# Problem static moment by sanmelody

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```									                        Key Solutions Assignment 13

Chapter 9, Problem 38.

Knowing that the shaded area is equal to 6000 mm 2 and that its
moment of inertia with respect to AA is 18  106 mm 4 ,
determine its moment of inertia with respect to BB for
d1  50 mm and d2  10 mm.

Chapter 9, Solution 38.

Given: A  6000 mm 2

          
I AA  18  106 mm4  I  6000 mm2 50 mm 
2

I  3  106 mm 4

        
I BB  I  Ad 2  3  106 mm4  6000 mm2  60 mm 
2

 24.6  106 mm4          I BB  24.6  106 mm 4
Chapter 9, Problem 41.

Determine the moments of inertia I x and I y of the area shown
with respect to centroidal axes respectively parallel and
perpendicular to side AB.

Chapter 9, Solution 41.

Determination of centroid:
y  0 by symmetry.

Part             Area               x  in.        
xA in 3
1             
3in0.75  2.25
2
1.5          3.375
2         6  0.75   4.50    0.375         1.6875
3         3  0.75   2.25     1.5           3.375
              9.00                          8.4375
xA 8.4375 in 3
x                      0.9375 in.
Α   9.00 in 2
Determination of I x :
1
Part : I x       3 in. 0.75 in.3  3 in. 0.75 in.3.375 in.2  25.734 in 4
12
1
Part : I x         0.75 in. 6 in.3  13.50 in 4
12
Part : (Same as Part ) I x  25.734 in 4
Entire Section: I x   25.734  13.50  25.734  in 4
 64.97 in 4                                I x  65.0 in 4
Determination of I y :
1
 0.75 in.3 in.3  0.75 in.3 in. 1.5  0.9375 in.
2
Part : I y                                                                 
12
 2.3994 in 4
1
 6 in. 0.75 in.3   6 in. 0.75 in.  0.9375  0.375 in.
2
Part : I y                                                                        
12
 1.6348 in 4
Part : (Same as Part ) I y  2.3994 in 4

Entire Section: I y   2.3994  1.6348  2.3994 in 4

 6.434 in 4                                I y  6.43 in4
Chapter 9, Problem 44.

Determine the moments of inertia I x and I y of the area shown
with respect to centroidal axes respectively parallel and
perpendicular to side AB.

Chapter 9, Solution 44.

Locate centroid:
x1  20 mm y1  45 mm A1   40 mm  90 mm 

 3600 mm2
1
x2  50 mm y2  51 mm A2        48 mm30 mm
2
 720 mm 2

xi Ai
Then    x 
Αi

 20 mm  3600 mm2   50 mm   720 mm2 

3600 mm2  720 mm2
 25.0 mm
yi Ai
And     y 
Ai

 45 mm   3600 mm2   51 mm   720 mm2 

3600 mm2  720 mm2
 46.0 mm
Now     I x   I x 1   I x 2

1
        
2
   40 mm  90 mm   3600 mm2 1 mm  
12
3


1
 36
3 1
2
                  2
   30 mm  24 mm   720 mm 2  59 mm  46.0 mm  

1
 36
3 1
2
                  2
   30 mm  24 mm   720 mm 2  46.0 mm  43 mm  


                                                                    
  2.430  106  3600  11.520  103  60.840  103  11.520  103  3240  mm4
                                                                        
I x  2.52  106 mm 4
Chapter 9, Problem 68.

Determine by direct integration the product of inertia of the
given area with respect to the x and y axes.

Chapter 9, Solution 68.

First note:             At x  a :      b  ka3

b         b
or k      3
; y  3 x3
a        a
Now                I xy   xydA

a b
 0 b 3    xydydx
x
a3

1 a  2 b2 6 
     x  b  a 6 x  dx
2 0 


a
b2  1 2  1 8
       x  6x 
2 2     8a  0

3 2 2
      ab
16
3 2 2
I xy       ab
16
Chapter 9, Problem 70.

Determine by direct integration the product of inertia of the
given area with respect to the x and y axes.

Chapter 9, Solution 70.
a
First note:       At x  a : b  ke a

b      b x
or k      ; y  ea
e      e

Now     I xy   xydA
x
b a
a ee
    
0 0
xydydx

1 b2 a 2ax
         xe dx
2 e2 0
a
                 
2                  
2x
1b  e 2 a

                x  1 
2 e2   2 2  a    
               
 a 
                 0


1 b2  a 2   2
           e 1  1 1 

2 e2  4  
 


8e2

a 2b2 2
e 1   
I xy  0.1419a2b2
Chapter 9, Problem 73.

Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.

Chapter 9, Solution 73.

Have                                                1   I xy 2
I xy  I xy

For each semicircle                          I xy  I xy  x y A    and           I xy   0 (symmetry)

Thus                                         I xy  x y A

A, mm 2                     x, mm                     y, mm                   Ax y , mm 4

                                                               160
1               1202    7200                 60                                           69.12  106
2                                                               

                                                           160
2               1202    7200                 60                                             69.12  106
2                                                               

                                                                                              138.24  106

or I xy  138.2  106 mm4 
Chapter 9, Problem 78.

Using the parallel-axis theorem, determine the product of inertia of the
area shown with respect to the centroidal x and y axes.

Chapter 9, Solution 78.

Have                                                      1   I xy 2
I xy  I xy

For each rectangle
I xy  I xy  x yA     and       I xy  0 (symmetry)

Then                     I xy   x yA   0.75 in. 1.5 in.  3 in. 0.5 in.
                  

  0.5 in.1.00 in.  4.5 in. 0.5 in.
                    

 1.6875  1.125  in 4  2.8125 in 4

or I xy  2.81 in 4 
Chapter 8, Problem 1.
\

Knowing that WA  25 lb and   30, determine (a) the smallest value
of WB for which the system is in equilibrium, (b) the largest value of WB
for which the system is in equilibrium.

Chapter 8, Solution 1.

FBD Block B:
Tension in cord is equal to W A  25 lb from FBD’s of block A and
pulley.
Fy  0:      N  WB cos30  0,              N  WB cos30

(a) For smallest WB , slip impends up the incline, and

F  s N  0.35WB cos30

Fx  0:      F  25 lb  WB sin 30  0

 0.35cos30  sin 30WB  25 lb
WB min  31.1 lb

(b) For largest WB , slip impends down the incline, and

F   s N   0.35 WB cos30

Fx  0:     Fs  WB sin 30  25 lb  0

 sin 30  0.35cos30WB  25 lb
WB max  127.0 lb 
Chapter 8, Problem 12.

The coefficients of friction are s  0.40 and k  0.30 between all
surfaces of contact. Determine the force P for which motion of the 30-kg
block is impending if cable AB (a) is attached as shown, (b) is removed.

Chapter 8, Solution 12.

Note that, since s  tan 1  s  tan 1  0.40   21.8  15, no motion
FBD top block:
will impend if P  0, with or without cable AB.
(a) With cable, impending motion of bottom block requires impending
slip between blocks, so F1   s N
Fy  0:          N1  W1 cos15  0,          N1  W1 cos15  189.515 N

F1   s N1   0.40 W1 cos15  0.38637W1
F1  75.806 N

FBD bottom block:            Fx  0:         T  F1  W1 sin15  0
T  75.806 N  50.780 N  126.586 N

            
W2   30 kg  9.81 m/s2  294.3 N

Fy  0 :          N 2  189.515 N  cos 15   294.3 N

  75.806 N  sin15  0
N2  457.74 N
F2   s N 2   0.40  457.74 N   183.096 N

Fx  0:           P  189.515 N    75.806 N  cos15
FBD block:
126.586 N  183.096 N  0
P  361 N
(b) Without cable, blocks remain together
Fy  0:       N  W1  W2  0                     N  196.2 N  294.3 N
 490.5 N
F   s N   0.40  490.5 N   196.2 N
Fx  0:           P  196.2 N  0                         P  196.2 N
Chapter 8, Problem 15.

A packing crate of mass 40 kg must be moved to the left along the floor
without tipping. Knowing that the coefficient of static friction between
the crate and the floor is 0.35, determine (a) the largest allowable value of
 , (b) the corresponding magnitude of the force P.

Chapter 8, Solution 15.

FBD:
For impending tip the floor reaction is at C.

         
W   40 kg  9.81 m/s2  392.4 N

For impending slip   s  tan 1  s  tan 1  0.35 

  19.2900
0.8 m               0.4 m
tan           ,      EG           1.14286 m
EG                  0.35

EF  EG  0.5 m  0.64286 m

EF            0.64286 m
(a)     s  tan 1          tan 1            58.109
0.4 m            0.4 m

 s  58.1
P         W
(b)              
sin19.29 sin128.820
P   392.4 N  0.424   166.379 N

P  166.4 N

Once slipping begins,  will reduce to k  tan 1 k .

Then  max will increase.

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