Beer & Johnston 9Ed Resolução do Capítulo 5 Forças Distribuídas (Centroides e Centro de Gravidade)

Prévia do material em texto

PROBLEM 5.1 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, inA , in.x , in.y 3, inxA 3, inyA 
1 8 6 48× = 4− 9 192− 432 
2 16 12 192× = 8 6 1536 1152 
Σ 240 1344 1584 
 
Then 
3
2
1344 in
240 in
xAX
A
Σ= =Σ or 5.60 in.X = W 
and 
3
2
1584 in
240 in
yAY
A
Σ= =Σ or 6.60 in.Y = W 
 
 
 
 
 
PROBLEM 5.2 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 
1 60 75 2250
2
× × = 40 25 90 000 56 250 
2 105 75 7875× = 112.5 37.5 885 900 295 300 
Σ 10 125 975 900 351 600 
 
Then 
3
2
975 900 mm
10 125 mm
xAX
A
Σ= =Σ or 96.4 mmX = W 
and 
3
2
351 600 mm
10 125 mm
yAY
A
Σ= =Σ or 34.7 mmY = W 
 
 
 
 
 
 
PROBLEM 5.3 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
For the area as a whole, it can be concluded by observation that 
 ( )2 24 in.
3
Y = or 16.00 in.Y = W 
 2, inA , in.x 3, inxA 
1 
1 24 10 120
2
× × = ( )2 10 6.6673 = 800 
2 
1 24 16 192
2
× × = ( )110 16 15.3333+ = 2944 
Σ 312 3744 
 
Then 
3
2
3744 in
312 in
xAX
A
Σ= =Σ or 12.00 in.X = W 
 
 
 
 
 
 
PROBLEM 5.4 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 21 22 462× = 1.5 11 693 5082 
2 ( )( )1 6 9 272− = − 6− 2 162 54− 
3 ( )( )1 6 12 362− = − 8 2 288− 72− 
Σ 399 567 4956 
 
Then 
3
2
567 mm
399 mm
xAX
A
Σ= =Σ or 1.421 mmX = W 
and 
3
2
4956 mm
399 mm
yAY
A
Σ= =Σ or 12.42 mmY = W 
 
 
 
 
 
PROBLEM 5.5 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 120 200 24 000× = 60 120 1 440 000 2 880 000 
2 
( )260 5654.9
2
π− = − 94.5 120 534 600− 678 600− 
Σ 18 345 905 400 2 201 400 
 
Then 
3
2
905 400 mm
18 345 mm
xAX
A
Σ= =Σ or 49.4 mmX = W 
and 
3
2
2 201 400 mm
18 345 mm
yAY
A
Σ= =Σ or 93.8 mmY = W 
 
 
 
 
 
PROBLEM 5.6 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, inA , in.x , in.y 3, inxA 3, inyA 
1 
( )29 63.617
4
π = 
( )
( )
4 9
3.8917
3π
− = − 3.8917 243− 243 
2 ( )( )1 15 9 67.52 = 5 3 337.5 202.5 
Σ 131.1 94.5 445.5 
 
Then 
3
2
94.5 in
131.1 in
xAX
A
Σ= =Σ or 0.721 in.X = W 
and 
3
2
445.5 in
131.1 in
yAY
A
Σ= =Σ or 3.40 in.Y = W 
 
 
 
 
 
PROBLEM 5.7 
Locate the centroid of the plane area shown. 
 
SOLUTION 
First note that symmetry implies X Y= 
 
 2, mmA , mmx 3, mmxA 
1 40 40 1600× = 20 32 000 
2 
2(40) 1257
4
π− = − 16.98 21 330− 
Σ 343 10 667 
 
Then 
3
2
10 667 mm
343 mm
xAX
A
Σ= =Σ or 31.1 mmX = W 
 and 31.1 mmY X= = W 
 
 
 
 
 
PROBLEM 5.8 
Locate the centroid of the plane area shown. 
 
SOLUTION 
First note that symmetry implies 0X = W 
 
 2, inA , in.y 3, inyA 
1 
( )24 25.13
2
π− = − 1.6977 42.67− 
2 
( )26 56.55
2
π = 2.546 144 
Σ 31.42 101.33 
 
Then 
3
2
101.33 in
31.42 in
yAY
A
Σ= =Σ or 3.23 in.Y = W 
 
 
 
 
PROBLEM 5.9 
For the area of Problem 5.8, determine the ratio 2 1/r r so that 13 /4.y r= 
 
SOLUTION 
 
 A y yA 
1 
2
12
rπ− 143
r
π 
3
1
2
3
r− 
2 
2
22
rπ 243
r
π 
3
2
2
3
r 
Σ ( )2 22 12 r rπ − ( )3 32 123 r r− 
 
Then Y A y AΣ = Σ 
or ( ) ( )2 2 3 31 2 1 2 13 24 2 3r r r r rπ× − = − 
2 3
2 2
1 1
9 1 1
16
r r
r r
π      − = −        
 
Let 2
1
rp
r
= 
[ ] 29 ( 1)( 1) ( 1)( 1)
16
p p p p pπ + − = − + + 
or 216 (16 9 ) (16 9 ) 0p pπ π+ − + − = 
 
 
 
 PROBLEM 5.9 CONTINUED 
Then 
2(16 9 ) (16 9 ) 4(16)(16 9 )
2(16)
p
π π π− − ± − − −= 
or 0.5726 1.3397p p= − = 
Taking the positive root 2
1
1.340r
r
= W 
 
 
 
 
 
PROBLEM 5.10 
Show that as 1r approaches 2,r the location of the centroid approaches that 
of a circular arc of radius ( )1 2 / 2.r r+ 
 
SOLUTION 
 
First, determine the location of the centroid. 
From Fig. 5.8A: 
( )
( ) ( )2 22 2 2 222
sin2 
3
y r A r
π
π
π
α αα
−= = −− 
 ( )2 2
2 cos
3
r π
α
α= − 
Similarly ( ) ( ) 21 1 1 122
2 cos 
3
y r A rππ
α αα= = −− 
( ) ( ) ( ) ( )
( )
2 2
2 2 1 12 2
2 2
3 3
2 1
2 cos 2 cosThen
3 3
2 cos
3
yA r r r r
r r
π π
π π
α αα αα α
α
   Σ = − − −   − −
= −
 
( )
2 2
2 1
2 2
2 1
and 
2 2
2
A r r
r r
π πα α
π α
   Σ = − − −      
 = − −  
 
( ) ( )2 2 3 32 1 2 1
3 3
2 1
2 2
2 1 2
Now
2 cos
2 3
2 cos 
3
Y A yA
Y r r r r
r rY
r r π
π α α
α
α
Σ = Σ
  − − = −    
−= −−
 
 
 
 PROBLEM 5.10 CONTINUED 
Using Figure 5.8B,Y of an arc of radius ( )1 21 is2 r r+ 
( ) ( )( )21 2 2
sin1
2
Y r r
π
π
α
α
−= + − 
 ( )1 2 2
1 cos( )
2
r r π
α
α= + − (1) 
( )( )
( )( )
2 23 3 2 1 2 1 2 12 1
2 2
2 1 2 12 1
2 2
2 1 2 1
2 1
Now 
r r r r r rr r
r r r rr r
r r r r
r r
− + +− = − +−
+ += +
 
2
1
Let r r
r r
= + ∆
= − ∆ 
Then ( )1 212r r r= + 
( ) ( )( ) ( )
( ) ( )
2 23 3
2 1
2 2
2 1
2 2
and 
3
2
r r r rr r
r rr r
r
r
+ ∆ + + ∆ − ∆ + − ∆− = + ∆ + − ∆−
+ ∆=
 
1 2In the limit as 0 (i.e., ), thenr r∆ → = 
3 3
2 1
2 2
2 1
1 2
3 
2
3 1 ( )
2 2
r r r
r r
r r
− =−
= × +
 
so that ( )1 2
2
2 3 cos
3 4
Y r r π
α
α= × + − or ( )1 2 2
1 cos
2
Y r r π
α
α= + − W 
Which agrees with Eq. (1). 
 
 
 
 
 
 
PROBLEM 5.11 
Locate the centroid of the plane area shown. 
 
SOLUTION 
First note that symmetry implies 0X = W 
 
2 2 2 in., 45r α= = ° 
( ) ( )
( ) 42 4
2 2 2 sin2 sin 1.6977 in.
3 3
ry
π
π
α
α= = =′ 
 2, inA , in.y 3, inyA 
1 ( ) ( )1 4 3 62 = 1 6 
2 ( )22 2 6.2834π = 2 0.3024y− =′ 1.8997 
3 ( ) ( )1 4 2 42− = − 0.6667 2.667− 
Σ 8.283 5.2330 
 
Then Y A yAΣ = Σ 
 ( )2 38.283 in 5.2330 inY = or 0.632 in.Y = W 
 
 
 
 
PROBLEM 5.12 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 (40)(90) 3600= 15− 20 54 000− 72 000 
2 
( ) ( )40 60 2121
4
π = 10 15− 6750 10 125− 
3 ( ) ( )1 30 45 6752 = 25.47− 19.099− 54 000− 40 500− 
Σ 6396 101 250− 21 375 
 
Then XA xA= Σ 
 ( )2 36396 mm 101 250 mmX = − or 15.83 mmX = − W 
and YA yA= Σ 
 ( )2 36396 mm 21 375 mmY = or 3.34 mmY = W 
 
 
 
 
PROBLEM 5.13 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 ( ) ( )2 40 80 21333 = 48 15 102 400 32 000 
2 ( ) ( )1 40 80 16002− = − 53.33 13.333 85 330− 21 330− 
Σ 533.3 17 067 10 667 
 
Then X A XAΣ = Σ 
 ( )2 3533.3 mm 17 067 mmX = or 32.0 mmX = W 
and Y A yAΣ = Σ 
 ( )2 3533.3 mm 10 667 mmY = or 20.0 mmY = W 
 
 
 
 
 
PROBLEM 5.14 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 ( )( )2150 240 24 0003 = 56.25 96 1 350 000 2 304 000 
2 ( ) ( )1 150 120 90002− = − 50 40 450 000− 360 000− 
Σ 15 000 900 000 1 944 000 
 
Then X A xAΣ = Σ 
 ( )2 315 000 mm 900 000 mmX = or 60.0 mmX = W 
and Y A yAΣ = Σ 
 ( )215 000 mm 1 944 000Y = or 129.6 mmY = W 
 
 
 
 
PROBLEM 5.15 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, inA , in.x , in.y 3, inxA 3, inyA 
1 ( ) ( )1 10 15 503 = 4.5 7.5 225 375 
2 ( )215 176.714
π = 6.366 16.366 1125 2892 
Σ 226.71 1350 3267 
 
Then X A x AΣ = Σ 
 ( )2 3226.71 in 1350 inX = or 5.95 in.X = W 
and Y A y AΣ = Σ 
 ( )2 3226.71 in 3267 inY = or 14.41 in.Y = W 
 
 
 
 
 
PROBLEM 5.16 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, inA , in.x , in.y 3, inxA 3, inyA 
1 ( ) ( )2 8 8 42.673 = 3 2.8 128 119.47 
2 ( ) ( )2 4 2 5.3333− = − 1.5 0.8− 8− 4.267 
Σ 37.33 120 123.73 
 
Then X A x AΣ = Σ 
 ( )2 337.33 in 120 inX = or 3.21 in.X = W 
and Y A y AΣ = Σ 
 ( )2 337.33 in 123.73 inY = or 3.31 in.Y = W 
 
 
 
 
 
 
PROBLEM 5.17 
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 
 
Note that xQ yA= Σ 
Then ( ) 21 5 1m 6 5 m3 2xQ    = × ×       or ( ) 3 31 25.0 10 mmxQ = × W 
and ( ) 2 22 2 1 1 12.5 m 9 2.5 m 2.5 m 6 2.5 m3 2 3 2xQ      = − × × × + − × × ×           
 or ( ) 3 32 25.0 10 mmxQ = − × W 
Now ( ) ( )1 2 0x x xQ Q Q= + = 
This result is expected since x is a centroidal axis ( )thus 0y = 
and ( )0 0x xQ y A Y A y Q= Σ = Σ = ⇒ = 
 
 
 
PROBLEM 5.18 
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 
First, locate the position y of the figure. 
 
