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/6 JMQRQQ? C9 MEEN 221 Summer 2002  Worksheet 5 Dr. A. Palazgolo (1) HWSS Due at Beginning of Next Class C115 (2, 11, 14, 29, 39, 89, 107) (2) Attendance Mandatory for 10:00 am 12:45 pm , unless excused by instructor.
Quizzes may be given at any time during this period. (3) Subjects: Centroids and Distributed Force Systems EXAM I ,, WEDS. a ,JUNE 6:00—8:00 PM ZACH 104B
EXTRA TIMEAIAYBE GIVEN (A) . Centers of Mass/Gravity and Volume
(a) Center of Mass  CM
Suppose "that a rigid body is represented by a collection of
particles With interaction and external forces. Each point on the rigid body may have a different acceleratiOn
(i.e., 5,4 for some arbitrary point A). How Can Newton's Law
2F: M be written, i e which value of if should be utilized? To answer this sum Newton's Law for all particles representing the
system the system _ ZMa =Zf. +.[ ~ (1) The ﬁrst sum on the RHS is zero by the nature of interaction
forces and Newton' 5 3rd Law The second sum on the RHS IS the resultant R of all external forces on the body ('0) Select the point G such" that ‘ ' M230 = R (3)
Where
M = Z M i = total mass of the body (4) Since the RHS of (l) and (3) are both K it must be true that their
LHS's are equal. Thus MEG = 2 Mid. ' (5)
Integrating twice in time yields M; =§Mia (6)
or ‘ 70 ﬁrm: +royi+erI€ <7) The summation in 1 may be expressed as an integral for a
continuous body, with Mi:>dM:>pdV,andFi:>F (8) _ This yields the following result: Deﬁne the point G as being the center of mass of the body then
the coordinates of G are 1 ' 1 1
rm, =—A7JXpdV, r0, =HJYNV, r02 =il—szdV (9) and it is true that
MEG = R = resultant of all external forces on the body (10) Center of Gravity — CG The center of gravity is the location where the entire weight of
the body could be concentrated and its moment about 0 would be the same as the resultant of the moments for weights of all the  differential masses that comprise the body. (C) F0. x (— mgé)= (raw? + rm] + r0212)" (— mgIE)
= J? x (1W: I(Xf+ ﬁ+Zl€)x (— pgdW) 3 rG.ij—ra.,mf= iXde/j— tide (11)
:> r” =l prdV, r67 =i [YpdV (12)
m V m V From (9) and (12) it is seen that the X and Y coordinates of the
CG and CM are identical. The Z coordinate of the CG is arbitrary although it is usually
taken as the Z coordinate of the CM. This result is consistent with the principle of transmissibility. Centroid of Volume CV .
The volume centroid is the location of the mass center if the density of the body was replaced by a constant, i.e. if (13) Eqs. 9 gives the centroid coordinates as r“ ;—I];V[XdV, rm, =%V[YdV, rcz =%V[ZdV  , (14) 4 p In analogous manner the centroid of an area A in the XY plane
' has coordinates gﬁlﬁ Three bodies with masses of 3, 6, and 7 kg are loeated at points
'ﬂ (4, 3, 1). (1, 3, 2), end (2. 2, 4), respeetively. Locate the
3 mass center of the system if the distances are measured in meters. Solution .§+12* :Locate the centroid of the ’ shaded triangular area shown 5" in Fig. 95—12 if b = 200 mm
and h = 300 mm.
QC)\/C)
i=0 _k—— ‘5f16.i Locate the centroid of the shaded area shown in Fig. P516.
é! 520 , Locate the centroid of the
'shaded area shown in Fig.
