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Unformatted text preview: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 52* Locate the center of mass
for the three particles shown
in Fig. P5—2 if mA = 26 kg, m8 = 21 kg, and mC = 36 kg. SOLUTION + mC = 26 + 21 + 36 = 83 kg IIIAXA + meB + 111ch = 26(200 cos 60°) + 21(200) cos 30°) + 36(200) = 962.69 kgmm ‘+
mAyA mByB + mcyc 26(200 sin 60°) + 21(—200 sin 30°) + 36(0) = 2403.33 kgmm giggégg = 28.956 mm 3 29.0 mm '29] ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 54 Locate the center of mass
for the four particles shown
in Fig. P54 if mA = 16 kg, = 24 kg, mC = 14 kg, and
= 36 kg. SOLUTION 16 + 24 + 14 + 36 = 90 k2 +
[BAXA meB + mcxc + meD 16(300) + 24(0) + 14(0) + 36(0) = 4800 kgmm + +
m y mByB + m V mDy A A C‘C D 16(0) + 24(0) + 14(500) + 36(0) 7000 kg~mm + +
mAzA mBzB + chC szD = 16(0) + 24(0) + 14(0) + 36(400) 14,400 kg'mm 273 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Locate the center of mass
for the five particles
shown in Fig. P5—6 if mA = 2 kg, mB = 3 kg, and :11E 2 kg. #SOLUTION : 9
c + mD + mE 2 + 3 + 4 + 3 + _ = + + ’ + +
. EAXA meB mch meD mExE 2(300) + 3(150) + 4(300) + 3(300) + 2(0) = 3150 kg'mm + + + +
mAyA mByB mcyc mDyD mEyE 2(240) + 3(400) + 4(400) + 3(0) + 2(200) 3680 kg'mm + + +
mAZA mBzB mczc + szD mEzE 2(0) + 3(0) + 4(270) + 3(270)+ 2(270) = 2430 kg'mm ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—12* Locate the centroid of the
shaded triangular area shown
in Fig. P5—12 if b = 200 mm
and h = 300 mm. Fig. P512 SOLUTION ’hor the differential element 3
of area shown in the sketch
at the right: _ h
y—gx
_ 2 d
dA — y dx — b x x ux
dMY : % dA = h—; x2 dx
2b
dM = x dA = h x2 dx
y b
_ b h2 2 _ h2 b 2 h2 x3 b hzb
MY— —§‘(dX——2 xdxz—ZB— =_6—
' 0 2b 2b 0 2b 0
b b 3 b 2
 22 2 2 _2x_ Lb
My  J0 b x dx — b [0 x dx — b[ 3]0 — 3
l
A — 2 bh
M 2
 _vmz 2 
Axc  My xc  A  bh/Z  3 b  3(200) — 133.3 mm Ans.
M 2
AyC = Mx yC = 31 = EB???  g = Egg = 100 0 mm Ans. ENGINEERING MECHANICS ~ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 514 Locate the centroid of the shaded area shown in Fig. P5_14o F%.P544 SOLUTION For the differential element
of area shown in the sketch
at the right: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—17* Locate the centroid of the
shaded area shown in Fig.
P5—17. SOLUTION For the differential element
of area shown in the sketch
at the right: 2
2X.
b2 x dy ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 519 Locate the centroid of the
shaded area shown in Fig.
P519. SOLUTION For the differential element
of area shown in the sketch
at the right: ~
n
u
I
I
I
V
I
I
I
I
n
I
I
I
I
I
u
a 5 2. _
I2 2x dx ‘ 5/2 5
3/2  x
I2 /5 x dx — /3 [ 5/2 5 x3/Z 5
I2 /5; dx = /5 [ 3/2]2 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—35 Locate the centroid of the shaded area shown
in Fig. P5—35. Flg. P53S SOLUTION The shaded area can be divided into
a rectangle with two circles removed.
The centroid for the composite area
is determined by listing the area,
the centroid location, and the first
moment for the individual parts in a
table and applying Eqs. 513. Thus, 1440
37.71
113.13
1289.16 M
_ Y _ 1289.16 _ .