 2, mmA , mmy 3, mmyA 
1 160 300 48 000× = 150 7 200 000 
2 150 80 16 000− × = − 160 2 560 000− 
Σ 32 000 4 640 000 
 
Then Y A y AΣ = Σ 
 ( )2 332 000 mm 4 640 000 mmY = 
or 145.0 mmY = 
 
 
 
 
 PROBLEM 5.18 CONTINUED 
 ( )
I I
6 3
:
155 115 (160 155) 80 115
2 2
 1.393 10 mm 
A Q yA= Σ
 = × + − × 
= × W
 
 ( ) ( )
II II
6 3
:
145 85 160 145 80 85
2 2
 1.393 10 mm 
A Q yA= Σ
 = − × − − ×  
= − × W
 
 ( )area I II Q 0x Q Q∴ = + = 
 Which is expected since xQ yA yA= Σ = and 0y = , 
 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 .xQ (a) Express xQ in terms of r and .θ (b) For what value of θ is 
xQ maximum, and what is the maximum value? 
 
SOLUTION 
 
( ) With and using Fig. 5.8 A,xa Q yA= Σ 
( ) ( ) ( )
( )
2
3 2 2 2
2 3
2
3 2
sin 1sin 2 cos sin
2
2 cos cos sin
3
x
r
Q r r r r
r
π
π
π
θ θ θ θ θθ
θ θ θ
 −    = − − × ×  −    
= −
 
 or 3 32 cos
3x
Q r θ= W 
( )b By observation, is maximum when xQ 0θ = W 
 and then 32
3x
Q r= W 
 
 
 
 
 
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 x axis of 
the red shaded areas shown, respectively, in parts a and b of the figure. 
Knowing that the force exerted on the bolt at A is 70 lb, determine the 
force exerted on the bolt at B. 
 
SOLUTION 
 
From the problem statement: xF Q∝ 
so that ( ) ( )A Bx xA B
F F
Q Q
= 
and 
( )
( )x BB Ax A
Q
F F
Q
= 
Now xQ yA= ∑ 
So ( ) ( ) 30.3757.5 in. in. 10 in. 0.375 in. 28.82 in
2x A
Q    = + × =     
( ) ( ) ( )( )
( ) ( )( )
0.375and 2 7.5 in. in. 1.625 in. 0.375 in.
2
 2 7.5 in. 1 in. 2 in. 0.375 in.
x xB AQ Q
   = + −    
 + −  
 
 3 3 328.82 in 8.921 in 9.75 in= + + 
 347.49 in= 
Then ( )3347.49 in 70 lb 115.3 lb28.82 inBF = = W 
 
 
 
 
PROBLEM 5.21 
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. 
 
SOLUTION 
 
 
First note that because the wire is homogeneous, its center of gravity will 
coincide with the centroid of the corresponding line. 
 , in.L , in.x , in.y 2, inxL 2, inyL 
1 16 8 0 128 0 
2 12 16 6 102 72 
3 24 4 12 96 288 
4 6 8− 9 48− 54 
5 8 4− 6 32− 48 
6 6 0 3 0 18 
Σ 72 336 480 
 
Then X L x LΣ = Σ 
 ( ) 272 in. 336 inX = or 4.67 in.X = W 
and Y L yLΣ = Σ 
 2(72 in.) 480 inY = or 6.67 in.Y = W 
 
 
 
 
PROBLEM 5.22 
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. 
 
SOLUTION 
 
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the 
corresponding line. 
 , mmL , mmx , mmy 2, mmxL 2, mmyL 
1 165 82.5 0 13 612 0 
2 75 165 37.5 12 375 2812 
3 105 112.5 75 11 812 7875 
4 2 260 75 96.05+ = 30 37.5 2881 3602 
Σ 441.05 40 680 14 289 
 
Then X L x LΣ = Σ 
 2(441.05 mm) 40 680 mmX = or 92.2 mmX = W 
and Y L yLΣ = Σ 
 2(441.05 mm) 14 289 mmY = 32.4 mmY = W 
 
 
 
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. 
 
SOLUTION 
 
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the 
corresponding line. 
 , mmL , mmx , mmy 2, mmxL 2, mmyL 
1 2 212 6 13.416+ = 6 3 80.50 40.25 
2 16 12 14 192 224 
3 21 1.5 22 31.50 462 
4 16 9− 14 144− 224 
5 2 26 9 10.817+ = 4.5− 3 48.67− 32.45 
Σ 77.233 111.32 982.7 
 
Then X L x LΣ = Σ 
 2(77.233 mm) 111.32 mmX = or 1.441 mmX = W 
and Y L yLΣ = Σ 
 2(77.233 mm) 982.7 mmY = or 12.72 mmY = W 
 
 
 
 
 
PROBLEM 5.24 
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. 
 
SOLUTION 
 
First note that because the wire is homogeneous, its center of gravity will 
coincide with the centroid of the corresponding line. 
By symmetry 0X = W 
 , in.L , in.y 2, inyL 
1 2 0 0 
2 ( )6π ( )2 6 3.820π = 72 
3 2 0 0 
4 ( )4π ( )2 4 2.546π = 32 
Σ 35.416 104 
 
Then Y L yLΣ = Σ 
 2(35.416 in.) 104 inY = or 2.94 in.Y = W 
 
 
 
 
 
 
PROBLEM 5.25 
A 750 g= uniform steel rod is bent into a circular arc of radius 500 mm 
as shown. The rod is supported by a pin at A and the cord BC. Determine 
(a) the tension in the cord, (b) the reaction at A. 
 
SOLUTION 
 
 
( )0.5 m sin 30First note, from Figure 5.8B:
/6
X
°= π 
 1.5 mπ= 
( )( )2Then mg0.75 kg 9.81 m/s
7.358 N
W =
=
=
 
 Also note that ∆ ABD is an equilateral triangle.Equilibrium then requires 
( ) ( )
( ) 0: 
1.5 0.5 m m cos30 7.358 N 0.5 m sin 60 0
A
BC
a M
Tπ
Σ =
    − ° − ° =      
 
 or 1.4698 NBCT = or 1.470 NBCT = W 
( )( ) 0: 1.4698 N cos60 0x xb F AΣ = + ° = 
 or 0.7349 NxA = − 
 ( )0: 7.358 N 1.4698 N sin 60 0y yF AΣ = − + ° = 
 or 6.085 NyA = thus 6.13 N=A 83.1° W 
 
 
 
 
 
 
PROBLEM 5.26 
The homogeneous wire ABCD is bent as shown and is supported by a pin 
at B. Knowing that 8 in.,l = determine the angle θ for which portion 
BC of the wire is horizontal. 
 
SOLUTION 
 
 
First note that for equilibrium, the center of gravity of the wire must lie 
on a vertical line through B. Further, because the wire is homogeneous, 
its center of gravity will coincide with the centroid of the corresponding 
line. 
Thus 0, which implies that 0BM xΣ = = or 0xLΣ = 
Hence 
( ) ( )2(6 in.) 8 in.6 in. 8 in.
2
ππ
 − × +    
 ( )6 in.8 in. cos 6 in. 0
2
θ + − =   
Then 4cos
9
θ = or 63.6θ = ° W 
 
 
 
 
 
PROBLEM 5.27 
The homogeneous wire ABCD is bent as shown and is supported by a pin 
at B. Knowing that 30 ,θ = ° determine the length l for which portion CD 
of the wire is horizontal. 
 
SOLUTION 
 
 
First note that for equilibrium, the center of gravity of the wire must lie on 
a vertical line through B. Further, because the wire is homogeneous, its 
center of gravity will coincide with the centroid of the corresponding line. 
Thus 0, which implies that 0BM xΣ = = or 0i ix LΣ = 
Hence ( ) ( ) ( )2 6 in. cos30 6 in. sin 30 6 in.ππ
 − ° + ° ×  
 
 ( ) ( ) in. cos30 in.
2
l
l
 + °  
 
 ( ) ( )6 in.in. cos30 6 in. 0
2
l + ° − =   
or 2 12.0 316.16 0l l+ − = 
1with roots 12.77 and 24.77.l = − 
Taking the positive root 
 12.77 in.l = W 
 
 
 
 
 
 
PROBLEM 5.28 
The homogeneous wire ABCD is bent as shown and is attached to a hinge 
at C. Determine the length L for which the portion BCD of the wire is 
horizontal. 
 
SOLUTION 
 
 
First note that for equilibrium, the center of gravity of the wire must lie on 
a vertical line through C. Further, because the wire is homogeneous, its 
center of gravity will coincide with the centroid of the corresponding line. 
Thus 0, which implies that 0CM xΣ = = 
or 0i ix LΣ = 
( ) ( ) ( ) ( ) ( )Hence 4 in. 8 in. 4 in. 10 in. 0
2
L L + − + − = 
or 2 2144 inL = or 12.00 in.L = W 
 
 
 
 
 
PROBLEM 5.29 
Determine the distance h so that the centroid of the shaded area is as close 
to line BB′ as possible when (a) 0.2,k = (b) 0.6.k = 
 
SOLUTION 
 
 
Then yAy
A
Σ= Σ 
( ) ( ) ( )
( )
2 2
or 
a ha ab kb a h
y
ba kb a h
 +  − −    = − − 
 ( )2 211
2 (1 )
a k kh
a k kh
− += − + 
Let 1 and hc k
a
ζ= − = 
Then 
2
2
a c ky
c k
ζ
ζ
+= + (1) 
Now find a value of ζ (or h) for which y is minimum: 
 
( ) ( )
( )
2
2
2
0
2
k c k k c kdy a
d c k
ζ ζ ζ
ζ ζ
+ − += =+ or ( ) ( )22 0c k c kζ ζ ζ+ − + = (2) 
 
 
 PROBLEM 5.29 CONTINUED 
Expanding (2) 2 22 2 0c c kζ ζ ζ+ − − = or 2 2 0k c cζ ζ+ − = 
Then 
( ) ( ) ( )22 2 4
2
c c k c
k
ζ − ± −= 
Taking the positive root, since 0h > (hence 0ζ > ) 
 
( ) ( ) ( )22 1 4 1 4 1
2
k k k k
h a
k
− − + − + −= 
(a) 0.2: k = ( ) ( ) ( )( )( )
22 1 0.2 4 1 0.2 4 0.2 1 0.2
2 0.2
h a
− − + − + −= or 0.472h a= W 
(b) 0.6: k = ( ) ( ) ( )( )( )
22 1 0.6 4 1 0.6 4 0.6 1 0.6
2 0.6
h a
− − + − + −= or 0.387h a= W 
 
 
 
 
PROBLEM 5.30 
Show when the distance h is selected to minimize the distance y from 
line BB′ to the centroid of the shaded area that .y h= 
 
SOLUTION 
From Problem 5.29, note that Eq. (2) yields the value of ζ that minimizes h. 
Then from Eq. (2) 
We see 
2
2 c k
c k
ζζ ζ
+= + (3) 
Then, replacing the right-hand side of (1) by 2ζ , from Eq. (3) 
We obtain ( )2
2
ay ζ= 
But h
a
ζ = 
So y h= Q.E.D. W 
 
 
 
 
 
 
PROBLEM 5.31 
Determine by direct integration the centroid of the area shown. Express 
your answer in terms of a and h. 
 
