Statics Chapter 5

PROBLEM 5.1

Locate the centroid of the plane area shown.

SOLUTION

A, in 2

x , in.

y , in.

xA, in 3

yA, in 3

1

8 × 6 = 48

−4

9

−192 192

432

2

16 × 12 = 192

8

6

1536 1536

1152 1152

Σ

240 240

1344 1344

1584 1584

Then

X =

Σ xA 1344 in 3 = Σ A 240 in 2

or X = 5.60 in. 

and

Σ yA 1584 in 3 = Y = Σ A 240 in 2

or Y = 6.60 in. 

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PROBLEM 5.2


SOLUTION

A, mm2

1

Then

and

1 2

× 60 × 75 = 2250

2

105 × 75 = 7875

Σ

10 125

x , mm

y , mm

xA, mm3

yA, mm3

40

25

90 000

56 250

112.5 112.5

37.5 37.5

885 900

295 300

975 900

351 600

Σ xA 975 900 mm3 X = = Σ A 10 125 mm 2

or X = 96.4 mm 

Σ yA 351 600 mm3 Y = = Σ A 10 125 mm 2

or Y = 34.7 mm 







































PROBLEM 5.3


SOLUTION

For the area as a whole, it can be concluded by observation that Y =

2 3

or Y = 16.00 in. 

( 24 in.)

A, in 2

1

2

Σ

1 2 1 2

x , in.

× 24 × 10 = 120

2

× 24 × 16 = 192

10 +

3

(10) = 6.667 1 3

312

(16 ) = 15.333

xA, in 3

800

2944 3744 3744

Σ xA

3744 in 3







































PROBLEM 5.4


SOLUTION

1

A, mm2

x , mm

y , mm

xA, mm3

yA, mm3

21 × 22 = 462

1.5

11

693

5082

( 6) (9 ) = −27

−6

2

162

−54

12) = −36 ( 6) (12

8

2

−288

−72

567

4956

2

−

1

3

−

1

Σ

2

2

399

Then

Σ xA 567 mm3 X = = Σ A 399 mm 2

or X = 1.421 mm 

and

Σ yA 4956 mm3 Y = = Σ A 399 mm 2

or Y = 12.42 mm 







































PROBLEM 5.5


SOLUTION

1 2

Σ

A, mm2

x , mm

y , mm

xA, mm3

yA, mm3

120 × 200 = 24 000

60

120

1 440 000

2 880 000

94.5

120

−534 600

−678 600

905 400

2 201 400

−

π

( 60 ) 2

2

= −5654.9

18 345

Σ xA 905 400 mm3 = Σ A 18 345 mm 2

Then

X =

and

Σ yA 2 201 400 mm 3 = Y = Σ A 18 345 mm 2

or X = 49.4 mm 

or Y = 93.8 mm 







































PROBLEM 5.6


SOLUTION

A, in 2

1

2

Σ

π

( 9) 4

1 2

x , in.

2

= 63.617

(15) ( 9) = 67.5

−4 ( 9 )

(3 )

= −3.8917

π

5

131.1

y , in.

x A, in 3

y A, in 3

3.8917

−243

243

3

337.5

202.5

94.5

445.5

Then

Σ xA 94.5 in 3 = X = Σ A 131.1 in 2

or X = 0.721 in. 

and

Y =

Σ yA 445.5 in 3 = Σ A 131.1 in 2

or Y = 3.40 in. 









































PROBLEM 5.7


SOLUTION

First note that symmetry implies X = Y

A, mm2

1 2

Σ

Then

40 × 40 = 1600

−

(40) 2

π

4

= −1257

343

Σ xA 10 667 mm 3 X = = Σ A 343 mm 2

x , mm

xA, mm3

20

32 000

16.98

−21 330 10 667

or X = 31.1 mm 









































PROBLEM 5.8


SOLUTION

X = 0 

First note that symmetry implies

y , in.

yA, in 3

= −25.13

1.6977

−42.67

= 56.55

2.546

A, in 2

1

2

−

π

2

2 π

(6)

2

2

Σ

Then

( 4)

31.42

Y =

Σ yA

=

144 101.33 101.33

101.33 in 3 2

or Y = 3.23 in. 









































PROBLEM 5.9

For the area of Problem 5.8, determine the ratio r2 /r 1 so that y = 3r 1/4.

