# Archimedes Principle

A crane is used to lower a vertical cylindrical pillar into a reservoir. The pillar has a mass of 4 tonnes and is submerged to a depth of 2.5m. It has a diameter of 1.1m.

The specific gravity of fresh water is $$1125kg/m^3$$, and $$9.81m/s^2$$.

Use a free body diagram to show the forces acting on the pillar when submerged and calculate the upthrust, the tension in the rope in air and the tension in the cable when submerged.

I have this for Upthrust:

$$Fbuoy = Pf \cdot Vd \cdot g$$ $$\pi r^2h$$ $$3.14 \cdot .55^2 \cdot 2.5 = 2.375m^3$$ $$1.125 \cdot 2.375 \cdot 9.81 = 26.2N$$

$$Upthrust = 26.2N$$

Tension in air

$$T = w - Fb$$ $$\pi r^2h$$ $$3.14 \cdot .55^2 \cdot 2.5 = 2.375m^3$$ $$Volume = 2.375m^3$$ $$w = m \cdot g$$ $$w = 4 \cdot9.81 = 39.24kN$$ $$w = 39.24kN$$ $$Fb = Pf \cdot Vd \cdot g$$ $$1.29 \cdot 2.375 \cdot 9.81 = 30N$$ $$T = w - Fb$$ $$T = 39.24 - 30 = 9.24N$$

$$Tension in air = 9.24N$$

Tension in water

$$T = w - Fb$$ $$\pi r^2h$$ $$3.14 \cdot .55^2 \cdot 2.5 = 2.375m^3$$ $$Volume = 2.375m^3$$ $$w = m \cdot g$$ $$w = 4 \cdot 9.81$$ $$w = 39.24kN$$ $$Fb = Pf \cdot Vd \cdot g$$ $$Fb = 1.125 \cdot 2.375 \cdot 9.81$$ $$T = w - Fb$$ $$T = 39.24 - 26.2 = 13.04N$$

$$Tension in the cable when sumerged = 13.04N$$

Are these correct?

If not could someone help?

I have been trying to complete Archimedes Principle but just getting a little stuck on the calculations.

Hi. Here is what the Free-body diagram should look like. When the cylinder is suspended in the air, the weight of the pillar must equal the vertical tension in the rope. When lowered, the bouyant force pushes the pillar upwards with a force equal to the weight of the volume of water displaced. This bouyancy force reduces the tension in the rope by the amounts shown.

Cheers!

• Thank you! So i had most of it correct. Just some of the units incorrect. – Kyle Anderson May 6 at 18:55