SOLUTION 
 
For the element of area (EL) shown hy x
a
= 
( )and 
1
dA h y dx
xh dx
a
= −
 = −  
 
( )
Then 
1
2
1
2
EL
EL
x x
y h y
h x
a
=
= +
 = +  
 
2
0
0
1Then area 1
2 2
a
a x xA dA h dx h x ah
a a
  = = − = − =      ∫ ∫ 
2 3
2
0
0
2 2
20 0
2 3
2
2
0
1and 1
2 3 6
1 1 1
2 2
1
2 33
a
a
EL
a a
EL
a
x x xx dA x h dx h a h
a a
h x x h xy dA h dx dx
a a a
h xx ah
a
   = − = − =         
     = + − = −             
 = − =   
∫ ∫
∫ ∫ ∫ 
2
Hence 
1 1
2 6
ELxA x dA
x ah a h
=
  =  
∫
 
 1
3
x a= W 
21 1
2 3
ELyA y dA
y ah ah
=
  =  
∫
 
 2
3
y h= W 
 
 
 
PROBLEM 5.32 
Determine by direct integration the centroid of the area shown. Express 
your answer in terms of a and h. 
 
SOLUTION 
 
 
For the element (EL) shown 
At 3, :x a y h h ka= = = or 3
hk
a
= 
Then 1/31/3
ax y
h
= 
1/3
1/3
Now dA xdy
a y dy
h
=
= 
1/3
1/ 3
1 1 , 
2 2EL EL
ax x y y y
h
= = = 
Then ( )1/3 4/31/3 1/30
0
3 3
4 4
h
h a aA dA y dy y ah
h h
= = = =∫ ∫ 
1/3 1/3 5/3 2
1/3 1/3 2/30
0
1 1 3 3and 
2 2 5 10
h
h
EL
a a ax dA y y dy y a h
h h h
   = = =      ∫ ∫ 
 1/3 7/3 21/3 1/30
0
3 3
7 7
h
h
EL
a ay dA y y dy y ah
h h
   = = =      ∫ ∫ 
Hence 23 3: 
4 10EL
xA x dA x ah a h = =  ∫ 
2
5
x a= W 
 23 3: 
4 7EL
yA y dA y ah ah = =  ∫ 
4
7
y h= W 
 
 
 
 
 
PROBLEM 5.33 
Determine by direct integration the centroid of the area shown. Express 
your answer in terms of a and h. 
 
SOLUTION 
 
 
For the element (EL) shown 
At 31, :x a y h h k a= = = or 1 3
hk
a
= 
 32a k h= or 2 3
ak
h
= 
Hence, on line 1 
3
3
hy x
a
= 
and on line 2 
1/3
1/3
hy x
a
= 
Then 
1/3 3 1/3 3
1/3 3 1/3 3
1 and 
2EL
h h h hdA x x dx y x x
a a a a
   = − = +      
1/3 3 4/3 4
1/3 3 1/3 30
0
3 1 1 
24 4
a
a h hA dA x x dx h x x ah
a a a a
   ∴ = = − = − =      ∫ ∫
1/3 3 7/3 5 2
1/3 3 1/3 30
0
3 1 8
357 5
a
a
EL
h hx dA x x x dx h x x a h
a a a a
   = − = − =      ∫ ∫ 
 1/3 3 1/3 31/3 3 1/3 30
1
2
a
EL
h h h hy dA x x x x dx
a a a a
  = + −    ∫ ∫ 
 
2 2/3 6 2 5/3 6
2
2/3 6 5/3 60
0
3 1 8
2 2 5 7 35
a
ah x x h x xdx ah
a a a a
   = − = − =         ∫ 
From 28:
2 35EL
ahxA x dA x a h = =  ∫ or 
16
35
x a= W 
and 28:
2 35EL
ahyA y dA y ah = =  ∫ or 
16
35
y h= W 
 
 
 
 
 
 
PROBLEM 5.34 
Determine by direct integration the centroid of the area shown. 
 
SOLUTION 
 
 
First note that symmetry implies 0x = W 
For the element (EL) shown 
 
2 (Figure 5.8B)EL
ry
dA rd r
π
π
=
=
 
Then ( )22
1
1
2
2 2
2 12 2
r
r
r
r
rA dA rd r r rππ π  = = = = −  ∫ ∫ 
and ( ) ( )22
1
1
3 3 3
2 1
2 1 22
3 3
r
r
EL r
r
ry dA rd r r r rππ
 = = = −  ∫ ∫ 
So ( ) ( )2 2 3 32 1 2 12: 2 3ELyA y dA y r r r rπ = − = −  ∫ 
 or 
3 3
2 1
2 2
2 1
4
3
r ry
r rπ
−= − W 
 
 
 
 
 
PROBLEM 5.35 
Determine by direct integration the centroid of the area shown. 
 
SOLUTION 
 
 
First note that symmetry implies 0x = W 
For the element (EL) shown 
 cos , siny R x Rθ θ= = 
 cos dx R dθ θ= 
 2 2cosdA ydx R dθ θ= = 
Hence
 ( )2 2 2 20
0
sin 2 12 cos 2 2 sin 2
2 4 2
A dA R d R R
αα θ θθ θ α α = = = + =  ∫ ∫ 
 
( )
( )
2 2 3 2
0
0
3
2
1 22 cos cos cos sin sin
2 3 3
cos sin 2sin
3
EL
Ry dA R d R
R
αα θ θ θ θ θ θ
α α α
 = = +  
= +
∫ ∫
 
But so ELyA y dA= ∫ 
( )
( )
3
2
2
cos sin 2sin
3
2 sin 2
2
R
y
R
α α α
α α
+
=
+
 
or 
( )
( )
2cos 22 sin
3 2 sin 2
y R
αα α α
+= + 
Alternatively, 
22 3 sinsin
3 2 sin 2
y R αα α α
−= + W 
 
 
 
 
PROBLEM 5.36 
Determine by direct integration the centroid of the area shown. 
 
SOLUTION 
 
 
For the element (EL) shown 
2 2by a x
a
= − 
( )
( )2 2
and dA b y dx
b a a x dx
a
= −
= − − 
( ) ( )2 21; 2 2EL EL bx x y y b a a xa= = + = + − 
Then ( )2 20a bA dA a a x dxa= = − −∫ ∫ 
To integrate, let 2 2sin : a cos , cosx a x a dx a dθ θ θ θ= − = = 
( )( )/20
/2
2 2
0
Then cos cos
2sin sin 1
2 4 4
bA a a a d
a
b a a ab
a
π
π
θ θ θ
θ θ πθ
= −
    = − + = −        
∫
 
( ) ( ) /23/22 2 2 2 20
0
3
1and 
2 3
1
6
a
EL
b b ax dA x a a x dx x a x
a a
a b
π    = − − = + −       
=
∫ ∫
 
( ) ( )
( )
2 2 2 2
0
2 2 3
2 2
2 20
0
2
1 
3 62 2
a
EL
a
a
b by dA a a x a a x dx
a a
b b xx dx ab
a a
 = + − − −  
 = = =   
∫ ∫
∫
 
 21: 1
4 6EL
xA x dA x ab a bπ  = − =    ∫ or ( )
2
3 4
ax π= − W 
 21: 1
4 6EL
yA y dA y ab abπ  = − =    ∫ or ( )
2
3 4
by π= − W 
 
 
 
 
 
PROBLEM 5.37 
Determine by direct integration the centroid of the area shown. Express 
your answer in terms of a and b. 
 
SOLUTION 
 
For the element (EL) shown on line 1 at 
 22, x a b k a= = or 2 2
bk
a
= 
 22 
by x
a
∴ = 
On line 2 at 31, 2x a b k a= − = or 2 3
2bk
a
−= 
3
3
2 by x
a
−∴ = 
2 3
2 3
2b bdA x x dx
a a
 = +   
3 3 4
2
2 20
0
2 2Then 
3 4
1 1 5
3 2 6
a
a b x b x xA dA x dx
x aa a
ab ab
   = = + = +      
 = + =  
∫ ∫
 
4 5
2 3 2
2 3 20
0
2
2 2 1 2and 
4 5 4 5
13
20
a
a
EL
b b b x xx dA x x x dx a b
aa a a
a b
    = + = + = +         
=
∫ ∫
2 3 2 3
2 3 2 30
2 2 2 5
2 3 7
2 3 4 20
0
2 5 2
1 2 2
2
1 2 2
2 52 7
1 2 13
10 7 70
a
EL
a
a
b b b by dA x x x x dx
a a a a
b b b xx x dx x
a a a a
b a ab
    = − +        
      = − = −               
 = − = −  
∫ ∫
∫ 
Then 25 13 :
6 20EL
xA x dA x ab a b = =  ∫ or 
39
50
x a= W 
 25 13:
6 70EL
yA y dA y ab ab = −  ∫ or 
39
175
y b= − W 
 
 
 
 
PROBLEM 5.38 
Determine by direct integration the centroid of the area shown. Express 
your answer in terms of a and b. 
 
SOLUTION 
 
 
At 0,x y b= = 
( )20b k a= − or 2bk a= 
Then ( )22 by x aa= − 
Now ( )22, 2 2EL EL
y bx x y x a
a
= = = − 
and ( )22bdA ydx x a dxa= = − 
( ) ( )2 32 20 0
1Then 
33
aa b bA dA x a dx x a ab
a a
 = = − = − =  ∫ ∫ 
( ) ( )
( ) ( ) ( )
2 3 2 2
2 20 0
4 2
3 2 2
2
2
2 2 5
2 2 40
0
2
and 2
2 1
4 3 2 12
1
52 2
1
10
a a
EL
a
a
EL
b bx dA x x a dx x ax a x dx
a a
b x aax x a b
a
b b by dA x a x a dx x a
a a a
ab
 = − = − +  
 = − + =   
   = − − = −      
=
∫ ∫ ∫
∫ ∫
 
Hence 21 1:
3 12EL
xA x dA x ab a b = =  ∫ 
1
4
x a= W 
 21 1:
3 10EL
yA y dA y ab ab = =  ∫ 
3
10
y b= W 
 
 
 
 
PROBLEM 5.39 
Determine by direct integration the centroid of the area shown. 
 
SOLUTION 
 
2
2
2
2
Have 
1 1
2 2
 1
EL
EL
x x
a x xy y
L L
x xdA ydx a dx
L L
=
 = = − +   
 = = − +   
 
Then 
22 2 3
2
2 20
0
81
2 33
L
L x x x xA dA a dx a x aL
L LL L
   = = − + = − + =     ∫ ∫ 
22 2 3 4
2
2 20
0
2
2 2
2
2 20
2 2 3 4
2 3 40
2 2 3 4
2 3
and 1
2 3 4
10
3
1 1
2
1 2 3 2
2
2 2
L
L
EL
L
EL
EL
x x x x xx dA x a dx a
L LL L
aL
a x x x xy dA a dx
L LL L
a x x x x dx
L L L L
a x x xx
L L L
    = − + = − +          
=
    = − + − +            
 = − + − +   
= − + − +
∫ ∫
∫ ∫
∫
25
2
4
0
11
55
L
x a L
L
  =  
 
Hence 28 10:
3 3EL
xA x dA x aL aL = =  ∫ 
5
4
x L= W 
 21 11:
8 5EL
yA y dA y a a = =  ∫ 
33
40
y a= W 
 
 
 
 
PROBLEM 5.40 
Determine by direct integration the centroid of the area shown. Express 
your answer in terms of a and b. 
 
SOLUTION 
 
 
For 1 at , 2y x a y b= = 22b ka= or 2
2bk
a
= 
Then 21 2
2by x
a
= 
By observation ( )2 2 2b xy x b ba a = − + = −   
Now 
and for 0 : 
ELx x
x a
=
≤ ≤ 
2 2
1 12 2
1 2 and 
2EL
b by y x dA y dx x dx
a a
= = = = 
For 2 :a x a≤ ≤ 
2 2
1 2 and 2
2 2EL
b x xy y dA y dx b dx
a a
   = = − = = −       
22
20
223
2
0 0
2Then 2
2 72
3 2 6
a a
a
aa
b xA dA x dx b dx
aa
b x a xb ab
aa
 = = + −  
    = + − − =         
∫ ∫ ∫
 
( ) ( ){ ( ) ( )
22
20
24 3
2
2
0 0
2 2 32 2
2
2and 2
2
4 3
1 12 2
2 3
7
6
a a
EL a
a a
b xx dA x x dx x b dx
aa
b x xb x
aa
a b b a a a a
a
a b
    = + −        
   = + −      
   = + − + −       
=
∫ ∫ ∫
 
 
 
 
PROBLEM 5.40 CONTINUED 
22 2
2 20 0
232 5 2
4
0
2
2 2 2
2
2 2
5 2 3
17
30
a a
EL
aa
a
b b b x xy dA x x dx b dx
a aa a
b x b a x
aa
ab
      = + − −           
    = + − −         
=
∫ ∫ ∫
 
Hence 27 7:
6 6EL
xA x dA x ab a b = =  ∫ x a= W 
 27 17:
6 30EL
yA y dA y ab ab = =  ∫ 
17
35
y b= W 
 
 
 
 
PROBLEM 5.41 
Determine by direct integration the centroid of the area shown. Express 
your answer in terms of a and b. 
 