SOLUTION

A

y

y A

π

4r 1

2

3π

− r 13

− r 12

1

π

2

Σ

2

(r 2

π

2 2

r 22

3

3

4r 2

2

3π

3 2

(r 3

)

− r 12

Y ΣA = Σy A

Then

2

2

3 2

r 23

)

− r 13









































PROBLEM 5.9 CONTINUED

Then or Taking the positive root

p =

−(16 − 9π) ± (16 − 9π) 2 − 4(16)(16 − 9π ) 2(16) p = −0.5726

p = 1.3397 r 2 r 1

= 1.340 









































PROBLEM 5.10

Show that as r 1 approaches r 2 , the location of the centroid approaches that of a circular arc of radius ( r1 + r 2 ) / 2.

SOLUTION

First, determine the location of the centroid. From Fig. 5.8A:

2y

2

=

3 2

=

Then

and

2

1y =

Similarly

Σ yA =

2

=

2

3 3

3

(r

3 2

π

(2

π

r 2

1r

− α )

A= 2

( 2 − ) 22r π

α

cos α

(2

− α )

π

cos α

(2

π

1A=

− α )

cosα 

2rπ 2

(

3

sin ( 2 − α ) r 2

( − α)  2 π

( 2 − ) 12r π

2 co cosα  − α ) 22r − 1 r − α ) 12r  3 ( − α ) ( 2  2 π

π

)

− r 13 cosα π Σ A

α

 

2

 π 

 

2












































Using Figure 5.8B, Y of an arc of radius

1 2

( r1 + r 2 )

Y =

=

Now

r23 − r 13 r22 − r 12

( r1 + r 2 )

2 1

(r1 + r 2 )

r22 − r 12

α

cos α π

α

+ r1 r2 + r12 r2 + r 1 r2 = r + ∆ r1 = r − ∆ 1 2

( r1 + r 2 )

2 2 ( r + ∆ ) + ( r + ∆ ) ( r − ∆ ) + ( r − ∆ ) = ( r + ∆ ) + ( r − ∆ )

= In the limit as ∆ → 0 (i.e., r1 = r 2 ), then

π

( r2 − r1 ) ( r2 + r1 )

r22

r = r23 − r 13

(2 − )

(2 − ) ( r2 − r1 ) ( r22 + r1 r2 + r12 ) = 2

Let

Then

sin ( 2 − α ) π

1

=

and

is

3r 2 + ∆ 2 2r

(1)











































PROBLEM 5.11


SOLUTION

X = 0 


r 2 = 2 2 in.,

α

45° = 45











































PROBLEM 5.12


SOLUTION











































PROBLEM 5.13


SOLUTION

A, mm2 2

( 40) ( 80)

x , mm 2133

y , mm

xA, mm3

yA, mm3











































PROBLEM 5.14


SOLUTION













































PROBLEM 5.15


SOLUTION











































PROBLEM 5.16


SOLUTION











































PROBLEM 5.17

The horizontal x axis is drawn through the centroid C of 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











































PROBLEM 5.18

The horizontal x axis is drawn through the centroid C of 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












































AI:

QI = Σ yA 155 115 − ( 80 × 115)  (160 × 155) + = 2 2   = 1.393 × 106 mm3

AII :

QII = Σ yA 145

=−

 85  (80 × 85)  2 

(160 × 145) − −

2 = −1.393 × 106 mm3



∴ ( Qarea ) x = QI + QII = 0 yA and y = 0 , Which is expected since Q x = Σ yA = yA since x is a centroidal axis.













































PROBLEM 5.19

The first moment of the shaded area with respect to the x axis is denoted by Q x . (a) Express Q x in terms of r of r and θ . (b) For what value of θ is Q x maximum, maximum, and what is the maximum value?











































PROBLEM 5.20

A composite beam is constructed by bolting four plates to four 2 × 2 × 3/8-in. angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal axis of











































PROBLEM 5.21

A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center center of gravity of the wire figure thus formed.











































PROBLEM 5.22

A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center center of gravity of the wire figure thus formed.











































PROBLEM 5.23

A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.





























































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































PROBLEM 5.53

A 15-mm-diameter hole is drilled in a piece of 20-mm-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process.

SOLUTION

The required volume can be generated by rotating the area shown about the y axis. Applying the second theorem of Pappus-Guldinus, we have

 5

 1   + 7.5  mm  ×  × 5 mm × 5 mm     3  2

V = 2π xA = 2π 

or V = 720 mm3 







































PROBLEM 5.54

Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design.

