I should be adding the weight but I believe I need to because the mgh is the kinetic energy added to the mass by the velocity obtained through the fall.

    Thanks for any feedback. :-)
  2. jcsd
  3. Feb 25, 2017 #2


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    What are you doing there? It looks like you are multiplying a force by a parallel distance (which should give the work done by the force) and adding a force. You cannot add energy and force.
  4. Feb 26, 2017 #3
    Thanks for the reply haruspex.

    I want the vertical force being applied up on the end of the swingarm lever at the axle, Fav, so that I can calculate the vertical force up on the bottom of the shock at Fsv.

    If the unsprung weight of the bike (my picture below doesn't show a rider but we can pretend he's there and part of the 100kg total weight) shown in green weighs 100kg, just sitting there on the ground I think Fav is 100 * 9.8 = 980N.

    Now if the bike is just starting to touch the ground after falling 1m, wouldn't I additionally add the force from the acceleration of the fall? In that case I'd be adding mgh which is 100 * 9.8 * 1 = 980N (N after converting the KE to Force). Now I have an Fav of 980N from the bike at rest + another 980N due to the fall.

    So, Fav at the point of impact, to be applied as vertical up force on the axel to calculate the force on Fsv is 1960N.

    Am I wrong? Is it not 1960N?

  5. Feb 26, 2017 #4


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    mgh is the kinetic energy, not a force. You cannot add and subtract terms that are dimensionally different.
    (As an aside, look what would happen if you were to convert to different units. You would get a quite different answer. That's a sure sign you've mixed up the dimensions.)
    The force resulting from the KE depends on how quickly the motion is brought to a stop. If I drop an egg 5cm onto concrete the forces will be much higher than if I drop it onto a rubber pad. Also, the force varies during the impact.

    For more on the subject see https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/.

    If we model the bicycle suspension as a spring of modulus k, the maximum force occurs at the maximum compression. If that is x, the energy absorbed is kx2/2 and the peak force is mg+kx, so we can write
    You can eliminate x between them, but you need to know k.
  6. Feb 26, 2017 #5
    I understand what you are saying. But I still don't understand how to apply it to my situation. Let me explain a little more.

    In the model, let's say the shock has 3" of travel, so the spring is going to be compressed 3 inches. I want to know the spring rate the spring needs to have, in this example, let's say to compress that spring 3 inches after the 100kg bike falls 1m.

    Spring Rate = Load / Travel.
    Spring Rate = Fsv/3.

    I know the angles and force fulcrum distances on the swing arm but in order to get Fsv I'll need to know Fav.

    So I'm not sure how to get Fav so I can calculate Fsv.

    My method above is how I've tried to get Fav, thinking I am looking for the force at that moment in time where a force at all begins to be applied to the axle. Sum of torque and force equations should tell me how much that force is after being leveraged through the swing arm to a greater force at Fsv. Then that force, Fsv, is applied to the spring and, with the right spring, compresses it 3 inches.

    If I'm wrong, can you help me understand what number I should be using for Fav in this calculation and how I would go about getting it?

    Also, I understand there are other relatively small forces at play here such as friction and such. In reality, I can change the spring weight to meet actual riding conditions. I just want to be sure I'm not so far off with the geometry of the frame and linkage that I can't get a spring weight even close to what I need. So, it's a good estimate. I'm not sending a rocket to mars.

    Last edited: Feb 26, 2017
  7. Feb 26, 2017 #6


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    Seems too much to me, but let's go with that.
    So that is x in my equations, so you can find kx from the first of them.
  8. Feb 26, 2017 #7
    x = 3 inches?

    You said "modulus k" above which implies to me k is the change the spring goes through so k would be 3 inches. Did you mean k is 3 inches or you really mean x is 3 inches? And the other one is the max force at the point the spring is fully compressed.
  9. Feb 26, 2017 #8


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    No, the modulus of a spring is its response to an applied load, specifically, applied load divided by change in length. So in SI units, N/m. If x is the change in length then the force is kx and the work done (from the relaxed state) is ½kx2.