SOLUTION 
 
 
2For y at , :x a y b= = 2a kb= or 2
ak
b
= 
Then 1/22
by x
a
= 
1/2 1/2
2
2
Now 
and for 0 : , 
2 2 2
EL
EL
x x
a y b x xx y dA y dx b dx
a a
=
≤ ≤ = = = = 
For ( ) 1/21 21 1: 2 2 2 2EL
a b x xx a y y y
a a
 ≤ ≤ = + = − +   
 
 ( ) 1/22 1 12x xdA y y dx b dxaa
 = − = − +   
( )( ) ( )
1/2 1/2
/2
0 /2
/2 3/2 2
3/2
0 /2
3/2 3/2
3/2
2
2
1Then 
2
2 2 1
3 3 2 2
2
3 2 2
1 1 
2 2 2 2
a a
a
aa
a
x x xA dA b dx b dx
aa a
b x xx b x
aa a
b a aa
a
a ab a a
a
 = = + − +   
  = + − +     
    = + −         
      + − − + −          
∫ ∫ ∫
13
24
ab
    
=
 
 
 
 
 
PROBLEM 5.41 CONTINUED 
( )
( ) ( )
1/2 1/2
/2
0 /2
/2 5/2 3 4
5/2
0 /2
5/2 5/2
5/2
3
3 2
1and 
2
2 2
5 5 3 4
2
5 2 2
1 1 
3 2 4
a a
EL a
aa
a
x x xx dA x b dx x b dx
aa a
b x x xx b
aa a
b a aa
a
ab a a
a
    = + − +            
  = + − +     
    = + −         
  + − − + −     
∫ ∫ ∫
2
2
1/2 1/2
/2
0
1/2 1/2
/2
/2 32 2 2
2
0 /2
2
71
240
 
2
1 1 
2 2 2
1 1 1
2 2 2 2 3 2
4
a
EL
a
a
aa
a
a
a b
b x xy dA b dx
a a
b x x x xb dx
a aa a
b b x xx
a a a a
b
            
=
 =   
    + − + − +            
       = + − −          
=
∫ ∫
∫
( )2 2 322
2
1
2 2 6 2 2
11
48
a a b aa
a a
ab
      + − − −             
=
 
Hence 213 71:
24 240EL
xA x dA x ab a b = =  ∫ 
17 0.546
130
x a a= = W 
 213 11:
24 48EL
yA y dA y ab ab = =  ∫ 
11 0.423
26
y b b= = W 
 
 
 
 
PROBLEM 5.42 
A homogeneous wire is bent into the shape shown. Determine by direct 
integration the x coordinate of its centroid. Express your answer in terms 
of a. 
 
SOLUTION 
 
 
First note that because the wire is homogeneous, its center of gravity 
coincides with the centroid of the corresponding line 
Have at 2, :x a y a a ka= = = or 1k
a
= 
Thus 21 2 and y x dy xdx
a a
= = 
( )
2 2
2 2
2
2 2 20
0
2
2
2Then 1 1
4 4 2 4 1 1 ln 1
2 4
 5 ln 2 5 1.4789
2 4
4 1
a
a
EL
dydL dx x dx
dx a
x x a xL dL x dx x
aa a a
a a a
xx dL x dx
a
   = + = +      
    ∴ = = + = + + + +    
= + + =
  = +  
∫ ∫
∫
( )
3/22
2
20
0
2
3/2 2
2 41
3 8
5 1 0.8484
12
a
a a x
a
a a
   = +        
= − =
∫
Then ( ) 2: 1.4789 0.8484ELxL x dL x a a= =∫ 0.574x a= W 
 
 
 
 
 
PROBLEM 5.43 
A homogeneous wire is bent into the shape shown. Determine by direct 
integration the x coordinate of its centroid. 
 
SOLUTION 
 
 
First note that because the wire is homogeneous, its center of gravity 
coincides with the centroid of the corresponding line 
Now cosELx r θ= and dL rdθ= 
Then [ ]7 /47 /4/4 /4 32L dL rd r rπππ πθ θ π= = = =∫ ∫ 
and ( )7 /4/4 cosELx dL r rdππ θ θ=∫ ∫ 
 [ ]7 /42 2 2/4 1 1sin 22 2r r r
π
πθ  = = − − = −   
23Thus : 2
2
xL xdL x r rπ = = −  ∫ 
2 2
3
x rπ= − W 
 
 
 
 
PROBLEM 5.44 
A homogeneous wire is bent into the shape shown. Determine by direct 
integration the x coordinate of its centroid. 
 
SOLUTION 
 
 
First note that because the wire is homogeneous, its center of gravity 
coincides with the centroid of the corresponding line 
Now 3cosELx a θ= and 2 2dL dx dy= + 
Where 3 2cos : 3 cos sinx a dx a dθ θ θ θ= = − 
3 2sin : 3 sin cosy a dy a dθ θ θ θ= = 
( ) ( )
( )
1/22 22 2
1/22 2
/ 2
0
Then 3 cos sin 3 sin cos
 3 cos sin cos sin
 3 cos sin
1 3 cos sin 3 sin
2
dL a d a d
a d
a d
L dL a d aπ
θ θ θ θ θ θ
θ θ θ θ θ
θ θ θ
θ θ θ θ
 = − +  
= +
=
 ∴ = = =   ∫
/ 2
2
0
3 
2
a
π
=
∫
 
( )/ 2 30
/2
2 5 2
0
and cos 3 cos sin
1 33 cos
5 5
ELx dL a a d
a a
π
π
θ θ θ θ
θ
=
 = − =  
∫ ∫
 
Hence 23 3 :
2 5EL
xL x dL x a a = =  ∫ 
2
5
x a= W 
 
 
 
PROBLEM 5.44 CONTINUED 
Alternative solution 
 
2/3
3 2
2/3
3 2
cos cos
sin sin
xx a
a
yy a
a
θ θ
θ θ
 = ⇒ =   
 = ⇒ =   
 
( )
( ) ( )
2/3 2/3 3/22/3 2/3
1/22/3 2/3 1/3
 1 or 
Then 
x y y a x
a a
dy a x x
dx
−
   ∴ + = = −      
= − −
 
( ) ( ) 1/22 21/22/3 2/3 1/3
Now 
and 1 1
ELx x
dydL dx a x x dx
dx
−
=
     = + = + − −         
 
1/3
1/3 2/3
1/ 30
0
3 3Then 
2 2
a
a aL dL dx a x a
x
 = = = =  ∫ ∫ 
and 
1/3
1/3 5/3 2
1/30
0
3 3
5 5
a
a
EL
ax dL x dx a x a
x
   = = =      ∫ ∫ 
Hence 23 3 :
2 5EL
xL x dL x a a = =  ∫ 
2
5
x a= W 
 
 
 
 
PROBLEM 5.45 
Determine by direct integration the centroid of the area shown. 
 
SOLUTION 
 
 
2 2Have cos cos
3 3
2 2sin sin
3 3
EL
EL
x r ae
y r ae
θ
θ
θ θ
θ θ
= =
= =
 
( )( ) 2 21 1and 
2 2
dA r rd a e dθθ θ= = 
Then
( )2 2 2 2 2 2 20
0
1 1 1 1 1 133.623
2 2 2 4
A dA a e d a e a e a
ππ θ θ πθ  = = = = − =  ∫ ∫ 
and 2 2 3 30 0
2 1 1cos cos
3 2 3EL
x dA ae a e d a e dπ πθ θ θθ θ θ θ = =  ∫ ∫ ∫ 
To proceed, use integration by parts, with 
3 3 and 3 
cos and sin
u e du e d
dv d v
θ θ θ
θ θ θ
= =
= = 
Then ( )3 3 3cos sin sin 3e d e e dθ θ θθ θ θ θ θ= −∫ ∫ 
Now let 3 3 then 3u e du e dθ θ θ= = 
 sin , then cosdv d vθ θ θ= = − 
Then ( )( )3 3 3 3sin sin 3 cos cos 3e d e e e dθ θ θ θθ θ θ θ θ θ− = − − − − ∫ ∫ 
So that ( )33 cos sin 3cos
10
ee d
θθ θ θ θ θ= +∫ 
( ) ( )3 33 3 3
0
1 sin 3cos 3 3 1239.26
3 10 30EL
e ax dA a e a
πθ πθ θ ∴ = + = − − = −  ∫ 
Also 2 2 3 30 0
2 1 1sin sin
3 2 3EL
y dA ae a e d a e dπ πθ θ θθ θ θ θ = =  ∫ ∫ ∫ 
 
 
 
 
PROBLEM 5.45 CONTINUED 
Using integration by parts, as above, with 
 3 3 and 3u e du e dθ θ θ= = 
 sin and cosdv d vθ θ θ= = −∫ 
Then ( ) ( )3 3 3sin cos cos 3e d e e dθ θ θθ θ θ θ θ= − − −∫ ∫ 
So that ( )33 sin cos 3sin
10
ee d
θθ θ θ θ θ= − +∫ 
( ) ( )3 33 3 3
0
1 cos 3sin 1 413.09
3 10 30EL
e ay dA a e a
πθ πθ θ ∴ = − + = + =  ∫ 
Hence ( )2 3: 133.623 1239.26ELxA x dA x a a= = −∫ or 9.27x a= − W 
 ( )2 3: 133.623 413.09ELyA y dA y a a= =∫ or 3.09y a= W 
 
 
 
 
 
PROBLEM 5.46 
Determine by direct integration the centroid of the area shown. 
 
SOLUTION 
 
Have 
1, sin
2EL EL
xx x y x
L
π= = 
and dA ydx= 
/22 2
/2
2 20
0
sin sin cos
L
L x L x L x LA dA x dx x
L L L
π π π
ππ π
 = = = − =  ∫ ∫ 
and 
/2
0 sin
L
EL
xx x dA x x dx
L
π = =   ∫ ∫ 
/22 3 3 3
2
2 3 2 3
0
2 2sin cos sin 2
L
L x L x L x L Lx x
L L L
π π π
ππ π π π
      = + − = −             
Also 
/2
0
1 sin sin
2
L
EL
x xy y dA x x dx
L L
π π = =   ∫ ∫ 
 
/22 3
2 3
0
1 2 2sin cos
2
L
L x L L xx x
L L
π π
ππ π
  = − −       
 ( ) ( )3 2 3 22 21 1 1 62 6 8 24 96L L L L ππ π    = − − = +          
 
 
 
 PROBLEM 5.46 CONTINUED 
Hence 
2
3
2 2 3
1:EL
L zxA x dA x Lπ π π
   = = −       ∫ 
 or 0.363x L= W 
 
2 3
2 2 2 3
1 2:
96EL
L LyA y dA y π π π π
   = = −       ∫ 
 or 0.1653y L= W 
 
 
PROBLEM 5.47 
Determine the volume and the surface area of the solid obtained by 
rotating the area of Problem. 5.2 about (a) the x axis, (b) the line 
165x = mm. 
 
