PYL 105

Archimedes' principle

Archimedes' principle says: A body wholly or partially immersed in a fluid is buoyed up by a force equal to the weight of the fluid it displaces.

Part 1: Density of water

• Place an empty graduated cylinder on a balance, determine its mass.
• Add some water, note the mass and the volume of the water.
• Add more water and take measurements for a total of five sets of masses and volume.
• Plot mass versus volume.
• Fit to a straight line and interpret the slope.

Part 2: Force as a function of volume displaced

• Connect a force sensor to the interface. Hang it from a ring stand.
• Tare (zero) the force sensor while nothing is hanging from it.
• Thread a string through the hole in the brass cylinders and hang it from the force sensor.
• Set up a hose connecting the spout on a flask to a graduated cylinder as shown below. (The path to the graduated cylinder must be entirely downhill.)

  

• Get some paper towels. Fill the flask to the point at which water spills over into the graduated cylinder. (Don't simply fill it up to the point at which you think is the spout level, you should actually observe some water spilling out.)
• You are going to lower the brass cylinder into the water, measuring the tension (force sensor reading) and volume of water displaced (graduated cylinder). You should have at least five measurements including an initial measurement with no water displaced and a final measurement with the cylinder fully immersed. Note: careless adjusting the clamp on the ringstand can result in an undesired lifting of the force sensor and cylinder. This will result in poor data. Loosen the clamp as little as possible to change the height. Practice a few times before taking your data.
• Plot Tension versus volume displaced. Fit it to a straight line. Determine what the slope should be theoretically. What is the percent error?

Part 3. The density of brass.

• Method 1. Calculated volume.
  
  • Measure the mass of the brass cylinder.
  • Measure the radius r and height h of the cylinder.
  • Calculate the density r_brass (mass/volume) of the cylinder in kg/m³. Recall that the volume of a cylinder is

    V_cyl = p r² h

• Method 2. Measured volume.
  Note it was the cylinder volume that displaced the water. Thus we could obtain the cylinder volume from the final water displacement reading from Part 2.
• Method 3. No volume.
  Neglecting any buoyant force due to air, the weight in air is given by

  F_weight = r_brass g V

  When the cylinder is submerged, there is an additional force, the buoyant force, acting. It is given by

  F_buoyant = - r_water g V

  where the minus sign indicates that the buoyant force is in the up direction while the weight was down. Derive an expression the density of brass r_brass that involves your initial (not submerged) force F_i and final (totally submerged) force F_f from above, but does NOT include volume. In this method you should not use the volume you calculated, but you can use the density of water (r_water = 1000 kg/m³).

The nice feature of this last technique is that it does not require any measurement or calculation of the volume. Apply the same technique to measure the density of one of the short hexagonal shaped weights. Also measure the density of one of the longer hexagonal shaped weights.

Discuss the relative merits of these techniques. What is hydrostatic weighing? What is it used for?

Part 4. Floating

If an object's density is less than water, it will float. That is, it will be positioned at the surface of the water with only some of its volume below the surface. If the object is in equilibrium then its weight must be balanced by the buoyant force (assuming no other forces are acting).

• Find the mass of a wooden block.
• Calculate its volume (length ´ width ´ height).
• Calculate the density in kg/m³.
• Place the block in the water and then determine the volume of the block that lies below the surface (length ´ width ´ submerged height).

  

• Next calculate the weight of the displaced water and compare it to the weight of the block.

One possible difficulty with this measurement is in determining the amount of the block that is below the surface. This is because fluids do not meet the surface of the block at a right angle. Typically the fluid-block interface will look roughly like one of the pictures below.

We say in the case on the top that the water "wets" the surface. In this case, if you measured the length of wetness to obtain the "submerged height," you would tend to overestimate the submerged volume. Those of you who have had chemistry may have seen/discussed this phenomenon in connection with the meniscus. Comparing the actual weight of the block to the calculated weight of the water displaced, do you think you underestimated or overestimated the displaced volume?