1. Fluids - Hydrostatics Physics 6B Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. Two Basic Concepts: Density and Pressure Density measures how “tightly packed” an object is. The definition is given by a formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3. Two Basic Concepts: Density and Pressure Density measures how “tightly packed” an object is. The definition is given by a formula: Important note: you can rearrange this formula to get m=ρV. You will need to do this a lot. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4. Two Basic Concepts: Density and Pressure Density measures how “tightly packed” an object is. The definition is given by a formula: Standard units for density are (you might be used to seeing in chemistry class) Important note: you can rearrange this formula to get m=ρV. You will need to do this a lot. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

5. Two Basic Concepts: Density and Pressure Density measures how “tightly packed” an object is. The definition is given by a formula: Standard units for density are (you might be used to seeing in chemistry class) One value you need to know is the density of water: Important note: you can rearrange this formula to get m=ρV. You will need to do this a lot. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6. Two Basic Concepts: Density and Pressure Density measures how “tightly packed” an object is. The definition is given by a formula: Standard units for density are (you might be used to seeing in chemistry class) One value you need to know is the density of water: Pressure is used for fluids mainly because fluids don’t retain their shape, so it is not as useful to measure the force on a fluid. Definition of pressure: Important note: you can rearrange this formula to get m=ρV. You will need to do this a lot. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

7. Two Basic Concepts: Density and Pressure Density measures how “tightly packed” an object is. The definition is given by a formula: Standard units for density are (you might be used to seeing in chemistry class) One value you need to know is the density of water: Pressure is used for fluids mainly because fluids don’t retain their shape, so it is not as useful to measure the force on a fluid. Definition of pressure: This takes into account the area as well as the force. An example will help clarify this definition: Important note: you can rearrange this formula to get m=ρV. You will need to do this a lot. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

8. A thumbtack has a point with a diameter of 1mm. The other end has diameter 1cm. A force of 10 Newtons is required to push it into the wall. Find the pressure on the wall, and the pressure on the person’s thumb. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

9. A thumbtack has a point with a diameter of 1mm. The other end has diameter 1cm. A force of 10 Newtons is required to push it into the wall. Find the pressure on the wall, and the pressure on the person’s thumb. Here are the calculations: Notice that the pressure on the wall is 100x larger because the area is 100x smaller. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

10. Pressure varies with depth in a fluid. The deeper you are under the surface, the larger the pressure will be. Here is a formula we can use: h is the depth under the surface of the fluid, and the gauge pressure is measured relative to the pressure at the surface (usually atmospheric pressure). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

11. Try this example: A cubical box 20.0 cm on a side is completely immersed in a fluid. At the top of the box the pressure is 105 kPa: at the bottom the pressure is 106.8 kPa. What is the density of the fluid? Pressure varies with depth in a fluid. The deeper you are under the surface, the larger the pressure will be. Here is a formula we can use: h is the depth under the surface of the fluid, and the gauge pressure is measured relative to the pressure at the surface (usually atmospheric pressure). Fluid Air 20cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

12. Try this example: A cubical box 20.0 cm on a side is completely immersed in a fluid. At the top of the box the pressure is 105 kPa: at the bottom the pressure is 106.8 kPa. What is the density of the fluid? Pressure varies with depth in a fluid. The deeper you are under the surface, the larger the pressure will be. Here is a formula we can use: h is the depth under the surface of the fluid, and the gauge pressure is measured relative to the pressure at the surface (usually atmospheric pressure). Fluid Air 20cm h top h bottom Calculate the pressures using our formula, then subtract and divide to get our answer: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

13. Try this example: A cubical box 20.0 cm on a side is completely immersed in a fluid. At the top of the box the pressure is 105 kPa: at the bottom the pressure is 106.8 kPa. What is the density of the fluid? Pressure varies with depth in a fluid. The deeper you are under the surface, the larger the pressure will be. Here is a formula we can use: h is the depth under the surface of the fluid, and the gauge pressure is measured relative to the pressure at the surface (usually atmospheric pressure). Fluid Air 20cm h top h bottom Calculate the pressures using our formula, then subtract and divide to get our answer: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