SOLUTION 
 
From the solution to Problem 5.2: 
( )2 area area10 125 mm , 96.4 mm, 34.7 mm AreaA X Y= = = 
From the solution to Problem 5.22: 
( )line line441.05 mm 92.2 mm, 32.4 mm LineL X Y= = = 
Applying the theorems of Pappus-Guldinus, we have 
(a) Rotation about the x axis: 
( )( ) 3 2lineArea 2 2 32.4 mm 441.05 mm 89.786 10 mmY Lπ π= = = × 
 
3 289.8 10 mmA = × W 
( )( ) 6 3areaVolume 2 2 34.7 mm 10125 mm 2.2075 10 mmY Aπ π= = = × 
 
6 32.21 10 mmV = × W 
(b) Rotation about 165 mm:x = 
( ) ( ) ( ) 5 2lineArea 2 165 2 165 92.2 mm 441.05 mm 2.01774 10 mmX Lπ π  = − = − = ×  
 
6 20.202 10 mmA = × W 
( ) ( ) ( ) 6 3areaVolume 2 165 2 165 96.4 mm 10 125 mm 4.3641 10 mmX Aπ π  = − = − = ×  
 
6 34.36 10 mmV = × W 
 
 
 
PROBLEM 5.48 
Determine the volume and the surface area of the solid obtained by 
rotating the area of Problem 5.4 about (a) the line 22y = mm, (b) the 
line 12x = mm. 
 
SOLUTION 
 
From the solution to Problem 5.4: 
2
area area399 mm , 1.421 mm, 12.42 mm (Area)A X Y= = = 
From the solution to Problem 5.23: 
line line77.233 mm, 1.441 mm, 12.72 mm (Line)L X Y= = = 
Applying the theorems of Pappus-Guldinus, we have 
(a) Rotation about the line 22 mm:y = 
( ) ( ) ( ) 2lineArea 2 22 2 22 12.72 mm 77.233 mm 4503 mmY Lπ π  = − = − =  
 
3 24.50 10 mmA = × W 
( ) ( ) ( )2 3areaVolume 2 22 2 22 12.42 mm 399 mm 24 016.97 mmY Aπ π  = − = − =  
 
3 324.0 10 mmV = × W 
(b) Rotation about line 12 mm:x = 
( ) ( ) ( ) 2lineArea 2 12 2 12 1.441 mm 77.233 mm 5124.45 mmX Lπ π  = − = − =  
 
3 25.12 10 mmA = × W 
( ) ( ) ( )2 3Volume 2 12 1.421 2 12 1.421 mm 399 mm 26 521.46 mmAπ π  = − = − =  
 
3 326.5 10 mmV = × W 
 PROBLEM 5.49 
Determine the volume and the surface area of the solid obtained by 
rotating the area of Problem 5.1 about (a) the x axis, (b) the line 
16 in.x = 
 
SOLUTION 
 
From the solution to Problem 5.1: 
2
area area240 in , 5.60 in., 6.60 in. (Area)A X Y= = = 
From the solution to Problem 5.21: 
line line72 in., 4.67 in., 6.67 in.L X Y= = = 
Applying the theorems of Pappus-Guldinus, we have 
( ) Rotation about the axis:a x 
( )( ) 2line2 2 6.67 in. 72 in. 3017.4 inxA Y Lπ π= = = 
 
23020 inA = W 
( )( )2 3area2 2 6.60 in. 240 in 9952.6 inxV Y Aπ π= = = 
 
39950 inV = W 
( ) Rotation about 16 in.:b x = 
( ) ( ) ( ) 216 line2 16 2 16 4.67 in. 72 in. 5125.6 inxA X Lπ π=  = − = − =  
 
2
16 5130 inxA = = W 
( ) ( ) ( )2 316 area2 16 2 16 5.60 in. 240 in 15 682.8 inxV X Aπ π=  = − = − =  
 
3 3
16 15.68 10 inxV = = × W 
 
 
 
 
 
PROBLEM 5.50 
Determine the volume of the solid generated by rotating the semielliptical 
area shown about (a) the axis ,AA′ (b) the axis ,BB′ (c) the y axis. 
 
SOLUTION 
 
Applying the second theorem of Pappus-Guldinus, we have 
(a) Rotation about axis :AA′ 
 ( ) 2 2Volume 2 2 2
abyA a a bππ π π = = =   
2 2V a bπ= W 
(b) Rotation about axis :BB′ 
 ( ) 2 2Volume 2 2 2 22
abyA a a bππ π π = = =   
2 22V a bπ= W 
(c) Rotation about y-axis: 
 
24 2Volume 2 2
3 2 3
a abyA a bππ π ππ
  = = =     
22
3
V a bπ= W 
 
 
 
PROBLEM 5.51 
Determine the volume and the surface area of the chain link shown, 
which is made from a 2-in.-diameter bar, if 3R = in. and 10L = in. 
 
SOLUTION 
First note that the area A and the circumference C of the cross section of the bar are 
2 and
4
A d C dπ π= = 
Observe that the semicircular ends of the link can be obtained by rotating the cross section through a horizontal 
semicircular arc of radius R. Then, applying the theorems of Pappus-Guldinus, we have 
( ) ( ) ( ) ( ) ( )side endVolume 2 2 2 2 2V V AL RA L R Aπ π= + = + = + 
 
( ) ( )2
3
2 10 in. 3 in. 2 in.
4
122.049 in
ππ   = +    
=
 
 
3122.0 inV = W 
( ) ( ) ( ) ( ) ( )side endArea 2 2 2 2 2A A CL RC L R Cπ π= + = + = + 
 ( ) ( )2 10 in. 3 in. 4 in.π π   = +    
 
2488.198 in= 
 
2488 inA = W 
 
 
 
PROBLEM 5.52 
Verify that the expressions for the volumes of the first four shapes in 
Figure 5.21 on page 261 are correct. 
 
SOLUTION 
 
 
 
 
 
 
Following the second theorem of Pappus-Guldinus, in each case a 
specific generating area A will be rotated about the x axis to produce the 
given shape. Values of y are from Fig. 5.8A. 
(1) Hemisphere: the generating area is a quarter circle 
Have 
242 2
3 4
aV yA aππ π π
  = =      
 
32or
3
V aπ= W 
(2) Semiellipsoid of revolution: the generating area is a quarter ellipse 
Have 
42 2
3 4
aV yA haππ π π
  = =      
 
22or
3
V a hπ= W 
(3) Paraboloid of revolution: the generating area is a quarter parabola 
Have 
3 22 2
8 3
V yA a ahπ π   = =      
 
21or
2
V a hπ= W 
 
 
(4) Cone: the generating area is a triangle 
Have 
12 2
3 2
aV yA haπ π   = =      
 
21or
3
V a hπ= W 
 
 
 
 
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 12 2 7.5 mm 5 mm 5 mm
3 2
V xAπ π     = = + × × ×         
 
3or 720 mmV = W 
 
 
 
 
 
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 CA of a 
belt is given by 
CA yL yLπ π= = Σ 
Where the individual lengths are the “Lengths” of the belt cross section 
that are in contact with the pulley 
( )
( ) ( )
1 1 2 2Have 2
2.5 2.5 mm2 60 mm
2 cos 20
60 2.5 mm 12.5 mm 
CA y L y Lπ
π
 = + 
      = −     °     
 + −   
 
 
3 2or 3.24 10 mmCA = × W 
Have ( )1 12CA y Lπ  =   
7.5 7.5 mm2 60 1.6 mm
2 cos 20
π     = − − ×         D 
 
3 2or 2.74 10 mmCA = × W 
 
 
 
Have ( ) ( )1 1 2 560 mm 5 mmCA y Lπ π ππ
 × = = − ×     
 
3 2or 2.80 10 mmCA = × W 
 
 
 
 
 
PROBLEM 5.55 
Determine the capacity, in gallons, of the punch bowl shown if 12 in.R = 
 
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 
 
 
( )
( )
1 1 2 2
2
3 3
3
3 3
2 2 2
1 1 1 1 3 2 sin 302 cos30
3 2 2 2 2 63
6
3 32
816 3 2 3
3 3 12 in. 3526.03 in
8
V xA xA x A x A
RR R R R
R R R
π π π
ππ π
π π
π
= = Σ = +
        = × × × +               ×    
 = + =   
= =
D D
 
3Since 1 gal 231 in= 
3
3
3526.03 in 15.26 gal
231 in /gal
V == 
 15.26 galV = W 
 
 
 
 
 
 
 
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 
30.101 lb/in , determine the weight of the shade. 
 
SOLUTION 
 
The weight of the lamp shade is given by 
W V Atγ γ= = 
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 
 ( )1 1 2 2 3 3 4 42 2 2A yL yL y L y L y L y Lπ π π= = Σ = + + + 
 ( ) ( ) ( )2 20.6 mm 0.60 0.752 0.6 mm mm 0.15 mm 1.5 mm2 2π
 + = + × +    
 ( ) ( )2 20.75 1.25 mm 0.50 mm 0.40 mm2
+ + × +   
 
( ) ( )2 2
2
1.25 1.5 mm 0.25 mm 1.25 mm
2
22.5607 in
+ + × +    
=
 
Then 
3 2 3lb/in0.101 22.5607 in in. 0.21362 lb
32
W ×= × = 
 0.214 lbW = W 
 
 
 
 
 
 
 
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 3690 kg/m , determine the weight of the waste 
wood resulting from the production of 5000 tops. 
 
SOLUTION 
All dimensions are in mm 
( ) ( )
waste blank top
2 6 3
blank
top 1 2 3 4
Have
550 mm 30 mm 9.075 10 mm
V V V
V
V V V V V
π π
= −
= × = ×
= + + +
 
Applying the second theorem of Pappus-Guldinus to parts 3 and 4 
( ) ( ) ( ) ( )
( )
( )
( )
2 2
top
2
2
6 3
6 3
529 mm 18 mm 535 mm 12 mm
4 122 535 mm 12 mm
3 4
4 182 529 mm 18 mm
3 4
5.0371 3.347 0.1222 0.2731 10 mm
8.8671 10 mm
V π π
ππ π
ππ π
π
π
   = × + ×      
  ×  + + ×       
  ×  + + ×       
= + + + ×
= ×
 
( ) 6 3waste
3 3
9.0750 8.8671 10 mm
0.2079 10 m
V π
π −
∴ = − ×
= × 
( ) ( )
waste wood waste tops
3 3 3 2
Finally 
690 kg/m 0.2079 10 m 9.81 m/s 5000 tops
W V g Nρ
π −
=
= × × × × 
 wasteor 2.21 kNW = W 
 
 
 
 
 
 
 
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 
top is given three coats of lacquer and that 1 liter of lacquer covers 12 m2. 
 
SOLUTION 
Referring to the figure in solution of Problem 5.57 and using the first theorem of Pappus-Guldinus, we have 
( ) ( )
( )
( )
surface top circle bottom circle edge
2 2
3 2
535 mm 529 mm
2 122 535 mm 12 mm
2
2 182 529 mm 18 mm
2
617.115 10 mm
A A A A
π π
ππ π
ππ π
π
= + +
   = +      
  ×  + + ×       
  ×  + + ×       
= ×
 
surface tops coats
3 2
2
Then # liters Coverage
1 liter617.115 10 m 5000 3
12 m
A N N
π −
= × × ×
= × × × × 
 or # liters 2424 L= W 
 
 
 
 
 
 
 
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 
weight of yellow brass is 30.306 lb/in . determine the weight of the 
escutcheon. 
 
SOLUTION 
The weight of the escutcheon is given by (specific weight)W V= 
where V is the volume of the plate. V can be generated by rotating the area A about the x axis. 
 
Have 3.0755 in. 2.958 in. 0.1175 in.a = − = 
and 
0.5sin 0.16745 R 9.5941
3
φ φ= ⇒ = = ° 
Then 2 26 9.5941 16.4059 or 8.20295 0.143169 radα α= − = = =D D D D 
The area A can be obtained by combining the following four areas, as indicated. 
 
