I need to design a pivot arm which is pulled up by a float (buoyancy force). In the upper positions of the lever arm, there's a mass trying to pull down. The reason that I'm presenting this problem is to consider all the variables (most significant) in order to build a low-cost prototype.

enter image description here

The variables that I consider are: 1.- Volume of the float, the material is polypropylene. I need a low density but high volume float in order to generate more buoyancy force, is that correct? 2.- The distance between the force applied and the pivot "d1". If I generate more torque, more will be the force generated by the buoyancy, right? 3.- The angle of the arm A.

This is my first attempt presenting this problem, so any suggestion will work. The goal is to achieve a bigger force than the 95 N generates by the "object x" in order to reach the 'sealed position' denoted by "d2".

  • 1
    $\begingroup$ You are correct about generating buoyancy. The buoyant force depends on the part of the float that is submerged. Then just ( :-) ) calculate the torque applied by the ObjectX: is it attached to Arm B, or is it free-standing so that it does not rotate about the pivot point? $\endgroup$ Mar 5, 2018 at 16:19
  • $\begingroup$ The object X is attached to the arm B. So when the water level goes down, the force of the object x generate a rotation of the arm A about the pivot point. When the water level goes up, the float will generate a rotation about the pivot point and reach the sealed position (d2), or at least that's what I'm intending to do. $\endgroup$ Mar 5, 2018 at 17:49
  • 1
    $\begingroup$ I think, then, that all you need is a buoyant object which is partially submerged at any water level -- i.e. regardless of the effective torque applied by the float and the massX, each of which is max when horizontally at same level as pivot, and minimum when vertically above or below the pivot (extremes which aren't reachable in your design because of the container wall). So long as that's the case, the system will always seal when the water level is high enough. If you want to control what "high enough' is, probably easier to make the massX threaded to adjust up & down arm B. $\endgroup$ Mar 5, 2018 at 19:31

2 Answers 2


Edit: Added the requested drawing + corrected formulae for the 4 bar setup

As Carl Witthoft mentioned, the buoyancy force $F_1$ from the block is proportional to the submerged/displaced volume ($V_{sub}$).

$$ F_1=\rho gV_{sub} $$

Let's the distance between the Arm A pivot and the intersection of Arms A and B be $d_3$

Summing the moments around Pivot A (for the 4 bar setup, see below):

$$F_1 sin( \theta ) \cdot d_1 + F_{objectx}sin(\theta)\cdot d_3=0$$

$$ F_{objectx}= \dfrac{-\rho g V_{sub}\cdot d_1}{d_3}$$

(I have negated the weight of the bars here)

So you can move object x, reducing the length $d_3$, or to increase the length $d_1$.

Other than that, you can only increase $V_{sub}$, which will increase anyway as the water rises. But you will have to adjust your geometry to make sure you reach the sealing force at your desired water height.

Adjusting for the desired water height:

  1. Calculate the required $V_{sub}$, where $F_{objectx} $ is the force that we need to both suspend the weight of object x (95.6N?) and seal the container.

  2. Make sure that your block volume > $V_{sub} $.

  3. Decide how high you want the water to go in order to create a seal ($h_{wmax}$). Adjust the geometry to ensure that when object x is in it's sealed position and the water is at $h_{wmax}$, $V_{sub} $ is as calculated.

I would suggest using a 4 bar linkage instead of your current set-up, so that object x does not rotate, and only transfers force vertically, assuming this was your design intention.

  1. Add an ARM C parallel to ARM A. It should be connected to the wall on a pivot above point A, and connected to ARM B with a pin, above ARM A.

  2. Change the connection between ARM A and ARM B to be a pin.

enter image description here

  • $\begingroup$ Thank you C.R. for your comments! Can you add a drawing showing the tips 1 and 2 about the 4 bar linkage that you suggested, please? $\endgroup$ Mar 8, 2018 at 13:22

enter image description here

In order to increase the buoyancy force over than the force applied by the weight of the object x, I'm considering increasing the distance "d1" or maybe decrease distance "d2", which is the distance between the force applied by the weight and the pivot point. So, the momentum in point A (pivot) generated by force of 95.6 N will be lower than the buoyancy force.

Any other suggestion?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.