SOLUTION

Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by A C

= π yL= π Σ yL

Where the individual lengths are the “Lengths” of the belt cross section that are in contact with the pulley Have

AC =

π

 2 ( y1 L1 ) + y2 L2 

    2.5 mm  2.5  60 − mm   = π 2  60   2    cos 20°     mm )  + ( 60 − 2.5 ) mm  (12.5 mm  or AC = 3.24 × 103 mm 2  Have

AC =

π

 2 ( y1 L1 )

   7.5 mm  7.5  mm  ×  = 2π  60 − 1.6 −   2     cos20  

or AC = 2.74 × 103 mm 2 

Have

AC =

π

(

y1 L1 ) =

π

  2× 5  60 − π  mm  (π × 5 mm )    or AC = 2.80 × 103 mm 2 







































PROBLEM 5.55

Determine the capacity, in gallons, of the punch bowl shown if R = 12 in.

SOLUTION

The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the second theorem of Pappus-Guldinus and using Fig. 5.8A, we have

V = 2π xA = 2πΣxA = 2π ( x1 A1 + x2 A2 )

  1 1   1 1 3 R R× = 2π  ×  ×  2  3 2   2 2 

   2 R sin 30 R cos 30  +  π   3 × 6  



    π 2   R   6     

 R3 R3  3 3 R 3 π = 2π  +  = 8  16 3 2 3  = Since

3 3 8

π

3 in.) = 3526.03 in in 3 (12 in

1 gal = 231 in 3 V =

3526.03 3526.03 in3 231in 231in 3/gal gal

15.26 gal = 15.26 V = 15.26 15.26 gal 







































PROBLEM 5.56

The aluminum shade for a small high-intensity lamp has a uniform thickness of 3/32 in. Knowing that the specific weight of aluminum is 0.101lb/in 3, determin determinee the weight weight of the the shade. shade.

SOLUTION

The weight of the lamp shade is given by W = γ V = γ At where A where A is the surface area of the shade. This area can be generated by rotating the line shown about the x axis. Applying the first theorem of Pappus-Guldinus, we have A = 2π yL = 2πΣ yL = 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 )

 0.6 mm 2 2  0.60 + 0.75  mm × ( 0.15 mm ) + (1.5 mm ) = 2π  ( 0.6 mm ) +   2    2 2 2  0.75 + 1.25  mm × ( 0.50 mm ) + ( 0.40 mm ) +  2   2 2  1.25 + 1.5  mm × ( 0.25 mm ) + (1.25 mm )  +  2   

22.5607 7 in 2 = 22.560 Then

3 W = 0.101 lb/in × 22.5607 in 2 ×

3 32

in. = 0.21362 lb W = 0.214 0.214 lb 







































PROBLEM 5.57

The top of a round wooden table has the edge profile shown. Knowing that the diameter of the top is 1100 mm before shaping and that the density of the wood is 690 kg/m3, determin determinee the the weight weight of of the the waste waste wood resulting from the production of 5000 tops.

SOLUTION

All dimensions are in mm Vwaste = Vblank − V top

Have

2

V blank = π ( 550 mm ) × ( 30 mm ) = 9.075π × 106 mm 3 Vtop = V1 + V2 + V3 + V4 Applying the second theorem of Pappus-Guldinus to parts 3 and 4 2 2 V top = π ( 529 mm ) × (18 mm )  + π (535 mm ) × (12 mm )









   π 4 × 12  2  m m 1 2 m m + 2π   535 + × ( )    3π   4       π 4 × 18  2  m m 1 8 m m + 2π   529 + × ( )    3π   4    = π ( 5.0371 + 3.347 + 0.1222 + 0.2731) × 106 mm3 = 8.8671π × 106 mm3 ∴ V waste = ( 9.0750 − 8.8671)π × 106 mm3 = 0.2079π × 10−3 m3 Finally

Wwaste = ρ wood Vwaste g N tops 3

(

3

3

)

2









































PROBLEM 5.58

The top of a round wooden table has the shape shown. Determine how many liters of lacquer are required to finish 5000 tops knowing that each 2 top is given three coats of lacquer and that 1 liter of lacquer covers 12 m .

SOLUTION

Referring to the figure in solution of Problem 5.57 and using the first theorem of Pappus-Guldinus, we have A surface =

A top circle + A bottom circle + A edge

2 2 = π ( 535 mm )  + π (529 mm )       π 2 × 12     mm  × (12 mm ) + 2π  535 +  π      2

 π 2 × 18     mm  × (18 mm )  + 2π  529 +  π      2 = 617.115π × 103 mm 2 Then

# liters =

suArface

× Coverage ×

= 617.115π × 10−3 m 2 ×

tN ops

×

1liter 12 m 2

cN oats

× 5000 × 3 or # li liters = 2424 L 









































PROBLEM 5.59

The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from yellow brass. Knowing that the specific 3 weight of yellow brass is 0.30 0.306 6 lb/in b/in . determine the weight of the escutcheon.