Applying the second theorem of Pappus-Guldinus and then using Figure 5.8A, we have 
2 2V yA yAπ π= = Σ 
 
 
 
 PROBLEM 5.59 CONTINUED 
 2, inA , in.y 
3, inyA 
1 ( )( )1 3.0755 1.5 2.3066
2
= ( )1 1.5 0.53 = 1.1533 
2 ( )23 1.28851α− = − ( ) ( )2 3 sin sin 0.609213
α α φα × + = –0.78497 
3 ( )( )1 2.958 0.5 0.7395
2
− = − ( )1 0.5 0.166673 = –0.12325 
4 ( )( )0.1755 0.5 0.05875− = − ( )1 0.5 0.252 = –0.14688 
 30.44296 inyAΣ = 
Then ( )3 32 0.44296 in 1.4476 inV π= = 
so that ( )3 31.4476 in 0.306 lb/in 0.44296 lbW = = 0.443 lbW = W 
 
 
 
 
 
 
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 axis. Applying 
the first theorem of Pappus-Guldinus, we have 
2A yLπ= 
( )22Now, since , at : 7.5x ky x a a k= = = 
or 56.25a k= (1) 
At ( ) ( )215 mm: 15 12.5x a a k= + + = 
or 15 156.25a k+ = (2) 
Then 
Eq. (2) 15 156.25: or 8.4375 mm
Eq. (1) 56.25
a k a
a k
+ = = 
 
1Eq. (1) 0.15
mm
k⇒ = 
20.15 and 0.3dxx y y
dy
∴ = = 
2
2Now 1 1 0.09dxdL dy y dy
dy
 = + = +   
So 2 andA yL yL ydLπ= = ∫ 
( )
12.5 2
7.5
12.5
3/22
7.5
2 1 0.09
2 1 2 1 0.09
3 0.18
A y y dy
y
π
π
∴ = +
  = +    
∫
 
 
21013 mm= 2or 1013 mmA = W 
 
 
 
 
 
 
 
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 
 
( )( )
( )( )
1 2
1
2
Resultant
( ) Have 40 lb/ft 18 ft 720 lb
1 120 lb/ft 18 ft 1080 lb
2
R R R
a R
R
= +
= =
= =
 
 or 1800 lbR = 
 The resultant is located at the centroid C of the distributed load x 
 Have ( ) ( )( )( ) ( )( )( )1: 1800 lb 40 lb/ft 18 ft 9 ft 120 lb/ft 18 ft 12 ft2AM xΣ = + 
 or 10.80 ftx = 
 
1800 lb
10.80 ft
R
x
=
=
W
 
( )b 0: 0x xF AΣ = = 
 0: 1800 lb 0, 1800 lby y yF A AΣ = − = = 1800 lb∴ =A W 
 ( )( )0: 1800 lb 10.8 ft 0A AM MΣ = − = 
 19.444 lb ftAM = ⋅ or 19.44 kip ftA = ⋅M W 
 
 
 
 
 
 
 
PROBLEM 5.62 
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 
 
( )( )
( )( )
I
II
( ) Have 300 N/m 6 m 1800 N
1 6 m 900 N/m 1800 N
3
a R
R
= =
= = 
 Then I II:yF R R RΣ − = − − 
 or 1800 N 1800 N 3600 NR = + = 
 ( ) ( )( ) ( )( ): 3600 N 3 m 1800 N 4.5 m 1800 NAM xΣ − = − − 
 or 3.75 mx = 
3600 NR = W 
 3.75 mx = 
(b) Reactions 
0: 0x xF AΣ = = 
 ( ) ( )( )0: 6 m 3600 N 3.75 m 0A yM BΣ = − = 
 or 2250 NyB = 2250 N=B W 
 0: 2250 N 3600 Ny yF AΣ = + = 
 or 1350 NyA = 1350 N=A W 
 
 
 
 
 
 
 
PROBLEM 5.63 
Determine the reactions at the beam supports for the given loading. 
 
SOLUTION 
 
( )( )
( )( )
( )( )
I
II
III
Have 100 lb/ft 4 ft 400 lb
1 200 lb/ft 6 ft 600 lb
2
200 lb/ft 4 ft 800 lb
R
R
R
= =
= =
= =
 
Then 0: 0x xF AΣ = = 
 ( )( ) ( )( ) ( )( ) ( )0: 2 ft 400 lb 4 ft 600 lb 12 ft 800 lb 10 ft 0A yM BΣ = − − + = 
or 800 lbyB = 800 lb=B W 
 0: 800 lb 400 lb 600 lb 800 0y yF AΣ = + − − − = 
or 1000 lbyA = 1000 lb=A W 
 
 
 
 
 
 
 
 
PROBLEM 5.64 
Determine the reactions at the beam supports for the given loading. 
 
SOLUTION 
 
( )( )
( )( )
I
II
Have 9 ft 200 lb/ft 1800 lb
1 3 ft 200 lb/ft 300 lb
2
R
R
= =
= = 
Then 0: 0x xF AΣ = = 
 ( )( ) ( )( ) ( )0: 4.5 ft 1800 lb 10 ft 300 lb 9 ft 0A yM BΣ = − − + = 
or 1233.3 lbyB = 1233 lb=B W 
 0: 1800 lb 300 lb 1233.3 lb 0y yF AΣ = − − + = 
or 866.7 lbyA = 867 lb=A WPROBLEM 5.65 
Determine the reactions at the beam supports for the given loading. 
 
SOLUTION 
 
( )( )I 1Have 200 N/m 0.12 m 12 N2R = = 
 ( )( )II 200 N/m 0.2 m 40 NR = = 
Then 0: 0x xF AΣ = = 
 0: 18 N 12 N 40 N 0y yF AΣ = + − − = 
or 34 NyA = 34.0 N=A W 
 ( )( ) ( )( ) ( )( )0: 0.8 m 12 N 0.22 m 40 N 0.38 m 18 NA AM MΣ = − − + 
or 2.92 N mAM = ⋅ 2.92 N mA = ⋅M W 
 
 
 
 
 
 
 
PROBLEM 5.66 
Determine the reactions at the beam supports for the given loading. 
 
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. 
 
( )( )IHave 1.8 m 2000 3600 NN/mR = = 
 ( )( )II 1 1.8 m 4500 N/m 4050 N2R = = 
Then 0: 0x xF AΣ = = 
 ( ) ( )( ) ( )( )0: 3 m 2.1 m 3600 N 2.4 m 4050 NB yM AΣ = − − + 
or 270 NyA = 270 N=A W 
 0: 270 N 3600 N 4050 N 0y yF BΣ = − + − = 
or 720 NyB = 720 N=B W 
 
 
 
 
 
 
 
 
PROBLEM 5.67 
Determine the reactions at the beam supports for the given loading. 
 
SOLUTION 
 
( )( )I 1Have 4 m 2000 kN/m 2667 N3R = = 
 ( )( )II 1 2 m 1000 kN/m 666.7 N3R = = 
Then 0: 0x xF AΣ = = 
 0: 2667 N 666.7 N 0y yF AΣ = − − = 
or 3334 NyA = 3.33 kN=A W 
 ( )( ) ( )( )0: 1 m 2667 N 5.5 m 666.7 NA AM MΣ = − − 
or 6334 N mAM = ⋅ 6.33 kN mA = ⋅M W 
 
 
 
 
 
 
 
PROBLEM 5.68 
Determine the reactions at the beam supports for the given loading. 
 
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. 
 
( )( )IHave 8 ft 100 lb/ft 800 lbR = = 
 ( )( )II 2 8 ft 600 lb/ft 3200 lb3R = = 
Then 0: 0x xF AΣ = = 
 ( )( ) ( )( )0: 11 5 ft 800 lb 4 ft 3200 lb 0AM BΣ = + − = 
or 800 lb=B W 
 0: 3200 lb 800 lb 800 lb 0y yF AΣ = − + + = 
or 1600 lb=A W 
 
 
 
 
 
 
PROBLEM 5.69 
Determine (a) the distance a so that the vertical reactions at supports A 
and B are equal, (b) the corresponding reactions at the supports. 
 
SOLUTION 
 
( )( ) ( )I 1( ) Have ft 120 lb/ft 60 lb2a R a a= = 
 ( )( ) ( )II 1 12 40 lb/ft 240 20 lb2R a a= − = − 
 Then ( )0: 60 240 2 0y y yF A a a BΣ = − − − + = 
 or 240 40y yA B a+ = + 
 Now 120 20y y y yA B A B a= ⇒ = = + (1) 
 Also ( ) ( ) ( ) ( )10: 12 m 60 lb 12 ft 12 ft 240 20 lb 03 3B y
aM A a a a         Σ = − + − + − − =                
 or 
2140 1080
3 9y
A a a= − − (2) 
 Equating Eqs. (1) and (2) 
2140 10120 20 80
3 9
a a a+ = − − 
 or 
240 320 480 0
3
a a− + = 
 Then 1.6077 ft, 22.392a a= = 
 Now 12 fta ≤ 1.608 fta = W 
( ) Haveb 0: 0x xF AΣ = = 
 Eq. (1) ( )120 20 1.61 152.2 lby yA B= = + = 
 152.2 lb= =A B W 
 
 
 
 
 
 
PROBLEM 5.70 
Determine (a) the distance a so that the vertical reaction at support B is 
minimum, (b) the corresponding reactions at the supports. 
 
SOLUTION 
 
( )( )I 1( ) Have ft 120 lb/ft 60 lb2a R a a= = 
 ( ) ( ) ( )II 1 12 ft 40 lb/ft 240 20 lb2R a a = − = −  
 Then ( ) ( ) ( )0: ft 60 lb 240 20 lb 8 ft 12 ft 03 3A y
a aM a a B     Σ = − − − + + =          
 or 
210 20 160
9 3y
B a a= − + (1) 
 Then 
20 20 0
9 3
ydB a
da
= − = or 3.00 fta = W 
( ) Eq. (1)b ( ) ( )210 203.00 3.00 1609 3yB = − + 
 150 lb= 150.0 lb=B W 
 and 0: 0x xF AΣ = = 
 ( ) ( )0: 60 3.00 lb 240 20 3.00 lb 150 lb 0y yF A    Σ = − − − + =    
 or 210 lbyA = 210 lb=A W 
 
 
 
 
 
 
 
PROBLEM 5.71 
Determine the reactions at the beam supports for the given loading when 
0 1.5 kN/m.w = 
 
 
SOLUTION 
 
( )( )I 1Have 9 m 2 kN/m 9 kN2R = = 
 ( )( )II 9 m 1.5 kN/m 13.5 kNR = = 
Then 0: 0x xF CΣ = = 
 ( )( ) ( )( ) ( )0: 50 kN m 1 m 9 kN 2.5 m 13.5 kN 6 m 0B yM CΣ = − ⋅ − − + = 
or 15.4583 kNyC = 15.46 kN=C W 
 0: 9 kN 13.5 kN 15.4583 0y yF BΣ = − − + = 
or 7.0417 kNyB = 7.04 kN=B W 
 
 
 
 
 
 
 
PROBLEM 5.72 
Determine (a) the distributed load 0w at the end D of the beam ABCD 
for which the reaction at B is zero, (b) the corresponding reactions at C. 
 
SOLUTION 
 
( ) ( ) ( )I 0 01Have 9 m 3.5 kN/m 4.5 3.5 kN2R w w = − = −  
 ( )( )II 0 09 m kN/m 9 kNR w w= = 
( ) Thena ( ) ( ) ( )( )0 00: 50 kN m 5 m 4.5 3.5 kN 3.5 m 9 kN 0CM w w Σ = − ⋅ + − + =  
 or 09 28.75 0w + = 
 so 0 3.1944 kN/mw = − 0 3.19 kN/mw = W 
 Note: the negative sign means that the distributed force 0w is upward. 
( )b 0: 0x xF CΣ = = 
 ( ) ( )0: 4.5 3.5 3.19 kN 9 3.19 kN 0y yF CΣ = − + + + = 
 or 1.375 kNyC = 1.375 kN=C W 
 
 
 
 
 
 
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 RR as shown. Knowing that P = 4 kN 
and 12 ,B Aw w= determine the values of wA and RR corresponding to 
equilibrium. 
 