P520. r Riley 528 6" Locate the mass centef of the
.hemisphere shown in Fig. P528
if the density p at any point P
is proportional to the distence ‘from the xyplane to point P. [> restart;
[> # Example Eh. Riley 5—28 . Mass Center ef a Hemisphere
' >‘26 := int(k*pi/m*z‘2*(2*R*zz*2),z=0..R); 3 knRs M  10m A
> m := int(k*pi*z*(2*R*z—z*2),z=0..R); 2f
E mz=jiknR4 é:
; 12 ®
{> 26;
I 18 25 _> #ENGR221 Centroids
> restart;
_ >
:>
:> #
' > AY:=J.n ,y=0..h*xlb); A:=int(AY,x=O..b);
H 2
' b
A —lhb
_ “2
‘ > XC1:=int(x,y=0..h*x/b); Kc:= int(Xc1,x=0..b)/A;
x2}:
Xc1:=——
b
X 'Zb
_ C' 3
r > Ycl:=int(y,y=0..h*xlb); Yc:= int(!'c1,x=0..b) IA;
1th2
Yc1.=‘ 2
b
1
Yc:="h
3 lllllfll V V V V #
AY:=J.n ,y=0 . .2.0*xx‘2/25.0) ; A:=int (AY,x=O. .50.); AY := 2.0 x — .04000000000 1:2
A := 8333333333
> Kc1:=int(x,y=0..2.0*xx*2/25.0); Kc:= int(Xc1,x=0. .50.)/A;
XcI := x (2.0 x — .04ooooooooo x2) Xc := 25 .00000000
> Yc1:=int(y,y=0 . .2.0*x—x*2/25.0) , ‘Ic:= int(Yc1,x=O. .50.)/A; z
YcI := .5000000000 (2.0 x — .04000000000 x2)
Y6 := 1000000000 :> :> :> # ’> Ax:=1nt( ,x=0..sqrt(40.*y)); A:=int(Ax,y=25.0..90.0);
AX := 6324555320 J; L A .= 3072953723 > Xc1:=int(x,x=0. .sqrt(40.*y)); XC:= int(Xc1,y=25.0..90.0)/A;
X6] := 2000000000 y [ Xc := 2432513039 [> Yc1:=int(y,x=0. .sqrt(40.*y)), Yc:= int(Yc1,y=25.0. .90.0)/A;
[ E (3/2)
YcI := 6.324555320 y Yc .= 6068894055 >
> Cg/foe 010w OF compowre HREA‘J‘ ﬁll I—a éO/e 1 1
=,_— ﬂ: —— a! */d d 
Xe ﬂdg"; 4c é’fX4/4ax/4v‘éx/4ﬂfxd/4f
=— M Xe +4 >42 +4434Xcﬂf 2)
4c
genera! C‘W‘E Id: 2/" (3’)
K6 {Z/Kc @J
yd=/Z_:—Z/4€ yet. (3] .3 a Y‘n'u '_ ,.
Wahfpfjéw. Sem1ciiciilar'éré‘a_ \ dr'a‘ntof a parabola
255; ~  ‘ iii?
538 Locate the centroid of 4‘ k
r ' the shaded area shown T
in Fig. P5—38. 1 f
q> W '25 mm
Fig. [538 “—45 Locate the centroid of
the shaded area. shown
in Fig. p5—45. (B) Distributed Loads on Beams g ‘ Mir, l IIIIIIlllll H
Agimgmm ‘ ' ‘ ' ' ' .. Let P = pressure acting on the beam then WOO = line density ofload = P(X)*t
= force per length The two load systems shown above are equivalent, in the sense that
they will have an equivalent effect on the other loads in the structure if R = L]'W(x)dx = total load on beam
0 :and d R
My =fodF=foW(x)dx=Rd z ﬂ’ 4; 23 0 ‘L
:> d=— W dx
p mix (30 4 +
4. ‘1 4? 4
4’: J’ u/dx .. J wdxfz‘ndx +fwa€x '1
o 0 4’1va X, I? X2. ,(7 3 A7
x = x=x¢ xIX ’ " 3
I 2 ’ 3:4, k‘“ L
z, ‘2 L: “2
A9: (fo’wdagfofmexz +01; wdxg — 4%M3
L 47“; L
:z [éﬂxi’dx hfxwdx +waqf},X,=X =9 afx=a€><l
4! 4,441 X4: Xil =b 4X= 06¢?
43
—_ ifdyg + f(4,+XAJW¢QXZ +0! (41.4.; ,Lx3)wd>(3§
= %[‘f*'€z % (way/a, +64%; “we;
r» F—— L —el :6 ,er. gag6)“ 574 Determine the resultant
i of the system of
distributed loads and
locate its line of action
with respect to the left
support for the beam
shown in Fig. P574. Fig. P574 Forces on Submer ed Surfaces
Consider the column of liquid shown below. W} F2, 0 T /
(P w» d I 5? .19 ‘_J;‘ 6;?111drfjr 4TH ,0
The pressure P must balance the weight of the column plus PM by
Newton's Law. So ‘ PA=P0A+Apgd
:> P—Po =pgd Thus pressure increases linearly with depth. 581* If the dam shown in Fig. P581
is 200 ft wide, determine the
magnitude of the resultant
force ﬂ exerted on the dag by
the water (1 = 62.4 lb/ft ) ‘ pressure. 587* The width of the rectangular
g!“ gate shown in Fig. P5—87 is 4
7 ft. Determine the magnitude of
the resultant force ﬁ exerted
on the gatg 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. Fig. P587 :’[> #Example Pb. 5 Riley (5—875
~[> dt := 3. 8: L := 6.: rhog :=62. 4 : b:= 4.:
) [> F :=rhog*b*int(dt+. 8666*x, x=0. .L); RI F:= 9584. 340480
 D:= rhog*b/F*int( (L—x)*(dt+. 8666*x), x=0. .6)
use the derivative symbol I
Error, attempting to assign to #note error. Do amt D which is protected [2» mm: rhog*b/F*int( (Lx)*(<:1t+. 8666*x) ,x= .6) , mﬂ/ DD := 25937685559 _' ...
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 Spring '08
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