SOLUTION

W = (specific weight)V

The weight of the escutcheon is given by

V is the volume of the plate. V can V can be generated by rotating the area A about the x where V is the x axis.

a = 3.0755 in i n. − 2.958 i n. n. = 0.1175 i n. n.

Have

sin φ =

and Then

0.5 3

⇒ φ = 0.16745 R = 9.5941°

2α = 26 − 9.5941 = 16.4059 





or

α = 8.20295 = 0.143169 rad 

The area A can be obtained by combining the following four areas, as indicated.










































A, in 2 1

1 2

( 3.0755) (1.5) =

y , in. 1

2.3066

3 2 ( 3) sin α

2

2

−α ( 3) = −1.28851

3

− ( 2.958) ( 0.5) = −0.7395

4

− ( 0.1755) ( 0.5) = −0.05875

3α

1

1

2

3

(1.5) = 0.5

yA, in 3 1.1533

× sin (α + φ ) = 0.60921

–0.78497

( 0.5) = 0.16667

–0.12325

1 2

( 0.5) = 0.25

–0.14688 0.44296 6 in3 Σ yA = 0.4429

Then so that

(

)

V = 2π 0.44296 in in 3 = 1.4476 in in 3

(

)

W = 1.4476 in i n 3 0.306 lb l b/in 3 = 0.44296 lb

W = 0.443 0.443 lb 









































PROBLEM 5.60

The reflector of a small flashlight has the parabolic shape shown. Determine the surface area of the inside of the reflector.

SOLUTION

First note that the required surface area A can be generated by rotating the parabolic cross section through 2π radians about the x the x axis. Applying the first theorem of Pappus-Guldinus, we have A= 2π yL x = ,ky2

Now, since

a = 56.25 k

or x=

At

Eq. (1)

:

a + 15 a

=

156.25k 56.25k

=x 0.15

2

y

and 2

Now

(2) a = 8. 8.4375 mm

or

Eq. (1) ⇒ k = 0.15

∴

2

(1)

a + 15 = 156.25k Eq. (2)

)

2 mm : a+ 15 = k(12. 5 ) ( a+ 15) mm

or Then

x =: a a = 7.k5 (

at

1 mm dx dy

= 0.3

y

 dx  2 dL = 1 +   dy = 1 + 0.09 y dy  dy 









































PROBLEM 5.61

For the beam and loading shown, determine ( a) the magnitude and location of the resultant of the distributed load, ( b) the reactions at the beam supports.

SOLUTION

Resultant

R = R1 + R2

(a) Have

R1 = ( 40 lb/ft ) (18 ft ) = 720 lb R2 =

or

1 2

lb/ft ) (18 ft ft ) = 1080 lb lb (120 lb R = 1800 1800 lb

C of the distributed load x The resultant is located at the centroid C of Have or

ΣM A: (1800 lb ) x = ( 40 lb/ft ) (18 ft ) (9 ft ) + x = 10.8 10.80 0 ft

1 2

(120 lb/ft ) (18 ft ) (12 ft )











































PROBLEM 5.62

For the beam and loading shown, determine (a ( a) the magnitude and location of the resultant of the distributed load, (b ( b) the reactions at the beam supports.

SOLUTION

(a) Have

RI = ( 300 N/m ) ( 6 m ) = 1800 N RII =

Then

1 3

N/m ) = 1800 N ( 6 m) ( 900 N/

Σ F y : − R = − RI − RII











































PROBLEM 5.63

Determine the reactions at the beam supports for the given loading.

SOLUTION

Have

RI = (100 lb/ft ) ( 4 ft ) = 400 lb RII =

1 2

lb/ft ) ( 6 ft ft ) = 600 lb lb ( 200 lb











































PROBLEM 5.64


SOLUTION

Have

RI = ( 9 ft ) ( 200 lb/ft ) = 1800 lb R

1

lb/ft ) ( 3 ftft ) ( 200 lb

300 lb lb











































PROBLEM 5.65


SOLUTION

Have

RI =

1 2

( 200 N/m ) ( 0.12 m ) = 12 N

RII = ( 200 N/ N/m ) ( 0.2 m) = 40 N Then

Σ F x = 0: Ax = 0 Σ F y = 0 : Ay + 18 N − 12 N − 40 N = 0











































PROBLEM 5.66


SOLUTION

First, replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a linear relation between load and distance, and the values at the end points are the same.