    Edit: actually, I have used the wrong term. I should call k the "spring constant", so as not to get confused with Young's Modulus, which is related but defined differently.
    Last edited: Feb 26, 2017
  10. Feb 26, 2017 #9
    I see this as a 2 part problem...
    The shock absorber and the spring both absorb energy in different ways.. the shock being linear in respect to position, but not in respect to velocity, while the spring is linear in respect to velocity, but not position (harder to move the more it's compressed).. In addition to this most gas charged shock absorbers also have a certain amount of "spring" to them

    I don't know the math behind it, but it really sounds like there's an infinite number of solutions depending on the shock used as that will change the energy absorbed.

    Correct me if I'm missing it, but it doesn't seem like you list the ratio of the swingarm pivot point to the wheel vs shock... that's going to be very important.

    Also, most suspension systems have spring preload, and a certain desired "sag" when the rider is on it.. Getting all this correct and to work together isn't going to be easy, and it'll change with riding style, rider weight, and personal preferences
  11. Feb 27, 2017 #10

    I'm not really sure how you are expecting me to apply this equation to getting Fav.

    "spring constant" is what I mentioned above in comment #5 as "spring rate".

    F = kx, yes. I'm talking about this equation at the shock's spring. I want to find k for a given F where F is the force pushing against the bottom of the spring/shock.
    k = F/x
    k = F/3

    F is the force applied to the bottom of the spring. To get that, I need to know Fav from the picture of the bike above because in reality there is a swing arm and series of levers in a linkage system connecting the shock to the axle force.

    Guys, I already know all the angles in the linkage. Yes, Rx7man, there are many possibilities of angles on the swing arm, ratios to pivot point, etc. But I know how to calculate all of that for any of those given scenarios using sums of torque, forces and trig. Given some value of Fav, I can tell you what the spring rate should be so the spring is compressed 3 inches by that force.

    In the picture above, which is simpler than my real world case, for sake of argument, it would be something like
    Ex: 3000N (Fav) vertical up applied to axel yields 12000N to bottom of shock (parallel to it's angle). 12000/3 = 4000. Then I convert to lbf and end up with some spring rate of 600lbf/in. These numbers are complete BS but you can see the relative increase or decrease in forces.

    So, how do I find Fav in order to start my trig equations in the swing arm?
    If the 100kg bike is sitting still on the ground, Fav is mg right?
    And if it just free fell 1m, Fav is mg + mgh?

    Or haruspex, are you saying rather than Fav after a 1m free fall being mg + mgh, it needs to also consider the movement the bike will go through in the suspension? That seems strange in this case since it's ultimately the spring rate I'm trying to find and without knowing a spring rate, I can't really say how far the KE of the fall will move through the suspension.

    Note: Fav is not just KE from a fall. It is the N force being applied to the end of the swing arm in a given case, even if the bike is sitting still on the ground. I want to be sure I'm applying the fall case KE correctly.
  12. Feb 27, 2017 #11


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    If a mass m falls distance h onto a verical spring of constant k and thereby depresses it a distance x then the energy equation is
    If x is known or estimated, this gives k= 2mg(h+x)/x2.
    The max force (isn't that rather more important than the average force?) is kx = 2mg(h+x/x.
  13. Feb 27, 2017 #12
    Actually, I think k is more important in this case. The springs are labeled by their "spring weight" which is the "spring rate" or "spring constant". It's the "average force" as lbf/in. I don't want to buy a 1200 lb/in spring. I want a 600lb/in spring that will be compressed 3 inches. For example, http://www.jensonusa.com/Springs/Fox-SLS-Rear-Spring-35-Stroke.


    k= 2mg(h+x)/x2
    3in = .07m
    k = (2(100)(9.8)(1+.07))/.072
    k = 2097.2/.0049
    k = 428000N?

    That looks totally wrong. In fact, we see the smaller the movement of the spring, the more ridiculous the force amount becomes, infinitely. Something is missing.
  14. Feb 27, 2017 #13


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    You misread my post. It's k=2mg(h+x)/x, not x2.
    It is quite correct that it should tend to infinity as x tends to 0. It takes an infinite force to to stop a motion instantly.
  15. Feb 27, 2017 #14
    You mean you mis-posted. ;-)

    So, really k is 29960N.
    If I understand, this is the force applied up on the bike as it hits the ground that causes the spring to be compressed 3 inches.