14. Here is the submerged box again. The pressure is exerted on the box in all directions by perpendicular forces, as shown. The idea that pressure in a fluid is applied in all directions is called Pascal’s Law Fluid Air F┴F┴ F┴F┴ F┴F┴ F┴F┴ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

15. Here is the submerged box again. The pressure is exerted on the box in all directions by perpendicular forces, as shown. The idea that pressure in a fluid is applied in all directions is called Pascal’s Law Pascal’s Law helps explain the idea of Buoyancy (why objects float or sink) Here’s the basic idea: since the pressure is larger at the bottom of the box, the upward force there is larger than the downward force on the top, creating a net force upward on the box. We call this the Buoyant Force. Note that the horizontal forces on the sides cancel out. Fluid Air F┴F┴ F┴F┴ F┴F┴ F┴F┴ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

16. Here is a formula for Buoyant Force: Very Important Note: the density in this formula is the density of the FLUID, not the submerged object. Fluid Air F┴F┴ F┴F┴ F┴F┴ F┴F┴ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

17. Here is a formula for Buoyant Force: Very Important Note: the density in this formula is the density of the FLUID, not the submerged object. Using the definition of density ( ), we can also see that the buoyant force is the WEIGHT of the displaced FLUID. Fluid Air F┴F┴ F┴F┴ F┴F┴ F┴F┴ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

18. Here is a formula for Buoyant Force: Very Important Note: the density in this formula is the density of the FLUID, not the submerged object. Using the definition of density ( ), we can also see that the buoyant force is the WEIGHT of the displaced FLUID. In the case of an object floating at the surface of a fluid, the buoyant force is also equal to the weight of the object. Fluid Air F┴F┴ F┴F┴ F┴F┴ F┴F┴ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

19. Here is a formula for Buoyant Force: Very Important Note: the density in this formula is the density of the FLUID, not the submerged object. Using the definition of density ( ), we can also see that the buoyant force is the WEIGHT of the displaced FLUID. In the case of an object floating at the surface of a fluid, the buoyant force is also equal to the weight of the object. One more rule of thumb: if an object is more dense than the fluid, it sinks; if the object is less dense, it floats. Fluid Air F┴F┴ F┴F┴ F┴F┴ F┴F┴ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

20. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

21. Water Air Here is a picture of the submerged brick. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

22. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. mg FTFT FBFB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

23. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

24. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

25. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB We know the weight, and we can find the buoyant force a couple of different ways: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

26. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB We know the weight, and we can find the buoyant force a couple of different ways: Option 1: Use the standard formula Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

27. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB We know the weight, and we can find the buoyant force a couple of different ways: Option 1: Use the standard formula The volume displaced is just the volume of the brick (it is fully submerged) and we can find that from the definition of density: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

28. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB We know the weight, and we can find the buoyant force a couple of different ways: Option 1: Use the standard formula The volume displaced is just the volume of the brick (it is fully submerged) and we can find that from the definition of density: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

29. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB We know the weight, and we can find the buoyant force a couple of different ways: Option 2: The buoyant force is the weight of the displaced fluid. Since the brick is fully submerged, and we know the densities, the weight of the fluid is just the weight of the brick times the ratio of the densities: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

30. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB We know the weight, and we can find the buoyant force a couple of different ways: Option 2: The buoyant force is the weight of the displaced fluid. Since the brick is fully submerged, and we know the densities, the weight of the fluid is just the weight of the brick times the ratio of the densities: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

31. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB Now that we have the buoyant force, we can calculate the tension Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

32. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air Here is a picture of the submerged brick. Draw a free-body diagram of the forces on the brick. Now we can write down Newton’s 2 nd law: mg FTFT FBFB Now that we have the buoyant force, we can calculate the tension: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

33. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air To answer the 2 nd question, just notice that when the string is cut, the tension force goes away. This gives us a new (simpler) free-body diagram. mg FBFB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

34. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air To answer the 2 nd question, just notice that when the string is cut, the tension force goes away. This gives us a new (simpler) free-body diagram. Write down Netwon’s 2 nd law again: mg FBFB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

35. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air To answer the 2 nd question, just notice that when the string is cut, the tension force goes away. This gives us a new (simpler) free-body diagram. Write down Netwon’s 2 nd law again: mg FBFB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

36. Example: A brick with mass 5kg and density 2800 kg/m 3 is tied to a string and lowered into a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool? Water Air To answer the 2 nd question, just notice that when the string is cut, the tension force goes away. This gives us a new (simpler) free-body diagram. Write down Netwon’s 2 nd law again: mg FBFB Note: the acceleration is negative because the brick is sinking Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

37. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

38. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? Same basic setup as the previous problem, but the block is floating. We can draw the picture and the free-body diagram again: Water Air Bottom of pool FBFB mg FTFT Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

39. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? Same basic setup as the previous problem, but the block is floating. We can draw the picture and the free-body diagram again: Newton’s 2 nd law: Water Air Bottom of pool FBFB mg FTFT Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

40. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? Same basic setup as the previous problem, but the block is floating. We can draw the picture and the free-body diagram again: Newton’s 2 nd law: Water Air Bottom of pool FBFB mg FTFT This time the buoyant force is greater than the weight of the block because the density of water is larger than the block’s density. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

41. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? Same basic setup as the previous problem, but the block is floating. We can draw the picture and the free-body diagram again: Newton’s 2 nd law: Water Air Bottom of pool FBFB mg FTFT This time the buoyant force is greater than the weight of the block because the density of water is larger than the block’s density. Using option 2 from the previous problem: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

42. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? Same basic setup as the previous problem, but the block is floating. We can draw the picture and the free-body diagram again: Newton’s 2 nd law: Water Air Bottom of pool FBFB mg FTFT Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

43. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? Water Air Bottom of pool FBFB mg To find the acceleration when the string is cut, again notice that the tension force goes away, then use Newton’s 2 nd law: a Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

44. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? FBFB mg To find the acceleration when the string is cut, again notice that the tension force goes away, then use Newton’s 2 nd law: Water Air Bottom of pool FBFB mg a Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

45. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? FBFB mg To find the acceleration when the string is cut, again notice that the tension force goes away, then use Newton’s 2 nd law: Water Air Bottom of pool FBFB mg a Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

46. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? FBFB mg To find the acceleration when the string is cut, again notice that the tension force goes away, then use Newton’s 2 nd law: Water Air Bottom of pool FBFB mg a Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

47. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

48. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

49. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg The buoyant force only has to be strong enough to balance out the weight of the block, so the block floats above the surface. If you split the block into 2 parts – the part above the surface and the part below, it is only the part below that is displacing water. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

50. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg The buoyant force only has to be strong enough to balance out the weight of the block, so the block floats above the surface. If you split the block into 2 parts – the part above the surface and the part below, it is only the part below that is displacing water. We can find the buoyant force from our standard formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

51. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg The buoyant force only has to be strong enough to balance out the weight of the block, so the block floats above the surface. If you split the block into 2 parts – the part above the surface and the part below, it is only the part below that is displacing water. We can find the buoyant force from our standard formula: Newton’s law becomes: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

52. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg The buoyant force only has to be strong enough to balance out the weight of the block, so the block floats above the surface. If you split the block into 2 parts – the part above the surface and the part below, it is only the part below that is displacing water. We can find the buoyant force from our standard formula: Newton’s law becomes: The mass of the block can be written in terms of the density of the wood, so that we can get a more general result Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

53. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg The buoyant force only has to be strong enough to balance out the weight of the block, so the block floats above the surface. If you split the block into 2 parts – the part above the surface and the part below, it is only the part below that is displacing water. We can find the buoyant force from our standard formula: Newton’s law becomes: This gives us a formula for the portion of the block that is under the surface. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

54. Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m 3 is tied to a string and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the water? For the last part, consider the forces on the block when it is floating at the surface. Water Air FBFB mg The buoyant force only has to be strong enough to balance out the weight of the block, so the block floats above the surface. If you split the block into 2 parts – the part above the surface and the part below, it is only the part below that is displacing water. We can find the buoyant force from our standard formula: Newton’s law becomes: The rest of the block is above the water, so the answer is 30% Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB