Gms Stainless
How to arrive at the size (volume) of Stainless steel ball (made of 1 mm thick plate) to float in kerosene?
The SS ball floating on Kerosene (Density 0.780) should be capable to push 500 gms load placed on top of it.
The floating ball should activate a pneumatic switch which requires at least 400 gms force to activate a cylinder when the kerosene level reaches a certain height.
The floating ball should activate a pneumatic switch which requires at least 400 gms force to activate a cylinder when the kerosene level reaches a certain height. Has the volume of kerosene in the tank plays a role in the size of the ball?
Buoyancy force of the ball has to balance the 3 gravity forces (or weights) of the 2 loads and its own.
According to Archimedes this force equals the weight of the displaced fluid.
Assume that when all system is in equilibrium, the steel ball with radius R is totally immersed into the kerosene. This will give us the minimum radius for such a ball you are looking for. A ball with radius greater than R will be partially immersed into the fluid.
Now you have to evaluate all the forces.
buoyancy B = V rho g
where V = 4/3 pi R^3 is volume of the ball ( and displaced kerosene), rho= .78 kg/m^3 is its density and g=9.8m/s^2 is the gravity constant.
Total gravity force is (m1+m2+m3)g
where m1=.5kg, m2=0.4kg and m3 is the mass of steel used to make the ball (the weight of air inside is neglected)
m3 = A h Rho where A=4piR^2 is ball's surface area, h=.01m its thickness and Rho=6980kg/m^3 is liquid iron's density.
I leave it to you to write down the equation and solve it for R.
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