Have

RI = (1.8 m ) ( 2000 N/m ) = 3600 N RII =

1 2

N/m ) = (1.8 m ) ( 4500 N/

4050 N













































PROBLEM 5.67


SOLUTION











































PROBLEM 5.68


SOLUTION

First, replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance, and the values at end points are the same.











































PROBLEM 5.69

Determine (a (a) the distance a so that the vertical reactions at supports A and B and B are equal, (b (b) the corresponding reactions at the supports.

SOLUTION











































PROBLEM 5.70

Determine (a (a) the distance a so that the vertical reaction at support B is minimum, (b (b) the corresponding reactions at the supports.

SOLUTION











































PROBLEM 5.71

Determine the reactions at the beam supports for the given loading when w0 = 1.5 kN/m.

SOLUTION













































PROBLEM 5.72

Determine (a (a) the distributed load w0 at the end D of the beam ABCD for which the reaction at B is zero, (b (b) the corresponding reactions at C .











































PROBLEM 5.73

A grade beam AB supports three concentrated loads and rests on soil and the top of a large rock. The soil exerts an upward distributed load, and the rock exerts a concentrated load R R as shown. Knowing that P that P 4 kN 1 and w B = 2 wA , determine the values of w A and R R corresponding to equilibrium. =











































PROBLEM 5.74

A grade beam AB supports three concentrated loads and rests on soil and the top of a large rock. The soil exerts an upward distributed load, and the rock exerts a concentrated load R R as shown. Knowing that w B 0.4w 0.4w A, determine (a (a) the largest value of P for which the beam is in equilibrium, (b (b) the corresponding value of w of w A

=











































PROBLEM 5.75

The cross section of a concrete dam is as shown. For a dam section of unit width, determine (a (a) the reaction forces exerted by the ground on the base AB of the dam, (b (b) the point of application of the resultant of the












































(b) Have

ΣM A = 0: X ( 5933 kN ) + ( 6 m ) (1589 kN ) kN ) − (12 m )( − ( 6 m) (1589 kN ) ( 2543 kN )





























































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































PROBLEM 5.112

A scratch awl has a plastic handle and a steel blade and shank. Knowing that the specific weight of plastic is 0.03 0.0374 74 lb/i lb/in n 3 and and of of ste steel el is 3 0.284 0.284 lb/in lb/in , locate locate the the cente centerr of gravit gravity y of the awl. awl.

SOLUTION

Y = Z = 0 

First, note that symmetry implies

xI =

5 8

in.) = ( 0.5 in

)  23π  ( 0.5 inin.)

(

3

0.3125 in in., W I = 0.0374 lb lb/in 3 

(

xII = 1.6 in in. + 0.5 in in. = 2.1 in in. W II = 0.0374 lb lb/in3

) (π ) ( 0.5 inin.)

2

in.) = ( 3.2 in

(

)  π 4  ( 0.12 inin.)

(

)  π 4  ( 0.12 inin.)

xIII = 3.7 in in. − 1 in. = 2.7 in in., W III = − 0.0374 lb lb/in3  xIV = 7.3 in in. − 2.8 in in. = 4.5 in in., W IV = 0.284 lb lb/in3  xV = 7.3 in in. +

1 4

in.) = ( 0.4 in

I II III IV V

Σ

Have

lb = 0.009791 lb

(

2

2

in.) = − 0.000846 0846 lb ( 2 in

2 in.) = ( 5.6 in

)  π 3  ( 0.06 inin.)

7.4 in in., W V = 0.284 lb lb/in3 

2

x , in. in.

xW , in i n ⋅ lb

0.009791 0.093996 −0.000846 0.017987 0.000428 0.12136

0.3125 2.1 2.7 4.5 7.4

0.003060 0.197393 −0.002284 0.080942 0.003169 0.28228

X( 0 .12136 lb ) = 0.28228 in.⋅ lb

0.017987 lb lb

in.) = ( 0.4 in

W , lb

XΣ W = Σ xW:

0.093996 lb lb

0.000428 lb lb

or X = 2.33 in. 







































PROBLEM 5.113

A bronze bushing is mounted inside a steel sleeve. Knowing that the density of bronze is 8800 8800 kg/m kg/m3 and and of of stee steell is 7860 7860 kg/m kg/m3, dete determ rmin inee the center of gravity of the assembly.

