Unformatted text preview: Fluids Buoyancy RHJansen Buoyancy is caused by the difference in pressure at the top of and bottom of a submerged object. Buoyancy F
A P= A P= A F RHJansen The pressure is greater at the bottom, but the area of the top and bottom are the same. The Force on the bottom of the object is greater than the force on the top. Buoyancy is the sum of this difference in force due to pressure. Buoyancy is always directed upward, and it is equal to Fbuoy = Buoyant Force
If an object is submerged in a liquid, add a force vector to the free body diagram. Isn't a gas a fluid? Shouldn't we add a force vector for buoyancy in air? The density of air is very small and generates negligible buoyancy on most objects, so buoyancy in air is ignored. There are exceptions. Balloons and blimps float in air. If an object floats in air include buoyancy for air. Fbuoy m Fg RHJansen Fluid Displaced
The fluid buoys the object up. This means you will need the density and volume of the fluid displaced. Fbuoy = The fluid displaced is the amount of fluid that had to get out of the way to make room for the object. Another way to say this: the volume of the fluid displaced is equal to the volume of the part of the object that is submerged in the fluid. RHJansen Example 1 a. Draw the FBD (step I) A mass m is floating on the surface of a liquid. Fbuoy Fg
What is it doing (step II)? You must always ask yourself this question. Standing Still. This means you can use the balance force approach. b. Write the sum of force equation. (step III) Fbuoy = c. Substitute known equations. (step IV) = RHJansen Example 2 a. Draw the FBD (step I) A mass m is submerged in a liquid and is neutrally buoyant. Fbuoy Fg
What is it doing (step II)? You must always ask yourself this question. Standing Still. This means you can use the balance force approach. b. Write the sum of force equation. (step III) Fbuoy = c. Substitute known equations. (step IV) = RHJansen Example 3
FResist a. Draw the FBD (step I) Fbuoy Fg A mass m is submerged in a liquid. It is released from rest and initially accelerates downward, against the resistance of water. Unlike air resistance which is often ignored, water resistance cannot be, as it is too strong. What is it doing (step II)? You must always ask yourself this question. Accelerating down. Use a sum of force equation, with down as positive. b. Write the sum of force equation. (step III) F =   c. Substitute known equations. (step IV) ma =   RHJansen FR has to be given or they will ask to solve it. Without calculus some erroneous oversimplifications have to be made, so this is not a likely scenario. If acceleration is in the formula you may end up doing kinematics as well. Example 4 a. Draw the FBD (step I) Fbuoy FResist Fg A mass m is submerged in a liquid. It is released from rest and initially accelerates upward, against the resistance of water. Unlike air resistance which is often ignored, water resistance cannot be, as it is too strong. What is it doing (step II)? You must always ask yourself this question. Accelerating up. Use a sum of force equation, with up as positive. b. Write the sum of force equation. (step III) F =   c. Substitute known equations. (step IV) ma =   RHJansen FR has to be given or they will ask to solve it. Without calculus some erroneous oversimplifications have to be made, so this is not a likely scenario. If acceleration is in the formula you may end up doing kinematics as well. Example 5
N a. Draw the FBD (step I) A mass m is submerged in a liquid and sitting on the bottom. Fbuoy Fg
A surface is involved in this problem. What is it doing (step II)? You must always ask yourself this question. Standing Still. This means you can use the balance force approach. b. Write the sum of force equation. (step III) N+ = c. Substitute known equations. (step IV) N + = RHJansen Example 6
T a. Draw the FBD (step I) A mass m is submerged in a liquid and suspended by a string. Fbuoy Fg
A string is involved in this problem. What is it doing (step II)? You must always ask yourself this question. Standing Still. This means you can use the balance force approach. b. Write the sum of force equation. (step III) T+ = c. Substitute known equations. (step IV) T + = RHJansen Example 7
Fs a. Draw the FBD (step I) A mass m is submerged in a liquid and suspended by a spring. Fbuoy Fg
A spring is involved in this problem. What is it doing (step II)? You must always ask yourself this question. Standing Still. This means you can use the balance force approach. b. Write the sum of force equation. (step III) Fs + = c. Substitute known equations. (step IV) kx + = RHJansen Example 8 a. Draw the FBD (Step I) An 80 kg mass floats with a portion of the mass sticking above the surface of the water. Fbuoy Fg
b. Determine the volume of fluid displaced. Step II: Step III: Stationary, use balanced force approach. Fbuoy = Step IV: = (1000 ) = (80 ) RHJansen Step V: V = 0.80 3 Example 9
N a. Draw the FBD (Step I) A 300 kg cube with 60 cm sides sits at the bottom of a swimming pool. Piece of advice: convert to meters before solving for volume. By doing that you use linear conversions and don't have to worry about cubing the conversion factor. Fbuoy Fg b. Determine the normal force. Step II: Step III: Step IV: Step V: Stationary, use balanced force approach. N+ = N + = N + (1000 ) ( 9.8) ( 0.6 ) = ( 300 ) ( 9.8) N = 823 RHJansen 3 Example 10 a. Draw the FBD (Step I) Fbuoy FResist Fg A 2.5 kg cube with 20 cm sides is released from rest at the bottom of a deep pool. We'll incorrectly assume that it accelerates uniformly (otherwise we'd need integral calculus to solve this one). It takes 4.0 s for the mass to reach a velocity of 0.5 m/s. b. Determine the acceleration of the cube. v = + 0 ( 0.5) = ( 0 ) + ( 4.0 ) a = 0.125 2 RHJansen Example 10 A 2.5 kg cube with 20 cm sides is released from rest at the bottom of a deep pool. We'll incorrectly assume that it accelerates uniformly Fbuoy (otherwise we'd need integral calculus to solve this one). It takes FResist F 4.0 s for the mass to reach a velocity g of 0.5 m/s. c. Determine the force of water resistance acting on the mass. Step II: Step III: Step IV: Step V: Accelerating up. Sum forces with up being positive. F =   ma =   ( 2.5) ( 0.125) = (1000 ) ( 9.8) ( 0.2 )  ( 2.5) ( 9.8)  3 FR = 53.6 RHJansen Example 11 An object weighs 130 N. When it is lowered completely into water (submerged) it appears to weigh only 80 N. Determine the buoyant force on the object. If it appears to weigh 50 N less in water than in air, then the water must be supporting 50 N of the objects weight. The buoyant force is 50 N. RHJansen ...
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This note was uploaded on 07/21/2009 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at Berkeley.
 Spring '08
 Packard
 Physics, Buoyancy

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