1. 7.5 Work and Pumping Liquids and Fluid Pressure

2. Review: Hooke’s Law: A spring has a natural length of 1 m. A force of 24 N stretches the spring to 1.8 m. a Find k : b How much work would be needed to stretch the spring 3m beyond its natural length?

3. Over a very short distance, even a non-constant force doesn’t change much, so work becomes: If we add up all these small bits of work we get:

4. A leaky 5 lb bucket is raised 20 feet The rope weighs 0.08 lb/ft. The bucket starts with 2 gal (16 lb) of water and is empty when it just reaches the top. Work: Bucket: Water:The force is proportional to remaining rope. Check:

5. A leaky 5 lb bucket is raised 20 feet The rope weights 0.08 lb/ft. The bucket starts with 2 gal (16 lb) of water and is empty when it just reaches the top. Work: Bucket: Water:

6. A leaky 5 lb bucket is raised 20 feet The rope weights 0.08 lb/ft. The bucket starts with 2 gal (16 lb) of water and is empty when it just reaches the top. Work: Bucket: Water: Rope: Check: Total:

7. 5 ft 10 ft 4 ft 5 ft 10 ft 10 0 4 ft dx I want to pump the water out of this tank. How much work is done? The force is the weight of the water. The water at the bottom of the tank must be moved further than the water at the top. Consider the work to move one “slab” of water:

8. 5 ft 10 ft 4 ft 5 ft 10 ft 10 0 4 ft dx I want to pump the water out of this tank. How much work is done? forcedistance

9. 5 ft 10 ft 4 ft 5 ft 10 ft 10 0 4 ft dx I want to pump the water out of this tank. How much work is done? A 1 horsepower pump, rated at 550 ft-lb/sec, could empty the tank in just under 14 minutes! forcedistance

10. 10 ft 2 ft 10 ft A conical tank is filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3. How much work is required to pump the oil to the rim? Consider one slice (slab) first:

11. 10 ft 2 ft 10 ft A conical tank if filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3. How much work is required to pump the oil to the rim?

12. 10 ft 2 ft 10 ft A conical tank if filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3. How much work is required to pump the oil to the rim?

13. What is the force on the bottom of the aquarium? 3 ft 2 ft 1 ft

14. If we had a 1 ft x 3 ft plate on the bottom of a 2 ft deep wading pool, the force on the plate is equal to the weight of the water above the plate. densitydepth pressure area All the other water in the pool doesn’t affect the answer!

15. What is the force on the front face of the aquarium? 3 ft 2 ft 1 ft Depth (and pressure) are not constant. If we consider a very thin horizontal strip, the depth doesn’t change much, and neither does the pressure. 3 ft 2 ft 2 0 density depth area It is just a coincidence that this matches the first answer!

16. 6 ft 3 ft 2 ft A flat plate is submerged vertically as shown. (It is a window in the shark pool at the city aquarium.) Find the force on one side of the plate. Depth of strip: Length of strip: Area of strip: densitydeptharea We could have put the origin at the surface, but the math was easier this way. 