SOLUTION

X = Z = 0 


W =

Now

( ρ g )V

(

)(

)

 π    0.036 2 − 0.0152 m 2  ( 0.008 m )    4   

(

yI = 4 mm , W I = 7860 kg/m3 9.81 m/s 2 

)

0.5188 887 7N = 0.51

(

)(

)

 π    0.0225 2 − 0.05 2 m 2  ( 0.02 m )    4   

yII = 18 mm, W II = 7860 kg/m3 9.81 m/s 2 

(

)

0.34065 5N = 0.3406

(

)(

)

 π    0.152 − 0.10 2 m 2  ( 0.028 m )    4   

yIII = 14 mm, W III = 8800 kg/m3 9.81 m/s 2 

(

0.23731 1N = 0.2373

Y ΣW = Σ yW

Have Y =

mm )( mm) ( 0.3406 N) + (14 mm mm) ( 0.2373 N) ( 4 mm ) ( 0.5189 N) + (18 mm 0 5189 N + 0 3406 N + 0 2373 N

)







































PROBLEM 5.114

A marker for a garden path consists of a truncated regular pyramid carved 3 from stone of specific weight 160 lb/ft . The pyramid is mounted on a steel base of thickness h. Knowing that the specific weight of steel is 3 490 lb/ft and that steel plate is available in 14 in. increments, specify the minimum thickness h for which the center of gravity of the marker is approximately 12 in. above the top of the base.

SOLUTION

First, locate the center of gravity of the stone. Assume that the stone is homogeneous so that the center of gravity coincides with the centroid of the corresponding volume. Have

y1 =

3 4

( 56 in.) =

V 1 =

42 in.,

1 3

(12 in.) (12 in.) ( 56 in.)

2688 in3 = 2688 y2 =

3 4

( 28 in.) =

V 2 = −

21 in.,

1 3

( 6 in.) ( 6 in.) ( 28 in.)

366 in3 = −366 V stone = 2688 in 3 − 366 in 3

Then

2352 in3 = 2352 Y =

and

=

Σ yV ΣV in.) ( 2688 in 3 ) + ( 21 in.) ( −366 in in 3 ) ( 42 in 2352 2352 in3

in. = 45 in. Therefore, the center of gravity of the stone is ( 45 − 28 ) in. = 17 in. in. above the base. 3

Wstone =

(

γ stoneV stonN e o=w

3

)(

3

)

 1 ft   160 lb/ft in.   12 in.

217.78 lb = 217.78 Wsteel =

γ

steelV steel

 1 ft  lb/ft3 (12 in.) (12 i n. = 490 lb/ft n.) h     12 in 

(

)

3

2352 in








































Then

Y marker =

= or

Σ yW in. = 12 in. ΣW  h  in.  ( 40.833 h ) lb  2 

in.) ( 217.78 lb ) +  − (17 in

( 217.78 + 40.833h ) lb h 2 + 24h − 53.334 = 0

With positive solution h = 2.0476 in. specify h = 2 in.  ∴ specify







































PROBLEM 5.115

The ends of the park bench shown are made of concrete, while the seat and back back are are woode wooden n boar boards. ds. Eac Each h piece piece of wood wood is 36 × 120 × 1180 mm. Knowing that the density density of concrete is 2320 kg/m3 and of of 3 wood is 470 470 kg/m kg/m , dete determ rmin inee the the x and y coordinates of the center of gravity of the bench.

SOLUTION

First, note that we will account for the two concrete ends by counting twice the weights of components 1, 2, and 3. W1 =

kg/m3 ) (9.81 m/ m/s 2 ) ( 0.480 m ) ( 0.408 m ) ( 0.072 m ) ( ρ c g ) V 1 = ( 2320 kg 320.9 9N = 320.

(

)(

)

W2 = − ( ρ c g )V 2 = − 2320 kg k g/m3 9.81 m/ m/s 2 ( 0.096 m ) ( 0.048 m ) ( 0.072 m ) 7.551 1N = −7.55 W3 =

k g/m3 ) (9.81 m/ m/s 2 ) ( 0.096 m ) ( 0.384 m ) ( 0.072 m ) ( ρ c g ) V 3 = ( 2320 kg 60.41N = 60.41N

W4 = W5 = W6 = W7 =

ρ wVboa rd

(

)(

)

k g/m3 9.81 m/ m/s 2 ( 0.120 m ) ( 0.036 m ) (1.180 m ) = 470 kg 23.504 04 N = 23.5










































