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Multiple leverage math

  1. Jan 23, 2013 #1
    So I am working on a cart and the last thing I can not figure out is the math to the braking system. It is relatively simple, though I have over complicated it. I am lost with how to add the multiple levers in play.

    So lets begin.
    Problem 1) What is the ratio of force between input and output?

    Problem 2) The braking apparatus I am designing has two sets of levers on either side identical to a scissor clamp. I can't even fathom how to figure that out,

    Problem 3) From the braking system, it will then be ran by cable to a simple hand lever at about 5:1, or 6:1.... How does that add up with the scissor clamp?

    Totally lost, I know I have probably not given all the required information, so please tell me what you need to help me figure this out. Also, I am not looking for someone to give me the exact answer, I like to figure things out as well, just have no idea how to actually figure it out.

    Thank you for any help.

    Attached Files:

  2. jcsd
  3. Jan 24, 2013 #2


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    In that scissor mechanism, the ratio of input to output is not the same throughout the range of action. The two levers rotate about the centre pivot and what counts is the 'perpendicular distance' of the line of the force and the pivot (you could say it's the radius around which the force is acting). As you pull on the cable and the jaws come together, the angle changes that the two links are pulling and that changes the mechanical advantage (the effective ratio that the lever gives you). Also, the two top links change their angle relative to the cable and that also affects the force on the levers.
    So you would need to know the spacing of the jaws then the brake blocks are actually in contact with the wheel rim. Once you know that, you can start on a rough calculation. (I say "rough" because the force from the block is distributed and you can't be sure where it is effectively acting. )
    First the levers: The output length would be measured to (say) the centre of the brake block from the pivot centre and the input length would be measured as the shortest line you can draw from the pivot to the centre line of the link (that will be a line at right angles to the link and through the pivot - when fully extended, this perpendicular line may not even intersect with the link itself but an extension of its mid line). The mechanical advantage will then be
    (input length) / (output length) and this may be a fraction less than one.
    Next the links: A force F pulling vertically upwards will produce a force F1 on the links of
    F/2Cos(θ/2) where θ is the angle between the two links (about 120 degrees, in the picture)
    That force on the cable is shared between the two links - hence the division by 2. The Cosine in the formula shows that, if the links lie in a straight line (θ is 180) you will get 'infinite' force and that drops to F/2 when the angle between the links is zero.
    The overall force F2 on your wheel will be F1 times the lever ratio. There is also the advantage of the hand lever to be considered - which you can also calculate using the length of the lever and the radius of the channel carrying the cable in the lever. Again, you have the problem of deciding exactly how effectively long the handle is and which bit of your hand is doing most. It will give you some idea, though. You could check the credibility of your final answer by finding what sort of weight you can lift with a strap, using your fingers.
  4. Jan 24, 2013 #3
    I probably should have mentioned this before, but I am a very simple person, never really could get the hang of multiplying the alphabet or symbols. I know and learn what I need too, but usually requires some crayons lol. That said, I am sure what you have said was very informative but not quite down to my layman level. So I have been studying this all day and this is what I have come up with.

    Starting with simple numbers for ease of conversation, and please correct where I am wrong. Starting at the front with the first hand lever, saying that it is a 10:1. So with 10# of input would be 100# output. From there we run down the cable to the brake levers. From my rough drawings I am figuring the two levers to lay at about 160 degrees, with about .125 of movement, figured this would allow for max leverage. Again, keeping this very simple I have figured those to be a 2:1 and I am guessing that with two levers that it will divide and not multiply (like I originally thought), the input force between the two levers. I am now concluding that with 100# input to the brake levers that it would equate to 100#s each lever, squeezing toward one another, totaling in 200# of combined braking/clamping/squeezing force.

    Sound about right? Where did I go wrong, where did I go right?

    Now I am not doing homework or building a spaceship, so exact numbers are not necessary. Just trying to figure how to stop the cart without ripping my shoulder out of socket, or flying off the rails. I suppose if anyone actually cared, I could give exact numbers and they could figure it out for grins and giggles lol.

    Again, thank you very much for the information and taking time to deal with my elementary questions.

  5. Jan 25, 2013 #4


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    I'm not sure why you asked your original question if you didn't want a way of working out the answer. If you want a more straightforward andswer to this problem you would be better on a cycling forum where people may have the results of measurements and also they will have much more practical experience.
    You can rely on the fact that the brake linkage you have was designed to take care of an average load for a wheel of that particular width. The actual force you will have to supply will depend as much on the pedal / hand lever as on anything else.
    You are right about multiplying the ratios together - but be prepared for some of the ratios to be less than, or near one.
    One point - the two halves of the calliper do not double the force by acting against each other. It's done that way just to avoid sideways forces on the wheel. Theuseful force you produce with the cable is halved, in fact. (as with all cycle brakes)
    Weight for weight and, bearing in mind that you will have four wheels (or at least a pair of them) braked then the effort you will need to stop you plus the engine etc on double the number of wheels wouls be about the same as for a bicycle. How do you balance the brake force to each side? That was always a problem for old motor cars before they used hydraulics,
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