Calculus 1 Lecture Notes, Section 5.6 by oxr14149

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									Calc 1 Lecture Notes                   Section 5.6                         Page 1 of 5

Section 5.6: Applications of Integration to Physics and Engineering
Big idea:You can integrate force to find net force…

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 (or Joules) or foot-pounds
            b
      W   F  x  dx (for a varying force)
            a


Practice:
   1. How much work is done by lifting 100.0 kg vertically 2.0 meters? State answer in both
       metric and English units.




    2. The force it takes to stretch a spring by x inches is given by the function
                  lb 
       F  x   15  x . How much work does it take to stretch the spring by 12 inches?
                  in 
Calc 1 Lecture Notes                 Section 5.6                            Page 2 of 5

  3. If the force you exert horizontally on a box to push it across the floor as a function of
                                                            x 
     distance x in meters is F  x    7500 N  532 N sin          . How much work does it
                                                            1.0 m  
     take to push it 15 meters?




  4. If a tank is made revolving the graph of the equation y = x2 (y in meters) about the y axis
     for 0  x  1, how much work does it take to fill the tank to the top with water?
Calc 1 Lecture Notes                           Section 5.6                    Page 3 of 5

Impulse
               …Is change in momentum
                     b
               J   f  t  dt  m  v  b   v  a  
                     a


Practice:

                                                                            t    
   1. If a baseball bat exerts a contact force of F  t   8500lb 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 1 Lecture Notes                   Section 5.6                                      Page 4 of 5

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 three masses would be:
    m x  m2 x2  m3 x3
x 1 1
      m1  m2  m3

If you chop up a solid mass into n thin slices in the x – direction, then the center of mass would
be:
     m x  m2 x2   mn xn
 x 1 1
        m1  m2   mn

If the linear density of the solid is (x), then any slice would have mass
mi  density  length
                       ,
mi    xi   x
                                           b


                                   M         x  xdx
so the center of mass would be x         a
                                            b
                                                          .
                                   m
                                              x  dx
                                           a



Practice:

   1. Find the mass and center of mass of an aluminum baseball that extends from
                                                               3   x  slug
      0  x  30 in. and has linear density   x   0.00468             .
                                                               16 60  inch
Calc 1 Lecture Notes                   Section 5.6                            Page 5 of 5

Hydrostatic Force:
A pressure P applied to an area A results in a force F on that area of:
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, we use either  = 1000 kg/m3 and g = 9.8 m/sec2, or we use g = 62.4 lb/in3.

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.

								
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