SOLUTION 
 
( )( )IHave 1.2 m kN/m 1.2 kNA AR w w= = 
 ( )II 1 11.8 m kN/m 0.45 kN2 2 A AR w w = =   
 ( )III 11.8 m kN/m 0.9 kN2 A AR w w = =   
Then ( ) ( ) ( ) ( )0: 0.6 m 1.2 kN 0.6 m 0.45 kN/mC A AM w w   Σ = − +    
 ( ) ( ) ( )( )0.9 m 0.9 kN/m 1.2 m 4 kN/mAw + −  
 ( )( ) ( )( )0.8 m 18 kN/m 0.7 m 24 kN/m 0− + = 
or 6.667 kN/mAw = 6.67 kN/mAw = W 
and ( )( ) ( )( )0: 1.2 m 6.67 kN/m 0.45 m 6.67 kN/my RF RΣ = + + 
 ( )( )0.9 m 6.67 kN/m 24 kN 18 kN 4 kN+ − − − 
or 29.0 kNRR = 29.0 kNRR = W 
 
 
 
 
 
 
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 RR as shown. Knowing that wB = 
0.4wA, determine (a) the largest value of P for which the beam is in 
equilibrium, (b) the corresponding value of wA. 
In the following problems, use γ = 62.4 lb/ft3 for the specific weight of 
fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S. 
customary units are used. With SI units, use ρ = 103 kg/m3 for the 
density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of 
concrete. (See the footnote on page 222 for how to determine the specific 
weight of a material given its density.) 
j 
SOLUTION 
 ( )( )IHave 1.2 m kN/m 1.2 kNA AR w w= = 
 ( )( )II 1 1.8 m 0.6 kN/m 0.54 kN2 A AR w w= = 
 ( )( )III 1.8 m 0.4 kN/m 0.72 kNA AR w w= = 
( ) Thena ( ) ( ) ( ) ( ) ( )0: 0.6 m 1.2 kN 1.2 m 1.8 m 0.54 kNA A R AM w R w   Σ = + +    
 ( ) ( ) ( )( )2.1 m 0.72 kN 0.5 m 24 kNAw + −  
 ( )( ) ( )2.0 m 18 kN 2.4 m 0P− + = 
 or 3.204 1.2 2.4 48A Rw R P+ − = (1) 
 and 0: 1.2 0.54 0.72 24 18 0y R A A AF R W W W PΣ = + + + − − − = 
 or 2.46 42R AR W P+ − = (2) 
 Now combine Eqs. (1) and (2) to eliminate :Aw 
( ) ( )3.204 Eq. 2 2.46 Eq. 1 0.252 16.488 2.7RR P− ⇒ = − 
 Since RR must be 0,≥ the maximum acceptable value of P is that for which 0,R = 
 or 6.1067 kNP = 6.11 kNP = W 
( ) Then, from Eq. (2):b 2.46 6.1067 42AW − = or 19.56 kN/mAW = W 
 
 
 
 
 
 
PROBLEM 5.75 
The cross section of a concrete dam is as shown. For a dam section of 
unit width, determine (a) the reaction forces exerted by the ground on the 
base AB of the dam, (b) the point of application of the resultant of the 
reaction forces of part a, (c) the resultant of the pressure forces exerted 
by the water on the face BC of the dam. 
In the following problems, use γ = 62.4 lb/ft3for the specific weight of 
fresh water and γc = 150 lb/ft3 for the specific weight of concrete if U.S. 
customary units are used. With SI units, use ρ = 103 kg/m3 for the 
density of fresh water and ρc = 2.40 × 103 kg/m3 for the density of 
concrete. (See the footnote on page 222 for how to determine the specific 
weight of a material given its density.) 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
The free body shown consists of a 1-m thick section of the dam and the 
triangular section BCD of the water behind the dam. 
Note: 1 6 mX = 
 ( )2 9 3 m 12 mX = + = 
 ( )3 15 2 m 17 mX = + = 
 ( )4 15 4 m 19 mX = + = 
( ) Now so thata W gVρ= 
 ( )( ) ( )( )( )3 21 12400 kg/m 9.81 m/s 9 m 15 m 1 m 1589 kN2W  = =   
 ( )( ) ( )( )( )3 22 2400 kg/m 9.81 m/s 6 m 18 m 1 m 2543 kNW  = =  
 ( )( ) ( )( )( )3 23 12400 kg/m 9.81 m/s 6 m 18 m 1 m 1271 kN2W  = =   
 ( )( ) ( )( )( )3 24 12400 kg/m 9.81 m/s 6 m 18 m 1 m 529.7 kN2W  = =   
 ( )( ) ( )( )( )3 3 21 1Also 18 m 1 m 10 kg/m 9.81 m/s 18 m2 2P Ap   = =     
 1589 kN= 
 Then 0: 1589 kN 0xF HΣ = − = 
 or 1589 kNH = 1589 kN=H W 
 0: 1589 kN 2543 kN 1271 kN 529.7 kNyF VΣ = − − − − 
 or 5933 kNV = 5.93 MN=V W 
 
 
 
 
 
 
 
 
 
 
 
PROBLEM 5.75 CONTINUED 
( ) Haveb ( ) ( )( )0: 5933 kN 6 m 1589 kNAM XΣ = + 
 
( )( ) ( )( )
( )( ) ( )( )
6 m 1589 kN 12 m 2543 kN
17 m 1271 kN 19 m 529.7 0
− −
− − = 
 or 10.48 mX = 10.48 mX = W 
 to the right of A 
(c) Consider water section BCD as the free body. 
 Have 0Σ =F 
 
 Then 1675 kN− =R 18.43° 
or 1675 kN=R 18.43°W 
 
Alternative solution to part (c) 
Consider the face BC of the dam. 
2 2Have 6 18 18.9737 mBC = + = 
 
6tan 18.43
18
θ θ= = ° 
( ) ( )( )( )3 2
2
and 1000 kg/m 9.81 m/s 18 m
176.6 kN/m
p g hρ= =
= 
( )( ) ( )21 1Then 18.97 m 1 m 176.6 kN/m2 2
1675 kN
R Ap  = =  
= 
1675 kN∴ =R 18.43° 
 
 
 
 
 
 
 
PROBLEM 5.76 
The cross section of a concrete dam is as shown. For a dam section of 
unit width, determine (a) the reaction forces exerted by the ground on the 
base AB of the dam, (b) the point of application of the resultant of the 
reaction forces of part a, (c) the resultant of the pressure forces exerted 
by the water on the face BC of the dam. 
 
SOLUTION 
 
The free body shown consists of a 1-ft thick section of the dam and the parabolic section of water above (and 
behind) the dam. 
Note ( )1 5 16 ft 10 ft8x = = 
( )2 116 6 ft 19 ft2x  = + =   
 ( )3 122 12 ft 25 ft4x  = + =   
 ( )4 522 12 ft 29.5 ft8x  = + =   
 
 
 PROBLEM 5.76 CONTINUED 
Now W Vγ= 
 ( ) ( )( ) ( )31 2150 lb/ft 16 ft 24 ft 1 ft 38,400 lb3W  = × =   
( ) ( )( ) ( )32 150 lb/ft 6 ft 24 ft 1 ft 21,600 lbW  = × =  
 ( ) ( )( ) ( )33 1150 lb/ft 12 ft 18 ft 1 ft 10,800 lb3W  = × =   
 ( ) ( )( ) ( )34 262.4 lb/ft 12 ft 18 ft 1 ft 8985.6 lb3W  = × =   
( ) ( )2 31 1Also 18 1 ft 62.4 lb/ft 18 ft 10,108.8 lb2 2P Ap  = = × × × =  
( ) Thena 0: 10,108.8 lb 0xF HΣ = − = 
 or 10.11 kips=H W 
 0: 38,400 lb 21, 600 lb 10,800 lb 8995.6 lb 0yF VΣ = − − − − = 
 or 79,785.6V = 79.8 kips=V W 
( )b ( ) ( )( ) ( )( ) ( )( )0: 79,785.6 lb 6 ft 38,400 lb 19 ft 21,600 lb 25 ft 10,800 lbAM XΣ = − − − 
 ( )( ) ( )( )29.5 ft 8985.6 lb 6 ft 10,108.8 lb 0− + = 
 or 15.90 ftX = 
 The point of application of the resultant is 15.90 ft to the right of A W 
(c) Consider the water section BCD as the free body. 
 Have 0Σ =F 
 
13.53 kips
41.6
R
θ
∴ =
= ° 
 On the face BD of the dam 
 13.53 kips=R 41.6°W 
 
 
 
 
 
 
PROBLEM 5.77 
The 9 12-ft× side AB of a tank is hinged at its bottom A and is held in 
place by a thin rod BC. The maximum tensile force the rod can withstand 
without breaking is 40 kips, and the design specifications require the 
force in the rod not exceed 20 percent of this value. If the tank is slowly 
filled with water, determine the maximum allowable depth of water d in 
the tank. 
 
SOLUTION 
 
Consider the free-body diagram of the side. 
( )1 1Have
2 2
P Ap A dγ= = 
Now ( )0: 9 ft 03A
dM T PΣ = − = 
maxThen, for :d 
( ) ( )( ) ( )( ) ( )3 3max max max19 ft 0.2 40 10 lb 12 ft 62.4 lb/ft 03 2d d d    × − =      
or 
3 3 3
max216 10 ft 374.4 d× = 
or 
3 3
max 576.92 ftd = max 8.32 ftd = W 
 
 
 
 
 
 
 
PROBLEM 5.78 
The 9 12-ft× side of an open tank is hinged at its bottom A and is held 
in place by a thin rod. The tank is filled with glycerine, whose specific 
weight is 380 lb/ft . Determine the force T in the rod and the reactions at 
the hinge after the tank is filled to a depth of 8 ft. 
 
SOLUTION 
 
Consider the free-body diagram of the side. 
( )1 1Have
2 2
P Ap A dγ= = 
 ( )( ) ( )( )31 8 ft 12 ft 80 lb/ft 8 ft 30,720 lb2  = =  
Then 0: 0y yF AΣ = = 
 ( ) ( )80: 9 ft ft 30,720 lb 03AM T  Σ = − =   
or 9102.22 lbT = 
 9.10 kips=T W 
0: 30,720 lb 9102.22 lb 0x xF AΣ = + − = 
or 21,618 lbA = − 
 21.6 kips=A W 
 
 
 
 
 
 
 
PROBLEM 5.79 
The friction force between a 2 2-m× square sluice gate AB and its 
guides is equal to 10 percent of the resultant of the pressure forces 
exerted by the water on the face of the gate. Determine the initial force 
needed to lift the gate that its mass is 500 kg. 
 
SOLUTION 
 
Consider the free-body diagram of the gate. 
( ) ( )( )( )2 3 3 2I I1 1Now 2 2 m 10 kg/m 9.81 m/s 3 m2 2P Ap   = = ×    
 58.86 kN= 
( ) ( )( )( )2 3 3 2II II1 1 2 2 m 10 kg/m 9.81 m/s 5 m2 2P Ap   = = ×    
 98.10 kN= 
( )I IIThen 0.1 0.1F P P P= = + 
 ( )0.1 58.86 98.10 kN= + 
 15.696 kN= 
Finally ( )( )20: 15.696 kN 500 kg 9.81 m/s 0yF TΣ = − − = 
 or 20.6 kN=T W 
 
 
 
 
 
 
PROBLEM 5.80 
The dam for a lake is designed to withstand the additional force caused 
by silt which has settled on the lake bottom. Assuming that silt is 
equivalent to a liquid of density 3 31.76 10 kg/msρ = × and considering 
a 1-m-wide section of dam, determine the percentage increase in the 
force acting on the dam face for a silt accumulation of depth 1.5 m. 
 
