1. Work and Fluid Pressure
Lesson 7.7

2. Work Definition The product of
The force exerted on an object
The distance the object is moved by the force
When a force of 50 lbs is exerted to move an object 12 ft.
600 ft. lbs. of work is done 50
12 ft

3. Hooke's Law Consider the work done to stretch a spring
Force required is proportional to distance
When k is constant of proportionality
Force to move dist x = k • x = F(x)
Force required to move through i th interval, x
W = F(xi) x a b x

4. Hooke's Law We sum those values using the definite integral
The work done by a continuous force F(x)
Directed along the x-axis
From x = a to x = b

5. Hooke's Law A spring is stretched 15 cm by a force of 4.5 N
How much work is needed to stretch the spring 50 cm?
What is F(x) the force function?
Work done?

6. Winding Cable Consider a cable being wound up by a winch
Cable is 50 ft long
2 lb/ft
How much work to wind in 20 ft?
Think about winding in y amt
y units from the top  50 – y ft hanging
dist = y
force required (weight) =2(50 – y)

7. Pumping Liquids
Consider the work needed to pump a liquid into or out of a tank
Basic concept: Work = weight x dist moved
For each V of liquid
Determine weight
Determine dist moved
Take summation (integral)

8. Pumping Liquids – Guidelinesa b r
Draw a picture with the coordinate system
Determine mass of thin horizontal slab of liquid
Find expression for work needed to lift this slab to its destination
Integrate expression from bottom of liquid to the top

9. Pumping Liquids Suppose tank has
r = 4
height = 8
filled with petroleum (54.8 lb/ft3)
What is work done to pump oil over top
Disk weight?
Distance moved?
Integral? 4 8
(8 – y)

10. Fluid Pressure
Consider the pressure of fluid against the side surface of the container
Pressure at a point
Density x g x depth
Pressure for a horizontal slice
Density x g x depth x Area
Total force

11. Fluid Pressure The tank has cross section of a trapazoid
Filled to 2.5 ft with water
Water is 62.4 lbs/ft3
Function of edge
Length of strip
Depth of strip
Integral
(-4,2.5)
(4,2.5)
2.5 - y
(-2,0)
(2,0)
y = 1.25x – 2.5 x = 0.8y + 2
2 (0.8y + 2)
2.5 - y