W , N

x , mm

y , mm

x W , mm mm ⋅ N

y W , mm mm ⋅ N

1

2 ( 320.4 ) = 641.83

312

−204

200 251.4

−130 933.6

2

2 ( −7.551) = −15.10

312

−384

−4711.8

5799.1

3

2 ( 60.41) = 120.82

84

192

10 148.5

23 196.5

4 5 6 7 8

23.504 23.504 23.504 23.504 23.504 865.06

228 360 442 124.7 160.1

18 18 18 328.3 139.6

5358.8 8461.3 10 388.5 2930.9 3762.9 236 590

423.1 423.1 423.1 7716.2 3281.1 −89 671

Σ Have

XΣ W = Σ x W:

X( 8 65.06 N ) = 236 590 mm mm ⋅ N or X = 274 mm 

Y ΣW = Σ yW : Y ( 8 65.06 N ) = −89 671 mm ⋅ N or Y = −103.6 mm 









































PROBLEM 5.116

Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A hemisphere.

SOLUTION

r and thickness dx. dx. Choose as the element of volume a disk of radius r and Then dV = π r 2 dx, x EL = x The equation of the generating curve is x2 + y2 = a2 so that r 2 = a 2 − x 2 and then dV =

(a

π

2

− x 2 ) dx

Component 1 a/2

V1 =

= and

a/2

∫ 0

11 24

π

(

 x3  a 2 − x 2 dx = π  a 2 x −  3  0

)

a3

π

a/2

∫1 x EL dV = ∫ 0

xπ a2 − x2 dx

(



)



a/2

 x2 x 4  = π a 2 −  2 4  0 = Now

x1 V 1 =

x ∫ 1 EL

7 64 dV:

a4

π

7  11 3  4  24 π a  = 64 π a  

x 1

21









































PROBLEM 5.116 CONTINUED a

and

∫2

xEL dV =

∫ a/2 x a

π

(

 2 x 2 x 4   a − x ) dx = π  a −   4  2 a/2 2

2

2 4   2 4  ( a )   2 ( a2 ) ( a2 )    2 ( a ) = π  a − − a −  2 4   2 4       

= Now

9

a4

π

64

x2 V2 =

∫ 2

xEL dV:

9  5 3 4 π a π a =   24  64

x2 

or x2 =

27 40

a









































PROBLEM 5.117

Determine by direct integration the values of for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A semiellipsoid of revolution.

SOLUTION

r and thickness dx. dx. Choose as the element of volume a disk of radius r and Then dV = π r 2 dx, xEL = x The equation of the generating curve is r2 =

a2 h

2

(h

2

x 2 h2

+

y2

= 1 so that

a2

− x 2 ) and then dV =

π

a2 h

2

(h

− x 2 ) dx

2

Component 1 V1 =

=

and

∫1

h/2

∫ 0

11

π

a2 h2

(h

− x ) dx =

2

2

π

a2 

h x − 2

h2 

a2h

π

24 x EL

dV =

h/2

∫ 0

= π =

 a2 2  xπ 2 ( h − x2 ) dx  h 

a2 

h

2

x2

h2 

7 64

a 2h 2

π

2

−

x4 



h/2

4 0

x3 



h/2

3 0










































Component 2 V2 =

=

=

and

∫2

x EL

h

∫ h/2

π

π

h  

π

a2 

h x − 2

h2 

x3 

h



3 h/2

 

3  

 2

3  



a 2h

π

h

∫ h/2

= π

= π

Now

)

− x dx = 2

3  3  h h ( ) h ( )   2 2 h 2    h h − − − ( )    2

24

=

(h

h2

2

 a 2 

5

dV =

a2

a2 

h

2

x2

h2 

 a 2 

2

−

x4 

h



4 h/2

2 4  2 4  h h h) h)   2 ( 2 ) ( ( ( 2 )  h − h − −  2   2 4 2 4  2

h  

9



 



a 2h 2

π

64

x2 V2 =

 a2 2  xπ 2 ( h − x2 ) dx  h 

∫ 2 x EL dV:

9  5 2  2 2 π a h = π a h   24  64

x2 

or x2 =

27 40

h











































PROBLEM 5.118

Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A paraboloid of revolution.