    A 100kg bike sitting on the ground exerts a force on the ground, and therefore the ground exerts the same force up of mg, which is 980N.

    So, dropping that same 100kg from 1m + .07m leads to 29960N being "exerted by the ground" up on the bike. A 30 time increase in force?

    Does this number make sense?

    If it does, I'm still confused by the x in the equation. Mainly... what about when there is no spring. I can't use x=0 but there is a force applied to the frame if it was just dropped 1m to a steal floor. Both floor and frame are so rigid but yet the force is not infinite. I suppose the steel gives a little and the frame bounces and x would represent that bounce movement?
  16. Feb 27, 2017 #15


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    No, I posted correctly the first time, but I confused you by posting the wrong correction to your post .:-(
    The peak force is 2mgh(h+x)x.
    In post #12 you wrote
    No, k=428000N/m. The peak force is about 30000N, which is entirely reasonable.
  17. Feb 27, 2017 #16
    The bike has tires, right?, if it's a typical tire it'll have about 1" or so of "spring" action to it before it bottoms out the tire.. how much force is exerted will depend on tire profile and tire pressure
  18. Feb 27, 2017 #17
    If the bike is sitting on the ground, Fav would be close to mg.. though you should really subtract the force due to the weight of the wheel assembly
  19. Feb 27, 2017 #18


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    The main point I am making is that mg+mgh is utter nonsense. It cannot be the correct formula for any physical entity because it is dimensionally inconsistent. It is like adding a distance and a speed. What units would the result be in?
  20. Feb 27, 2017 #19
    Rx7man, thanks for confirming. Yes, in my real calculations I do consider Fav to be calculated with m = "sprung mass", the mass above the suspension. I should say in my example 100kg is the sprung mass.

    So, if we can ignore the unsprung mass for now and Fav is mg (or close enough we can simplify it as such) then what would Fav be when instead the bike has just touched the ground at the end of a 1m fall? Note at this point, there is no movement yet through the suspension. It's just before impact.

    My attempt goes:
    The mass of the bike is 100kg so the force pushing up against it sitting still is 100*9.8 = 980N. But now it's not sitting still, it additionally has KE = mgh = 980J. That 980J will be converted to 980N of force and added to the first 980. So there is 980 of normal force (the mass has simply from acceleration of gravity on the ground) plus another 980 due to the velocity of the fact it's just fallen 1m. haruspex, this is why I said mg + mgh, because even sitting still without a fall, mg is being applied to the axle. mgh is the additional force from the KE due to the fall.

    Now, at this moment in time, the force pushing up on the axle is 1960N. Let's say that works through the swing arm lever to result in a bigger force at the shock attachment point as 8000N. The shock is angled, we'll estimate 40 degrees so that 8000N vertical applies a force to the bottom of the shock of cos40 * 8000N = 6128N. 6128N = 1377lbf Then, if the shock has 3in of travel the spring constant will need to be 1377lb/3in = 459lbf/in. I need a "450lb spring" to absorb this fall.

    The real linkage, as I've stated many times now is more complicated than this so I need a starting point for the math... a force I can push on the first lever with, which pushes on the next lever, etc, until yielding a force applied to the bottom of the shock, at which point I can calculate the spring rate by how much it will move and the force applied.

    Make sense?
  21. Feb 27, 2017 #20


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    How can I persuade you that is nonsense? Energy is energy, force is force. The one does not magically transmute into the other.
    Consider how that would play out in different units. The old cgs system used centimetres, grams, seconds, ergs and dynes. By your reasoning, 980 ergs could turn into 980 dynes. 1 erg = 10-7J, 1 dyne = 10-5N, so that would mean 980*10-7J turns into 980*10-5N, conflicting with your claim.
  22. Feb 27, 2017 #21
    Sorry haruspex. Thanks for bearing with me.

    I'm not a physicist by any means so I admit I could be wrong. Even I'm likely wrong. In fact the assumption I'm wrong is why I've come to this forum. But I'm not understanding the connection with your equations and how they apply to my case of needing Fav. Knowing the work done by the spring in an isolated model is not helpful to me because I need to calculate the load put into the spring in differently configured suspension linkages. All of those linkages start with Fav into trig math through a series of levers. If I know Fav, I can find the load put on the spring.

