PHYS 201 Lab Report 5

PHYS 201L Section 004 Archimedes’ Principle Oluwaseyi Adelakun & Anna Sandwell I. Table of Data. Object m air (g) m water (deg) ρ 1 (g/cc) ρ 2 (g/cc) Percent difference Composition Guess Object 1 (silver cylinder) 71 45.5 2.84 2.78 2.14% Aluminum Object 2 (gold cylinder) 219.3 193 8.76 8.34 4.91% Brass Object 3 (rusted cube) 249 216.5 8.36 7.66 8.74% Iron II. Calculations. ρ 1 = mass / volume ρ 1 [Object 1] = (71 g) / π(1.25cm) 2 (5.1cm) = 2.84 g/cc ρ 1 [Object 2] = (219.3 g) / π(1.25cm) 2 (5.1cm) = 8.76 g/cc ρ 1 [Object 3] = (249 g) / (3.1cm) 3 = 8.36 g/cc ρ 2 = m air / m air - m water ρ 2 [Object 1] = (71 g) / (71 g - 45.5 g) = 2.78 g/cc ρ 2 [Object 2] = (219.3 g) / (219.3 g - 193 g) = 8.34 g/cc

ρ 2 [Object 3] = (249 g) / (249 g - 216.5 g) = 7.66 g/cc Percent Difference = |( ρ 1 - ρ 2 ) / ½ (ρ 1 - ρ 2 )| x 100% % difference [Object 1] = (|2.84 g/cc - 2.78 g/cc) / ½ (2.84 - 2.78)| x 100% = 2.14% % difference [Object 2] = |(8.76 g/cc - 8.34 g/cc) / ½ (8.76 - 8.34)| x 100% = 4.91% % difference [Object 3] = |(8.36 g/cc - 7.66 g/cc) / ½ (8.36 - 7.66)| x 100% = 8.74% III. Graphs & Diagrams. Make two free-body diagrams showing all the forces acting on object (a) when it is hanging from the string not in the water and (b) when it is hanging from the string and submerged in the water. IV. Questions & Calculations. 1. What is Archimedes’ Principle? How is it related to the concept of hydrostatic pressure? Archimedes’ principle plainly states that “the upward buoyant force that is exerted on a body immersed in a fluid, whether partially or fully submerged, is equal to the weight of the fluid that

The presence of the pen will affect the reading of the balance because the pen will displace the water slightly causing the weight to increase. The weight of the pen will push the water level higher and the weights will be combined. V. Results & Conclusions. Results: The results showed that the percent difference for the first and second objects were 2.14% and 4.91%, respectively. The expected densities and measured densities were very close, however, the percent difference for the densities for the third object was 8.74%. The first object was identified as aluminum, the second as brass, and the third as iron. Conclusions: In conclusion, the identity of an unknown object can be determined by utilizing the relationship between mass and volume to find density. By using the known density of water, the displacement of weight of the object after being placed in water can lead to the density of the objects. Other factors in this experiment were a tension force from the string, the gravitational force on the object, and the force of buoyancy in the water. Potential errors could have arised from a faulty measurement system if the scale was inaccurate.