SOLUTION 
 
 
 
 
 
 
First, determine the force on the dam face without the silt. 
Have ( )1 12 2w wP Ap A ghρ= = 
 ( )( ) ( )( )( )3 3 21 6 m 1 m 10 kg/m 9.81 m/s 6 m2   =     
 176.58 kN= 
Next, determine the force on the dam face with silt. 
( )( ) ( )( )( )3 3 21Have 4.5 m 1m 10 kg/m 9.81 m/s 4.5 m2wP  ′  =     
 99.326 kN= 
 ( ) ( )( ) ( )( )( )3 3 2I 1.5 m 1 m 10 kg/m 9.81 m/s 4.5 msP   =     
 66.218 kN= 
 ( ) ( )( ) ( )( )( )3 3 2II 1 1.5 m 1 m 1.76 10 kg/m 9.81 m/s 1.5 m2sP   = ×    
 19.424 kN= 
Then ( ) ( )I II 184.97 kNw s sP P P P′ ′= + + = 
The percentage increase, % inc., is then given by 
( )184.97 176.58% inc. 100% 100% 4.7503%
176.58
w
w
P P
P
−′ −= × = × = 
% inc. 4.75%= W 
 
 
 
 
 
 
 
 
PROBLEM 5.81 
The base of a dam for a lake is designed to resist up to 150 percent of the 
horizontal force of the water. After construction, it is found that silt 
(which is equivalent to a liquid of density 3 31.76 10 kg/msρ = × ) is 
settling on the lake bottom at a rate of 20 mm/y. Considering a 1-m-wide 
section of dam, determine the number of years until the dam becomes 
unsafe. 
 
SOLUTION 
From Problem 5.80, the force on the dam face before the silt is deposited, is 176.58 kN.wP = The maximum 
allowable force allowP on the dam is then: 
( )( )allow 1.5 1.5 176.58 kN 264.87 kNwP P= = = 
Next determinethe force P′ on the dam face after a depth d of silt has settled. 
 
Have ( ) ( ) ( )( )( )3 3 21 6 m 1 m 10 kg/m 9.81 m/s 6 m2wP d d ′  = − × −    
 ( )24.905 6 kNd= − 
 ( ) ( ) ( )( )( )3 3 2I 1 m 10 kg/m 9.81 m/s 6 msP d d  = −    
 ( )29.81 6 kNd d= − 
 ( ) ( ) ( )( )( )3 3 2II 1 1 m 1.76 10 kg/m 9.81 m/s m2sP d d  = ×    
 
28.6328 kNd= 
( ) ( ) ( ) ( )2 2 2I II
2
4.905 36 12 9.81 6 8.6328 kN
3.7278 176.58 kN
w s sP P P P d d d d d
d
 ′ ′= + + = − + + − + 
 = + 
 
 
 
 
 PROBLEM 5.81 CONTINUED 
Now required that allowP P′ = to determine the maximum value of d. 
( )23.7278 176.58 kN 264.87 kNd∴ + = 
or 4.8667 md = 
Finally 
3 m4.8667 m 20 10
year
N−= × × 
or 243 yearsN = W 
 
 
 
 
 
 
 
PROBLEM 5.82 
The square gate AB is held in the position shown by hinges along its top 
edge A and by a shear pin at B. For a depth of water 3.5d = m, 
determine the force exerted on the gate by the shear pin. 
 
SOLUTION 
 
First consider the force of the water on the gate. 
Have ( )1 12 2P Ap A ghρ= = 
Then ( ) ( )( )( )2 3 3 2I 1 18 m 10 kg/m 9.81 m/s 1.7 m2P = 
 26.99 kN= 
 ( ) ( )( )( )2 3 3 2II 1 18 m 10 kg/m 9.81 m/s 1.7 1.8cos30 m2P = × ° 
 51.74 kN= 
Now ( ) ( )I II1 20: 03 3A AB AB AB BM L P L P L FΣ = + − = 
( ) ( )1 2or 26.99 kN 51.74 kN 0
3 3 B
F+ − = 
or 43.49 kNBF = 
 4.35 kNB =F 30.0° W 
 
 
 
 
 
 
 
PROBLEMS 5.83 AND 5.84 
Problem 5.83: A temporary dam is constructed in a 5-ft-wide fresh 
water channel by nailing two boards to pilings located at the sides of the 
channel and propping a third board AB against the pilings and the floor of 
the channel. Neglecting friction, determine the reactions at A and B when 
rope BC is slack. 
Problem 5.84: A temporary dam is constructed in a 5-ft-wide fresh 
water channel by nailing two boards to pilings located at the sides of the 
channel and propping a third board AB against the pilings and the floor of 
the channel. Neglecting friction, determine the magnitude and direction 
of the minimum tension required in rope BC to move board AB. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
First, consider the force of the water on the gate. 
Have ( )1 12 2P Ap A hγ= = 
( )( ) ( )( )3I 1So that 1.5 ft 5 ft 62.4 lb/ft 1.8 ft2P   =     
 421.2 lb= 
 ( )( ) ( )( )3II 1 1.5 ft 5 ft 62.4 lb/ft 3 ft2P   =     
 702 lb= 
5.83 Find the reactions at A and B when rope is slack. 
 ( ) ( )( ) ( )( )0: 0.9 ft 0.5 ft 421.2 lb 1.0 ft 702 lb 0AM BΣ = − + + = 
or 1014 lbB = 
1014 lb=B W 
 ( ) ( )4 40: 2 421.2 lb 702 lb 0
5 5x x
F AΣ = + + = 
or 449.28 lbxA = − 
Note that the factor 2 (2 )xA is included since xA is the horizontal force 
exerted by the board on each piling. 
 ( ) ( )3 30: 1014 lb 421.2 lb 702 lb 05 5y yF AΣ = − − + = 
or 340.08 lbyA = − 
563 lb∴ =A 37.1°W 
 
 
 
 
 
 
 
 
 
 
 
 
PROBLEMS 5.83 AND 5.84 CONTINUED 
5.84 Note that there are two ways to move the board: 
 1. Pull upward on the rope fastened at B so that the board 
 rotates about A. For this case 0→B and BCT is 
 perpendicular to AB for minimum tension. 
 2. Pull horizontally at B so that the edge B of the board moves 
 to the left. For this case 0yA → and the board remains 
 against the pilings because of the force of the water. 
 
Case (1) ( )( )0: 1.5 0.5 ft 421.2 lbA BCM TΣ = − + 
 ( )( )1.0 ft 702 lb 0+ = 
or 608.4 lbBCT = 
 
 
Case (2) ( )( ) ( )( )0: 1.2 ft 2 0.5 ft 702 lbB xM AΣ = − − 
 ( )( )1.0 ft 421.2 lb 0− = 
or 2 643.5 lbxA = − 
( )40: 643.5 lb 421.2 lb
5x BC
F TΣ = − − + 
 ( )4 702 lb 05+ = 
or 255.06 lbBCT = 
( )min 255 lbBC∴ =T W 
 
 
 
 
 
 
 
 
PROBLEMS 5.85 AND 5.86 
Problem 5.85: A 2 3-m× gate is hinged at A and is held in position by 
rod CD. End D rests against a spring whose constant is 12 kN/m. The 
spring is undeformed when the gate is vertical. Assuming that the force 
exerted by rod CD on the gate remains horizontal, determine the 
minimum depth of water d for which the bottom B of the gate will move 
to the end of the cylindrical portion of the floor. 
Problem 5.86: Solve Problem 5.85 if the mass of the gate is 500 kg. 
 
SOLUTION 
 
 
 
 
First, determine the forces exerted on the gate by the spring and the 
water when B is at the end of the cylindrical portion of the floor. 
Have 
1sin 30
2
θ θ= ∴ = ° 
Then ( )1.5 m tan 30spx = ° 
and sp spF kx= 
 ( )( )12 kN/m 1.5 m tan 30= ° 
 10.39 kN= 
Assume 2 md ≥ 
Have ( )1 12 2P Ap A g hρ= = 
Then ( )( ) ( )( )( )3 3 2I 1 2 m 3 m 10 kg/m 9.81 m/s 2 m2P d  = −    
 ( )29.43 2 kNd= − 
( )( ) ( )( )( )3 3 2II 1 2 m 3 m 10 kg/m 9.81 m/s 2 2cos30 m2P d  = − + °    
 ( )29.43 0.2679 kNd= − 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
PROBLEMS 5.85 AND 5.86 CONTINUED 
5.85 Find mind so that gate opens, 0.W = 
Using the above free-body diagrams of the gate, we have 
 ( )20: m 29.43 2 kN3AM d   Σ = −     
 ( )4 m 29.43 0.2679 kN3 d   + −     
 ( )( )1.5 m 10.39 kN 0− = 
or ( ) ( )19.62 2 39.24 0.2679 15.585d d− + − = 
58.86 65.3374d = 
or 1.1105 md = 1.110 md = W 
5.86 Find mind so that the gate opens. 
( )( )29.81 m/s 500 kg 4.905 kNW = = 
Using the above free-body diagrams of the gate, we have 
 ( )20: m 29.43 2 kN3AM d   Σ = −     
 ( )4 m 29.43 0.2679 kN3 d   + −     
 ( )( )1.5 m 10.39 kN− + 
 ( )( )0.5 m 4.905 kN 0− = 
or ( ) ( )19.62 2 39.24 0.2679 18.0375d d− + − = 
or 1.15171 md = 1.152 md = W 
 
 
 
 
 
 
 
PROBLEMS 5.87 AND 5.88 
Problem 5.87: The gate at the end of a 3-ft-wide fresh water channel is 
fabricated from three 240-lb, rectangular steel plates. The gate is hinged 
at A and rests against a frictionless support at D. Knowing that d = 2.5 ft, 
determine the reactions at A and D. 
Problem 5.88: The gate at the end of a 3-ft-wide fresh water channel is 
fabricated from three 240-lb, rectangular steel plates. The gate is hinged 
at A and rests against a frictionless support at D. Determine the depth of 
water d for which the gate will open. 
 
SOLUTION 
 
5.87 Note that in addition to the weights of the gate segments, the water exerts pressure on all submerged 
 surfaces ( ).p hγ= 
( )( )3 20.5Thus, at 0.5 ft, 62.4 lb/ft 0.5 ft 31.2 lb/fth p= = = 
 ( )( )3 22.52.5 ft, 6.24 lb/ft 2.5 ft 156.0 lb/fth p= = = 
( )( ) ( )21 1Then 0.5 ft 3 ft 31.2 lb/ft 23.4 lb2P  = =  
 ( )( ) ( )22 2 ft 3 ft 31.2 lb/ft 187.2 lbP  = =  
 ( )( ) ( )23 1 2 ft 3 ft 31.2 lb/ft 93.6 lb2P  = =  
 ( )( ) ( )24 1 2 ft 3 ft 156 lb/ft 468 lb2P  = =  
and ( )( ) ( )( ) ( ) ( )( )10: 4 2 ft 240 lb 1 ft 240 lb 2 0.5 ft 23.4 lb 1 ft 187.2 lb3AM D
  Σ = − + + − + × −     
 ( )( ) ( )( )2 12 ft 93.6 lb 2 ft 468 lb 03 3− − = 
or 11.325 lbD = 11.33 lb∴ =D W 
 
 
 
 PROBLEMS 5.87 AND 5.88 CONTINUED 
0: 11.32 23.4 93.6 468 0x xF AΣ = + + + + = 
or 596.32 lbxA = − 
 0: 240 240 240 187.2 0y yF AΣ = − − − + = 
or 532.8 lbyA = 800 lb∴ =A 41.8°W 
5.88 At ( ) ( ) 2 322 ft, 2 lb/ft where 62.4 lb/ftdh d p dγ γ−= − = − = 
 ( ) 2ft, lb/ftdh d p dγ= = 
( ) ( ) ( ) ( )231 1 21 1 3Then 2 ft 3 ft lb/ft 2 ft 2 lb2 2 2dP A p d d dγ γ−   = = − × − = −    
(Note: For simplicity, the numerical value of the densityγ will be substituted into the equilibrium equations 
 below, rather than at this level of the calculations.) 
( )( ) ( ) ( )2 2 2 2 ft 3 ft 2 ft 6 2 lbdP A p d dγ γ−    = = − = −    
( )( ) ( ) ( )3 3 21 1 2 ft 3 ft 2 ft 3 2 lb2 2dP A p d dγ γ−