# Calculus 2 Lecture Notes_ Section 56

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```					Calc 2 Lecture Notes                    Section 5.6                         Page 1 of 8

Section 5.6: Applications of Integration to Physics and Engineering
Big idea: You can integrate small pieces of work to find the total work, small changes in
momentum to find the total change in momentum, small pieces of mass to find the total mass and
center of mass, and small pieces of force exerted by a fluid to find the total force exerted by a
fluid.

Big skill: You should be able to compute work, impulse, center of mass, and hydrostatic force.

Work
      W = Fd (for a constant force)
 units of work are Newton-meters (N-m, also called Joules) in the metric system or
foot-pounds in the English system
      If the force varies over the distance applied, we can chop up the path into little pieces
over which the force is roughly constant, then add up all the little amounts of work done.
n
W   F  ci  x
i 1
                 n
W  lim  F  ci  x
n 
i 1
b
      W   F  x  dx (for a varying force)
a

      Forces you should know:
 Gravity force near the surface of a planet: Fg  mg (g  9.8 m/s2 on the surface of
the earth)
 Gravitational force between two masses m 1 and m 2 whose center of masses are
mm                          N  m2
separated by a distance r: Fg  G 1 2 2 ( G  6.67300 1011        )
r                           kg 2
 Spring force when the spring is stretched by a distance x from its equilibrium
position: Fs   kx
 Friction force when two surfaces are pushed together with a normal force Fn :
F f   Fn
 Electrostatic force between two charges q1 and q2 whose centers are separated by
1 q1q2                               C2
a distance r: Fe        2 (  0  8.8541878176 1012        )
4 0 r                              N  m2
Practice:
1. Compute the work done by lifting 100.0 kg vertically 2.0 meters. State answer in both
metric and English units.
Calc 2 Lecture Notes                Section 5.6                            Page 2 of 8

2. The force it takes to stretch a spring by x inches is given by the function
 lb 
F  x   15  x . Compute the work it takes to stretch the spring by 12 inches.
 in 

3. Compute the work required to lift a 15 ton space module 800 miles above the surface of
the earth, given that the radius of the earth is about 4,000 miles.
Calc 2 Lecture Notes                   Section 5.6                               Page 3 of 8

4. Suppose the force you exert horizontally on a box to push it across the floor as a function
                     x 
of distance x in meters is F  x    7500 N  532 N sin          . Compute the work it
                     1.0 m  
takes to push the box 15 meters.

x2
5. If a tank is made revolving the graph of the equation y        (x and y measured in
1.0 m
meters) about the y axis for 0 m  x  1 m, how much work does it take to fill the tank to
the top with water?
h2

Note: These type of problems can be solved using the formula W    gA  y  ydy , where 
h1

is the density of the fluid, g is acceleration due to gravity, A(y) is the area of thin slices of the
fluid as a function of height y, and h1 and h2 are the lower and upper heights of the tank.
Calc 2 Lecture Notes                             Section 5.6                  Page 4 of 8

Impulse
   …Is change in momentum
F  ma
d
 F t   m        v  t  
dt         
    mv  t    F  t  dt
t2

 mv   F  t  dt
t1
t2

   J   F  t  dt  mv  m  v  t2   v  t1  
t1

Practice:
     t    
7. If a baseball bat exerts a contact force of F  t   6500lb sin             for
 0.0008sec 
0  t  0.0008 sec, what is the speed of the ball after impact if it was initially moving at
130 ft/sec and has a mass of 0.01 slug?
Calc 2 Lecture Notes                  Section 5.6                            Page 5 of 8

8. A crash test is performed on a 200 slug vehicle, and the force of the wall on the front
bumper is shown in the table below. Graph the force vs. time, and then estimate the
impulse and the initial speed of the vehicle.

t (s)               0          0.1         0.2          0.3         0.4          0.5            0.6
F (lb)              0        8,000      16,000       24,000      15,000        9,000              0
Calc 2 Lecture Notes                   Section 5.6                                Page 6 of 8

Center of Mass:
The center of mass is an extremely important concept in physics because many problems
involving forces on solids (which are comprised of gadzillions of atoms) can be reduced to
simply considering the force on one point located at the center of mass.
For example, when doing basic gravitational calculations between the sun and earth, one
does not need to add up the gravitational attractions between every atom in the sun and every
atom in the earth. Instead, one can treat the entire mass of each body as being concentrated at
one point, the center of mass.
Another example of the importance of finding the center of mass is that freely spinning
objects will spin about their center of mass.
A third example of the importance of finding the center of mass is that a planar object
will balance when placed on a pivot at its center of mass.

The center of mass of a discrete group of points is the (literally) weighted average of their
locations:

For the point masses to the left, the center of
mass is:

x
 5 kg 1 m   10 kg  3 m   17 kg  5 m 
5 kg  10 kg  17 kg
120 kg  m
x
32 kg
x  3.75 m

A general formula for the center of mass of        A more general formula for n masses would
three masses would be:                             be:
m x  m2 x2  m3 x3                                                                n
x 1 1
m1  m2  m3                                   m x  m2 x2  ...  mn xn      m x     i i
x 1 1                           i 1

m1  m2  ...  mn              n

m
i 1
i

To find the center of mass of a solid aligned along the x-axis, you can think of chopping it up
into more and more thin slices along the x – direction, and then the center of mass would be:
n
lim  mi xi
n 
x          i 1
n
lim  mi
n 
i 1
Calc 2 Lecture Notes                    Section 5.6                                          Page 7 of 8

The question is: how do we represent the masses mi? Answer: If the linear density of the solid is
(x) (units would be something like kg per meter), then any slice would have mass

mi  density  length
mi    xi   x
n                  n                     b

lim  mi xi
n 
lim    xi  xi  xi
n 
   x  xdx
and the center of mass would be x            i 1
n
          i 1
n
   a
b
.
lim  mi           lim    xi  xi              x  dx
n                n 
i 1                   i 1              a

Practice:

9. Find the mass and center of mass of an aluminum baseball bat that extends from
 3    x  slug
0  x  30 in. and has linear density   x   0.00468                .
 16 60 in.  in.

10. Find the center of mass of a 3 – 4 – 5 right triangle.
Calc 2 Lecture Notes                    Section 5.6                              Page 8 of 8

Hydrostatic Force:
A fluid under a pressure P will apply a force F to an area A according to the formula:
F = PA

At a depth of d below the surface of a fluid, the pressure is P = gd, where  is the density of the
fluid. For water, use  = 1000 kg/m3 and g = 9.8 m/sec2 for metric units, use g = 62.4 lb/ft3 for
English units.

If a surface, like a dam, has to withstand the force of water pushing against it over a given depth
of water, we divide the face of the dam up into thin horizontal rectangles, compute the force of
water on each rectangle, then integrate the forces to get the total force on the dam.

Practice:

1. Find the total hydrostatic force that a semi-circular faced dam must withstand if its radius
is 75 feet.
h2

Note: These type of problems can be solved using the formula W    gW  y  ydy , where 
h1

is the density of the fluid, g is acceleration due to gravity, W(y) is the width of thin slices of
the fluid as a function of depth y, and h1 and h2 are the lower and upper depths of the fluid.

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