    Maybe if we take smaller steps so we understand each other.

    Do you agree with me that I need Fav in order to find the load ultimately put on the shock spring?
  23. Feb 27, 2017 #22


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    It depends what you consider to be the knowns.
    If you start with the known characteristics of the absorber and of the tyre, and the drop height, you could not calculate Fav without involving the absorber characteristics. If, by some experimentation, you were to find the overshoot past the equilibrium position during the landing, you might be able to estimate Fav independently of the absorber characteristics, but if you then change the absorber that calculation will not be valid.
    Let's consider the absorber. As Rx7man posted, it essentially consists of a spring plus a damper. If the compression extent is x at time t, the force opposing motion it generates will be a sum of terms that are functions of x and of dx/dt.
    Any frictional component of the damper will contribute a constant term.
    An ideal spring component will contribute a kx term.
    There may be some air compression, contributing a ((1-x/L)-1) term. (In adiabatic compression, PVγ is constant.)
    There may be an air drag term, maybe of the form dx/dt or (dx/dt)2.
    So in general it is rather complicated.
    Having figured that out, you could use geometry to relate x, dx/dt and the opposing force of the damper to the angle (θ, say) that the rear wheel arm makes to the seatpost, dθ/dt, and Fav. Finally, you should add some springiness in the tyre itself.

    For the geometry step, you need to think about components of motion - it's not trivial.
  24. Feb 27, 2017 #23
    I understand all these variables are there but like I said I can change the spring weight to best fit actual conditions. I'm just trying to get in the ballpark. I don't want to build so much leverage into the swingarm and linkage that a 10lb spring is required to compress the full travel or the opposite, so little leverage I end up needing a 1500lb spring. I can get a spring from 250 to 400lb easily so getting somewhere in there is good.

    The resistance from the damper is small relative to that of the spring. Same with tires. In a 1m fall, the tires aren't going to give more than 1 or 2mm whereas the spring is going to give 76mm. So, I feel I can safely ignore the affects of the tires and damper to get a calculation close enough to get a spring to start with. Then if it's too soft I'll increase it but I'll be in range so I can obtain a spring that works.

    Let me try another approach here which is quite simpler.

    Let's say I just want to know how much Fav needs to be in order to compress the spring 25%. 25% is the sag I want, which means it is reached while the rider is sitting on the bike, not moving, on the ground. Now if the rider and bike weight on the rear wheel are 100kg, is Fav 980N? (We're ignoring the front wheel) Or does Fav still have to consider the .25 * 76mm = 19mm sag compression distance in the spring?

    If I modeled this in the real world I could measure 100kg at the rear wheel on a scale. An upward force must oppose that weight and that upward force would then be 980N so seems Fav is 980N.
  25. Feb 27, 2017 #24


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    As I hinted, the relationship between the compression force on the spring and Fav depends on the geometry. As the spring compresses, the geometry changes a little. So let θ be the angle the rear wheel arm makes to the suspension rod and Φ the angle it makes to the horizontal, measured when at max compression, say,. Do you know how to takes moments about the pivot point for the rear wheel arm?
  26. Feb 27, 2017 #25
    Yes, I think I know what you mean.

    So, if Fav is 980N, that is the vertical up force. Then I can calculate the force at Fav perpendicular to the swing arm and by finding torques at Fsv and Fav I can find force at Fsv. I'll consider the angle the shock has to the swing arm to get the actual force into the shock. I know how to do this and have built a spreadsheet for it. I have the angles at 0, 25%, 50%, 75% and 100% compressed. In my real world case there are multiple calculations because there is another multiplier lever between the swing arm at Fsv and the bottom of the shock, which I've left out of this example to simplify the conversation. I'm pretty confident I know what's going on in those calculations.

    You mention "moments" around the swing arm. I can reason you mean I should calculate that force on the shock using angles at different points from fully extended to the 25% compressed, as I've done at 0, 25, 50 and 75% but in this case multiple points from 0 to 25%. Say I calculated the force on the shock at 4 different points from 0 to 25% compression, then would I simply average these to get the force to be used as the force on the shock in calculating the spring rate?
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