Physics Lab - Archimedes' Principle

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1 Archimedes’ Principle Lab Online Purpose The Purpose of this activity is to show some basic properties of the buoyant force. Namely that the buoyant force is a function of the density of the fluid, that the buoyant force is equal to the weight of the fluid displace by a submerged or floating object, and that the apparent weight of a submerged object is the difference between its weight and the buoyant force acting on it. Theory Archimedes' Principle states that the upward buoyant force exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces; it is also applicable to gases: 𝐹 ஻ = 𝑚 ௙ 𝑔 There are 2 ways to measure buoyancy: direct and displacement. Direct measurement is the difference between the actual weight of the object (W o ) and its apparent weight (W a ) when fully submerged. Displacement measurement utilizes the fact that the volume of fluid displaced (V f ) is equal to the volume of the object (V o ) that is submerged. Recall that density (ρ) = m/V, such that: 𝐹 ஻ = ൫𝜌 ௙ 𝑉 ௙ ൯𝑔 = ൫𝜌 ௙ 𝑉 ௢ ൯𝑔 From the free body diagram: 𝐹 ஻ − 𝐹 ௚ = 𝑔൫𝜌 ௙ 𝑉 ௢ − 𝑚 ௢ ൯ = −𝑚 ௢ 𝑎 where, solving for acceleration we find 𝑎 = 𝑔 ൬ 𝜌 ௙ 𝜌 ௢ − 1൰ where it can be observed that 𝜌 ௙ > 𝜌 ௢ ⇒ +𝑎 𝑂𝑏𝑗𝑒𝑐𝑡 𝑤𝑖𝑙𝑙 𝑓𝑙𝑜𝑎𝑡 𝜌 ௙ = 𝜌 ௢ ⇒ 𝑎 = 0 𝑂𝑏𝑗𝑒𝑐𝑡 𝑖𝑠 𝑖𝑛 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝜌 ௙ < 𝜌 ௢ ⇒ −𝑎 𝑂𝑏𝑗𝑒𝑐𝑡 𝑤𝑖𝑙𝑙 𝑠𝑖𝑛𝑘 If the object is only partially submerged, then the volume of fluid displaced is the volume of the part of the object actually submerged. For example, a cylinder has a volume V c = πr 2 l = A c l, where A c = πr 2 is the cross sectional area and l is the length. Assume the cylinder is submerged by a depth d, then V sub = A c d such that: 𝐹 ஻ = 𝜌 ௙ 𝑔𝑉 ௦௨௕ = 𝜌 ௙ 𝑔𝐴 ௖ 𝑑 It is important to note that the buoyant force and depth of the object are directly related such that buoyant force will increase with depth until the object is fully submerged.

2 Setup 1. Go to the following website: https://phet.colorado.edu/sims/html/density/latest/density_en.html 2. You should now see the following. 3. Click the Intro option on the left. 4. You should see the following:

3 Procedure 1. In the grey box near the top left of your screen make sure that Styrofoam is selected. a. Record the Styrofoam block’s mass in kg in the Table. b. Record the Styrofoam block’s volume in L in the Table. 2. The fluid in the tank is water. The fluid Density to Water is 1.00 kg/L. a. Record the Fluid Density in the Table. b. Remove the Styrofoam block from the ‘tank’ and place in on the ground. c. Now record the initial volume, V i , of the fluid in the Table. 3. Click and drag the Styrofoam block and place it in the fluid. a. When it comes to a rest, record the new fluid volume, V, in the Table. 4. Now repeat steps 1 through 3 for Wood, Ice, Brick, and Aluminum.

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4 Analysis of Archimedes’ Principle Lab Online Name Melissa Fernandez Ayala Group# NA Course/Section PHY 1951 011 Instructor Christopher Dunn, TA: Amilcar Torres Quijano Table Styrofoam Wood Ice Brick Aluminum Mass (kg) 0.75 kg 2.00 kg 4.60 kg 10.00 kg 13.50 kg Weight (N) 7.36 N 19.62 N 45.13 N 98.01 N 132.3 N Volume (L) 5.00 L 5.00 L 5.00 L 5.00 L 5.00 L Volume (m 3 ) 0.005 m 3 0.005 m 3 0.005 m 3 0.005 m 3 0.005 m 3 Density (kg/m 3 ) 150 kg/m 3 400 kg/m 3 920 kg/m 3 2000 kg/m 3 2700 kg/m 3 Apparent Weight (N) 48.96 N 83.25 N Water Density (kg/L) 1 1 1 1 1 Density (kg/m 3 ) 1000 1000 1000 1000 1000 𝑉 ௜ (𝐿) 100.00 100.00 100.00 100.00 100.00 𝑉 ௜ (m 3 ) 0.100 0.100 0.100 0.100 0.100 𝑉 (m 3 ) 0.10075 0.102 0.10459 0.105 0.105 ∆𝑉 (m 3 ) 0.00075 0.002 0.00459 0.005 0.005 Weight of displaced water (N) 7.36 N 19.62 N 45.03 N 49.05 N 49.05 N

5 Styrofoam Calculations Complete Table, show any calculations in the space provided (10 points)

6 Wood Calculations Complete Table, show any calculations in the space provided (10 points)

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7 Ice Calculations Complete Table, show any calculations in the space provided (10 points)

8 Brick Calculations Complete Table, show any calculations in the space provided (10 points)

9 Aluminum Calculations Complete Table, show any calculations in the space provided (10 points)

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10 1. Draw a free body diagram of the wood block floating in the water , then write the force summation equation from that free body diagram . (10 points) 2. What is the apparent weight of the wood block when it is floating? (5 points) 3. Calculate the buoyant force acting on the wooden block when it is floating. (5 points)

11 4. Draw a free body diagram of the aluminum block for when it was submerged in the water, and then write the force summation equation for that free body diagram. (10 points) 5. Calculate the apparent weight of the aluminum block when it is in the water. (5 points)

12 6. Calculate the buoyant force acting on the aluminum block when it is in the water. (5 points) 7. Why did the wood block float in the fluid while the aluminum block sank ? (5 points) The wood block floated in the fluid while the aluminum block sank because the apparent weight of the wood block is zero, meaning the object is floating because the buoyant force exactly counteracts the objects weight (they’re the same value, which is proven in the free body diagram). The apparent weight is equal to the difference between the buoyant force and the objects weight, if it is zero then the values are equal and the object floats – this was the case for the wood as shown in the calculations. As per the aluminum, the apparent weight is a positive value and displays how the objects weight is greater than the buoyant force. This is why the object sank (the heavier object counteracts the buoyant force). 8. Briefly explain what apparent weight is . (5 points) Apparent weight refers to the force exerted on an object in a gravitational field (like water, a free fall, or even an elevator) that gives a perception of its weight. Since this is a perception of its weight, it can be different from the actual weight of the object, in this case, due to buoyant force, when an object is submerged in water its apparent weight will be reduced. This concept is based on Archimedes’ Principle, which states that an object partially or completely submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. This buoyant force acts in the opposite direction of gravity. The apparent weight of the object is the difference between the actual weight and the buoyant force.

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