Experiment 9: Archimede's Principle

1. A meter stick
2. pivot
3. hangers
4. cylindrical object
5. rectangular object
6. a beaker
7. a string
8. water

Theory

Because of this bouyant force, an object immersed in fluid weighs less than an object weighs in air. The bouyant force ( B ) is equal to the difference between the weight of the object in air ( W_air ) and the weight of the object in the fluid ( W_fluid ).

B = W_air - W_fluid . . . . . ( 1 )

Also, according to Archimede’s principle, the bouyant force is equal to the weight of the displaced fluid. Thus, if the volume of the object is V and the density of the fluid is ρ_fluid , the bouyant force ( B ) is also given by

B = ρ_fluid V|g| . . . . . ( 2 )

Comparing equations ( 1) and ( 2 ), the following expression for the volume of the object can be obtained.

V = ( W_air - W_fluid ) ⁄ ( ρ_fluid |g| ) . . . . . ( 3 )

Procedure

1. To determine the volume of a rectangular metal object by using Archimede’s principle and by measuring its dimensions and then compare

   1. Measure the masses of the two hangers by means of a balance.

      hanger 1 mass ( m₁ ) = kg = 0.022 kg
      
      hanger 2 mass ( m₂ ) = kg = 0.022 kg

   2. Put the meter stick in the pivot. (Use one of the hangers without the suspending wires as your pivot.)

   3. Put the pivot-meter stick system on the fulcrum and balance it to determine the location ( x_cg ) of its center of gravity and .

      center of gravity ( x_cg ) = m = 0.5 m

   4. Put hanger 1 on the x₁ = 0.4 m mark and determine the perpendicular distance ( r_⊥1 = x_cg - x₁ ) between hanger 1 and the center of gravity

      hanger 1 distance ( r_⊥1 ) = x_cg - x₁ = m

   5. Attach the rectangular object to a string and the string to hanger 1. Suspend the rectangular object inside an empty beaker. Put hanger 2 on the other side of the fulcrum and balance the meter stick by moving hanger 2 across the meter stick making sure the rectangular object doesn’t touch the beaker. Determine the location ( x₂^air ) of hanger 2 and the perpendicular distance ( r_⊥2^air = x₂^air - x_cg ) between hanger 2 and the center of gravity.

      hanger 2 location ( x₂^air ) = m
      
      hanger 2 distance ( r_⊥2^air ) = x₂^air - x_cg = m

   6. Since the system is in rotational equilibrium, the net torque acting on it should be zero. If the weight of the rectangular object in air is W_air , then from condition of rotational equilibrium we get

      ( W_air + m₁ |g| ) r_⊥1 = m₂ |g|r_⊥2^air

      or

      W_air = ( m₂ |g|r_⊥2^air ⁄ r_⊥1 ) - m₁ |g|

      Using the mass of hanger 1 ( m₁ ) obtained in procedure (1.), the mass of hanger 2 ( m₂ ) obtained in procedure (1.), the perpendicular distance ( r_⊥1 ) of hanger 1 obtained in procedure (3.), and the perpendicular distance ( r_⊥2 ) of hanger 2 obtained in procedure (4.), calculate the weight ( W_air ) of the rectangular object in air.

      weight in air ( W_air ) = ( m₂ |g|r_⊥2^air ⁄ r_⊥1 ) - m₁ |g| = N

   Now pour water into the glass slowly so that the rectangular object is immersed in the water completely. Because of the bouyant force due to the water there will be a non-zero torque and it will not be balanced with the earlier location of hanger 2. Balance it again by changing the location of hanger 2. Determine the location ( x₂^water ) of hanger 2 and the perpendicular distance ( r_⊥2^water = x₂^water - x_cg ) between hanger 2 and the center of gravity.

   hanger 2 location ( x₂^water ) = m
   
   hanger 2 distance ( r_⊥2^water ) = x₂^water - x_cg ) = m

   Once again since it is balanced the net torque is zero and the weight ( W_water ) of the rectangular object in water may be calculated by a process similar to that of procedure (6.).

   weight in water ( W_water ) = ( m₂ |g|r_⊥2^water ⁄ r_⊥1 ) - m₁ |g| =

   According to Archimede’s principle, the bouyant force ( B ) which is equal to the difference between the weight in air and the weight in water, is equal to the weight of displaced water. So if the volume of the rectangular object is V and the density of water is ρ_water = 1000 kg ⁄ m ³, then

   Vρ_water |g| = W_air - W_water

   Therefore the volume of the rectangular object ( V ) may be calculated from

   V = ( W_air - W_water ) ⁄ ( ρ_water |g| )

   Using the fact that the density of water ( ρ_water ) is equal to 1000 kg ⁄ m ³ , the weight in air ( W_air ) obtained in procedure (6.) and the weight in water ( W_water ) obtained in procedure (8.), calculate the volume of the rectangular object ( V ).
   volume ( V ) = ( W_air - W_water ) ⁄ ( ρ_water |g| ) =
   Measure the height ( h ), the width ( w ), and the length ( l ) of the rectangular object by means of a vernier caliper.
   height ( h ) = m = 0.016 m
   
   width ( w ) = m = 0.01 m
   
   length ( l ) = m = 0.1
   Use the formula for the volume of a rectangular object to calculate its volume.
   volume ( V ) = hwl =
   Compare the volume obtained from archimede’s principle (procedure 9) with the volume obtained from the dimensions (procedure 10) by calculating the percentage error.

   % error = | { volume (8.) - volume (9.) } ⁄ { volume (9.) }| * 100% =

2. To determine the volume of the cylindrical object by using Archimede’s principle and by measuring its dimensions and then compare

   1. Repeat procedures (I.1.) through (I.9.) for the cylinderical object to find the following values

      air hanger 2 distance ( r_⊥2^air ) = x₂^air - x_cg = m = 0.21 m
      
      weight in air ( W_air ) = ( m₂ |g|r_⊥2^air ⁄ r_⊥1 ) - m₁ |g| = N
      
      water hanger 2 distance ( r_⊥2^water ) = x₂^water - x_cg ) = m = 0.19 cm
      
      weight in water ( W_water ) = ( m₂ |g|r_⊥2^water ⁄ r_⊥1 ) - m₁ |g| = N
      
      volume ( V ) = ( W_air - W_water ) ⁄ ( ρ_water |g| ) = m ³

   Measure the diameter ( d ) and the height ( h ) of the cylindrical object by means of a vernier caliper:
   height ( h ) = = 0.023 m
   
   diameter ( d ) = = 0.016 m
   Calculate the volume by using the formula for a cylindrical object.
   volume ( V ) = π ( d ⁄ 2 ) ²h =
   Compare the volumes obtained from Archimede’s principle (procedure 1) and the volume obtained from dimensions procedure 2) by calculating the % error:

   % error = |{ volume (1.) - volume (2.) } ⁄ { volume (2.) } * 100% = %