C j Ci y C — 3 — §TZT§E — 6.00 in. Ans. >
X
ll
[‘1
>
><
U
2
X —_—.._._.—_—_.— ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 537 Locate the centroid of
the shaded area shown in Fig. P5~37. SOLUTION The shaded area can be divided into
three rectangles. The centroid for
the composite area is determined by
listing the area, the centroid location, and the first moment for
the individual parts in a table and applying Eqs. 5—13. Thus, ___§£_ 
YC  A — 48  4.00 in. (Note symmetry about y axis) 326. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—40 Locate the centroid of
the shaded area shown
in Fig. P5—40. “SOLUTION The shaded area can be divided into
a rectangle, and a quarter circle,
with a square removed. The centroid
for the composite area is determined
by listing the area, the centroid
location, and the first moment for
the individual parts in a table and
applying Eqs. 513. Thus, 2 2 = %n(150)2 = 17,671 mm 4r _ 4(150) 3n = 63.66 mm, 113.66 mm My yCi Mx (mmZ) (mm) (mm3) (mm) (mma) __
7500 75 . 562,500 25 187,500
17,671 63.66 1,124,936 113.66 2,008,486 5625 37.5 210,938 87.5 —492,188
19,546 1,476,498 1,703,798 = 1,476,498 :
19,546 = 1,703,798 :
19,546 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 544 Locate the centroid of
the shaded area shown
in Fig. P5—44. SOLUTION The shaded area can be divided into
a square with a circle and a quarter
circle removed. The centroid for
the composite area is determined by
listing the area, the centroid
location, and the first moment for
the individual parts in a table and
applying Eqs. 5—13. Thus, = an: n(60)2 = 11,310 mm 2 2: ln(1zo)2 11,310 x _ _ 4(120) _
 240 3“  189.07 mm
4(120) 240 3" ' 189.07 mm Ai XCi My yCi Mx 3 (mmz) (mm) (mma) (mm! (mm ) 1 57,600 120 6,912,000 120 6,912,000
2 —11,310 100 1,131,000 80 904,800
3 11,310 189.07 2,138,382 189.07 2,138,382
2 34,980 3,642,618 3,868,818 = 3,642,618 _
34,980 = 3,868,818 =
34,980 ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 569 Determine the resultant
g of the system of
distributed loads and
locate its line of action
with respect to the left
support for the beam
shown in Fig. P5—69. SOLUTION 200(3) 1.5 ft. 150(3) 4.5 ft. 100(3) 7.5 ft. 2F = A1 + A2 + A3 = 600 + 450 + 300 = 1350 lb = 1350 lb 1 A + A x + A x 1XC‘1 2 C2 3 C3 = 600(1.5) + 4.50(4.5) + 300(7.5) = 5175 ft'lb '"' "'"UL  . 3.56 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 572 Determine the resultant
ﬁ of the system of
distributed loads and
locate its line of action 25kNm1
with respect to the left
support for the beam
shown in Fig. 95—72. WE ‘ ‘ ‘ L—nglv‘im Fig. P572
SOLUTION 2.5(4) = 10 RN 4 m %(2.5)(4) = 5 MI 2 + gm = 4.667 In SF = A1 + A2 = 10 + = 15.00 kN ¢ + Azxc2= 10(4) + 5(4.667) = 63.335 kNm 357 ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 5—85* The width of the rectangular gate
.shown in Fig. P5—85 is 8 ft.
Determine the magnitude of the
resultant force R exerted on the 3
gate by the water (7 = 62.4 lb/ft )
pressure and the location of the
center of pressure with respect to the hinge at the bottom of the gate. SOLUTION p,, = 7d = 62.4(7) = 436.8 lb/ftz max
1 Vps = Epmax %(436.8)(7)(8) R hw 12230.4 lb 2 12.23 kip Ans. %(7) = 2.33 ft Ans. The width of the rectangular
gate shown in Fig. P586 is 2 m. Determine the magnitude
of the resultant force ﬂ
exerted on the gate 9y the
water (0 = 1000 kg/m ) pressure
and the location of the center
of pressure with respect to the
hinge at the top of the gate. SOLUTION p x: pgh 1000(9.807)(3) = 29,421 N/m2  l
R — V — 2p hw ps max  §<29.421)(3)(2) 88,263 N a 88.3 RN 2 _
1 + 3(h) — 1 + ...
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This note was uploaded on 03/25/2008 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.
 Spring '08
 McVay

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