SOLUTION

r and thickness dx. dx. Choose as the element of volume a disk of radius r and Then dV =

r 2 dx, xEL = x

π

The equation of the generating curve is x = h − r2 =

a2 h

(h − x)

and then dV =

π

a2 h

( h − x ) dx

Component 1 V1 =

=

h/2

∫ 0

π

π

a2 h

( h − x ) dx

a2 

x2 

h 

2

 hx −



h/2

h a

2

y2 so that












































Component 2 V2 =

=

=

h

∫ h/2

π

π

a2 h

( h − x ) dx =

x2 

h 

2 h/2

 hx −



 a 2 

2  2  h h)    h  ( 2 )  (  − h  − h ( h ) −  2   2 2  

h   1

π

h

a2 



 



 

a 2h

π

8

h

and

∫2

x EL

dV =

h

∫ h/2

= π

= Now

 a2  a2  x2 x3  xπ ( h− x) dx = π  h −  h  2 3  h  h/2

 a2 

1 12

3  2 3  h 2 h h) h)   ( 2 ) ( ( ( 2 )  − h − −  h   2 2 3 3 

h  





 

a 2h 2

π

x2 V2 =

∫ 2 x EL dV:

1 1 2  π a h = π a2 h2  8  12

x2 

2













































PROBLEM 5.119

Locate the centroid of the volume obtained by rotating the shaded area about the x the x axis.

SOLUTION

y = 0 


z = 0  r and thickness dx. dx. Choose as the element of volume a disk of radius r and Then dV =

  

Now r = b 1 −

π r

2

dx, x EL = x

x2 

 so that

a 2 

 dV = π b 2  1 

x2 

2

 dx












































Then

xV =

x ∫ EL

dV:

1 8 2 2 2 π ab  = π a b  15  16

x

or x =

15 6

a











































PROBLEM 5.120

Locate the centroid of the volume obtained by rotating the shaded area about the x the x axis.

SOLUTION

y = 0 


z = 0  r and thickness dx. dx. Choose as the element of volume a disk of radius r and Then dV = Now r = 1 −

1 x

π r

2

dx, x EL = x

so that 2

1  dV = π 1 −  dx  x  2 1   = π 1 − + 2  dx  x x 













































PROBLEM 5.121

Locate the centroid of the volume obtained by rotating the shaded area about the line x = h.

SOLUTION

x = h 


z = 0  r and thickness dx. dx. Choose as the element of volume a disk of radius r and Then 2












































and

∫

y EL dV =

∫ 0

= π = π = π =

 h2

a

yπ

2  a

h2 a

a

∫

2 0

(

a−

a2 − y2

2

(

a y + a a − y 3 a 2  h2 



a 2 − y 2 − y 3 dy

2

2 2



dy

)

( 2a y − 2ay

h2 

)

2

2

2

3/2

)

a

 − y  4 0 1

4

3/2   1 4 2 2 2 2 a a a a a − − ( ) ( )   4   3 a 2   

1 12

π a

2 2

h











































PROBLEM 5.122

Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x the x axis.

SOLUTION


y = 0 

z = 0  Choose as the element of volume a disk of radius r and r and thickness dx. dx.












































Then

∫

2a    sin π ax   2a  x  x sin πa x  x dV =   x − 2π  − ∫  − 2π a  dx   2 a a  a 2   2a   a2 π x    2a   a   1 2 2  = π b  2a   − a    −  x + 2 cos   a   2π  2   4   2  a   3   1 a2 1 a2  2 2  = π b2  a 2  −  ( 2a ) + 2 − ( a ) + 2   4 2π 2π     2   4 2 π b EL

1  3 = π a 2b 2  − 2   4 π  2 2











































PROBLEM 5.123

Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y axis. ( Hint: Hint: Use a thin cylindrical shell of radius r and r and thickness dr as dr as the element of volume.)

SOLUTION


x = 0  z = 0 

Choose as the element of volume a cylindrical shell of radius r and r and thickness dr . 1

(

)( )( )














































Use integration by parts with u = r

dv = sin 2

π r

2

dr











































PROBLEM 5.124

Show that for a regular pyramid of height h and n sides ( n = 3, 4, 4, … ) the centroid of the volume of the pyramid is located located at at a distance h / 4 above the base.











































PROBLEM 5.125

Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R radius R..











































PROBLEM 5.126

The sides and the base of a punch bowl are of uniform thickness t . If t << R and R = 350 mm, determine the location of the center of gravity of (a (a) the bowl, (b (b) the punch.












































Now Then

x2 + y2 = R2 V =

dV =

so that 0

∫ −

3/2 R

π

(

(R

π

2

)

− y 2 dy 0

1   R − y dy = π  R2 y − y3  3 −  2

2

)

3/2 R

























































































































































































































































































































































































































































































































































































































































































































































Statics Chapter 5

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