# Problem center of gravity by sanmelody

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```									                               Assignment 8 Key Solution

Chapter 5, Problem 7

Locate the centroid of the plane area shown.

Solution 7.

A,in 2                x ,in.        xA,in 3

 16 
2
4 16 
1                   201.06            6.7906   1365.32
4                      3

2        8  8    64           4             256

            137.06                              1109.32

xA 1109.32
Then         X               in.                             or      X  8.09 in.
A   137.06
and          Y  X by symmetry                                  or Y  8.09 in.
Chapter 5, Problem 10.
Show that as r1 approaches r2 , the location of the centroid approaches that
of a circular arc of radius  r  r2  /2.
1

Chapter 5, Solution 10.

First, determine the location of the centroid.

2 sin 2                          
From Fig. 5.8A:                         y2      r2                                            A2      2    r22
3    2
                     
2    cos 
     r2 
3    2  
                   
cos 
Similarly                               y1 
2
r1                                            A1     2    r12
3  
2                 
cos                                                          cos 
Then                                  yA 
2
r2                            2    r22   2 r1                        2    r12 
3    2  
                                     3                 
2
                   


2 3
3

r2  r13 cos          
               
and                                     A      r22      r12
2           2    

     r22  r12
2


                       
Now                                 Y A   yA

                 2 3
 2

Y     r22  r12  

       
3

r2  r13 cos                        
2 r23  r13 cos 
Y 
3 r22  r12   
2
Chapter 5, Problem 18.

The horizontal x axis is drawn through the centroid C of the area shown
and divides the area into two component areas A1 and A2. Determine the
first moment of each component area with respect to the x axis, and
explain the results obtained.

Solution 18.

A, mm 2          y, mm    yA, mm3
1       80  20   1600     90      144 000

2        20 80   1600     40       64 000

          3200                      208 000

yA 208 000
Then          Y                65.000 mm
A   3200
Now, for the first moments about the x-axis:
Area I

QI  yA  25  80  20   7.5  20  15   42 250 mm3,          or QI  42.3  103 mm3
Area II

QII  yA   32.5  20  65  42 250 mm3 ,
or QII  42.3  103 mm3

Q area  QI  QII  0                      y  0 and Qarea   yA
Note that                       which is expected as                         since x is a
centroidal axis.
Chapter 5, Problem 26.

The homogeneous wire ABCD is bent as shown and is supported by a pin
at B. Knowing that l  200 mm, determine the angle  for which
portion BC of the wire is horizontal.

Chapter 5, Solution 26.

The wire supported only by the pin at B is a two-force body. For equilibrium the center of gravity
of the wire must lie directly under B. Also, because the wire is homogeneous the center of gravity
will coincide with the centroid. In other words, x  0, or xL  0.

2 150 mm                     200 mm                         150 mm      
xL                    150 mm   
                         200 mm    200 mm         cos  150 mm   0
                             2                              2        

or
5000
cos 
11250
or        63.6
Chapter 5, Problem 37.

Determine by direct integration the centroid of the area shown. Express

Chapter 5, Solution 37.

At                  x  0, y  b

b  k  0  a
2                     b
or       k 
a2
b
2
x  a
2
Then                y
a

 x  a2
y   b
Now                 xEL  x, yEL           
2 2a2

2
x  a dx
b          2
and                     dA  ydx 
a
b                 b               a 1
2
x  a  dx  2  x  a    ab
a               2                 3
Then      A   dA   0                                      0 3
a                3a            

 xELdA   0 x  a 2  x  a  dx   a 2  0  x  2ax  a x dx
a      b          2         b  a 3        2   2
and
                  
b  x4 2 3 a2 2        1 2
      4  3 ax  2 x   12 a b
2                
a                 
a
 x  a 2  2  x  a 2 dx   4   1          5
b                b                  b2
 5  x  a 
a
 yELdA   0                    a                 2a
2a 2                                                0
1 2
      ab
10
1      1 2                                            1
Hence xA   xELdA: x  ab    ab                                      x      a 
 3  12                                                4

1      1 2                                          3
yA   yELdA: y  ab    ab                                   y       b 
 3  10                                             10
Chapter 5, Problem 43.

A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.

Solution 43.

First note that because the wire is homogeneous, its center of gravity
coincides with the centroid of the corresponding line

Now            xEL  a cos3                and              dL            dx2  dy 2

Where              x  a cos3  : dx  3a cos 2  sin  d

y  a sin 3  : dy  3a sin 2  cos d
1/2

                                                             
2                                      2
Then        dL   3a cos2  sin  d                 3a sin 2  cos d
                                                                    

                         
1/2
 3a cos sin  cos2   sin 2                             d
 3a cos sin  d
 /2
 /2                         1        
 L   dL  0            3a cos sin  d  3a  sin 2  
2        0
3
     a
2
 /2
 xELdL  0 a cos   3a cos sin  d 
3
and
 /2
 1                            3 2
 3a 2   cos5                       a
  5       0                   5

3  3                                                        2
Hence                xL   xELdL : x  a   a 2                                             x      a 
2  5                                                        5
Alternative solution
2/3
 x
x  a cos3   cos 2    
 a
2/3
 y
y  a sin 3   sin 2    
 a
2/3           2/3
x             y
             
3/2
                        1         or             y  a 2/3  x 2/3
a            a
Chapter 5, Problem 91.

Determine the y coordinate of the centroid of the body shown.

Chapter 5, Solution 91.

Labeling the two parts of the body as follows:

V       y       yV

1 2       h     1 2 2
1        a h            a h
2         2     4

1 2       h    1
2        a h             a 2h 2
6         4    24


2 2            7
a h             a 2h 2
3              24

Then Y 
yV

   7  a 2h 2
24                                   or Y 
7
h 
V          2  a 2h
3                                            16
Chapter 5, Problem 100.

Sheet metal of uniform thickness is used to fabricate a portion of
the flashing for a roof. Locate the center of gravity of the flashing
knowing that it is composed of the three elements shown.

Chapter 5, Solution 100.

Labeling the five parts of the body as follows, and noting that the center of gravity coincides with
the centroid of the area due to the uniform thickness.

       4  150                         
z5    300 
3 
   236.34,   A5         1502  11 250     35 343
                                        2

A mm 2             x , mm   y, mm        z , mm     xA,106 mm3 yA, 106 mm3 zA, 106 mm3

1    600  400   240000    300      200            0          72             48            0

2    300  400   120000    600      200          150         72             24           18

3    120  280    33600 600       140           240      20.160        4.7040      8.0640

4    600  300   180000    300      400          150         54             72            27

5           35 343            240      400          236.34    2.7000          4.5       2.6588

          471 057                                             169.358         125.159      28.583
Therefore:

xA 169 358 000
X                    359.53 mm                                  or X  360 mm 
A    471 057

yA 125159 000
Y                   265.70 mm                                   or Y  266 mm 
A    471057

zA  28 583 000
Z                      60.678 mm                             or Z   60.7 mm 
A     471 057

Chapter 5, Problem 109.

A thin steel wire of uniform cross section is bent into the shape
shown, where arc BC is a quarter circle of radius R. Locate its center
of gravity.

Chapter 5, Solution 109.
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of
the line

L, mm               x , mm         y, mm     z , mm    xL, mm 2    yL, mm 2    zL, mm 2

1          300                  0              150       0           0       45 000        0
2          280                 140              0        0        39 200        0          0
3          260                 230              0       120       59 800        0        31 200
                  3  2  300  360       600      480
4       300  150                                           54 000     90 000      72 000
2                  5                            

        1311.24                                                 153 000     135 000     103 200

Then
xL 153 000
X                                                 or X  116.7 mm
L 1311.24
yL 135 000
Y                                                  or Y  103.0 mm
L 1311.24
zL 103 200
Z                                                  or Z  78.7 mm
L 1311.24

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