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Unformatted text preview: OOOOOOOOOOO Statics Final Exam Review TOPICS covered on revrew:
Adding vectors
Moments 3—D Moments 3—D Equrllbrlum
Resultants (wrenches)
Trusses Internal Forces
Centrords Area Moments of lnertla
Friction Non—Ideal Pulleys Problems taken from "Engineering Mechanlcs: Statlcs" by J. L. Merlam and LG Kralge (6th ed.)
Anal from "ESM 2 I 04 Sample Common Final Exam" retrieved at Kooferscom Good luckl Notehall Page 1 Vector Components and Projections Determine the scalar components {alRi and {MFﬂb of the Force R along the nonrectangular
axes a and b. Also determine (5) the orthogonal projeotion P3 of R onto axis a. 30" 3:102:11 Problem 2/018
htt : edu en.wie .co edu en mared ass' nment test int.univ Scalar components are the two vectors along the axes that Wlll add together to equal the given vector. To fmd scalar components, construct a parallelogram Wlth the axes / ’
(may need to be extended) as the sides and the given vector as the diagonal. / / /
/0
We can then use the law of slnes to ﬁnd the magnitude of the components. Lif/ I03 N
IJHO 3‘? / Law of smes: Ho / o = e: = c /\/
sinA sin B sin [7 / l03 _ lLa _ (Lb Smblo ‘ SlTxllO — $\VL30 Ra=l5l,Rb=50.l The PFOJBCtIOI’I of a vector onto a aXIs Is the vector times the COSH’IB of the angle between the vector
and the axns. It Is the right triangle the vector makes Wltl’l the axe, Wltl’l the vector as the hypotenuse. q
x< // \1\
AM 103
l03Co§ 30: aqua Pa=59.2 Notehall Page 2 Adding Vectors The magnitude of the resultant at the 2 force vectere is most nearly m“ N a. 1200 N b. 1 130M 3. 1091 N d. tsan 411'? 50”“
e. 451 N To add two vectors place the tall of the second on the head of ﬁrst; the resultant Is the vector
from the tall of the ﬁrst to the head of the second. Note that the two added vectors forms a triangle Wltl’l the resultant—we can use the law of cosines to solve for r. l 70 booN Law of cosmes: 1 N _ : c1 = a2 + b2 — 25m cesff) / 160 L10 ll—LO
N / r a = 5063 wooa — Msboxmo) Coat I H0) 700 r r= I30 Sld’e note: To subtract vectors, add the subtracted vector #7 the oppos/te o’lrect/on. Notehall Page 3 Moments The BﬂH force F’ is applied a’r paint (3. The mumem 0' me force I’ — H” N
F' about Ihe point A is meal nearly.
a. (16.44 —73. IjiNm H
b. (7‘8. H + 16.4jwm i " a
(49.3i —78. j)Nm  I 13'
(73. li+49.3j)Nm I: m ,_
—5 .TkNm “c
6 .TkNm —23.9I~LNm
ZBHRNm mango gun Moments are force Interactions that cause rotation.
To find the moment about a pomt:
I . Break each force Into components
2. Multiply the each component by the moment arm
3. Sum moments; we usually deﬁne counter—clockWIse as positive and clockWIse as negatlve Ux’l +\\9\sm 30) L80 5m :10) ‘UJLCos 30) K8004» aoy—  2ny
M=25.9 kNm ThlS process Is equnvalent to taking the cross product er Notehall Page 4 3D Moments The manlent of the ESQN force abaul point A is mast nearly I'
gm  1'5!"___ F=250N
(0i+0j—3000k)Nm ______F (iii — 225“ j + Uk )Nm
[ﬁi — 2250 j — AIM0mm” 12 m _
I0i+2250j+3000kwm " ' "     I
(BHIHHI—[lj—EZSUMNIn "' . , 1
[—3000i+ﬂj+2250kWIn 2 I: 17 E“ In 3—D we get three different moment equatlons:
I . The moment In the Y—Z plane (Max)
2. The moment In the XZ Plane (May)
3. The moment In the XY Plane (Maz) M ax
y
'F.
7.
0L
thxxzo z (my = Juno): —5k950
2 Wm: : ~ lamsosuaooo Notehall Page 5 3—D Equilibrium A 5x3 ft unilerrn sign weighs 320 lb and is supported by a bell
and sealrel jeinl at A and by 1WD cables. The force in cable ED is mesi nearly a. 142.6 It:
b. W35 lb
B c. 92.0 lb
d. T'E‘LE lb
_ e. 225 lb
I w H‘ l . Freebodydiagram drawn in blue above. 2. Unknown: BD, CE, Ay, A2, Ax (5)
We can only write three equations, but if we use moment equations aboutAz and Ay we can eliminate three variables and solve for the system. 3. We need to break BD and CE into components so we can find the moments Produced by each force. We
know the direction of the forces (along the cables) but we don't know the magnitudes. Therefore we need
to rewrite the forces as the magnitudes of BD and CE times each force's respective unit vector. (5"): lw148.’q’_‘5) C/E:[CE.[<—QJ 31% > W1 W22 4. Now we can solve the system With the appropriate 3D moments. : M9,: Mime 8C}. BD)=O CE= .162 6D
7 1T0”: WED—5 :— l gala ‘l' _l_l...
zMorz Limo) Bl7 ) MIR—$13) =—H(3203*&E_ED +§R_ BD :0 :7 159: Him
11—03 «hos 5. ED: I 42.6 lb Notehall Page 6 Resultants Replace the two forces acting on the frame by a wrench.
Write the moment associated with the wrench as a vector and
specify the coordinates of the point P in the y—z plane through
which the line of action of the wrench passes. Note that the
140”: force is parallel to the xaxis. Problem 2/157
htt : edu en.wie .co edu en shared assi nment test int.uni A resultant represents several forces With an equwalent force. A couple is the moment produced by two noncollineartorces. A wrench is a resultant force With a parallel resultant couple.
Equwalent system problems such as this reqwre two pictures With the original tree body diagram and the free body diagram of the edUIvalent system. i. Draw free—body diagram of system and equwalent system ‘10 lb /1L{O (2+ Niki». List unknowns: R, Y, Z, M (3)
We have three moment equations and l equation relating R to the two known vectors, so we're good to go. ll:<lLiO,o,o> +< oJo,—oio>= <1qu OJ0l0> ’F= ' <1LlO,O;0O>
’liLiohoio‘ ZMVle’lOLR7): Z. n m 31 =7 Z217 iiobitl
zm:o= M '59 — qOU) =7 m= [ON
lbwl
zmz=iLioUi)= m<:ﬂ9)— Mom
[bloil’l ['ljig7 " 7  Y =_l5L[O :7 \(:7.78
M=107(7.76): 833 Ni: 63% (mo/o; qo; = (700.6,oJLiso.q>
llolai Notehall Page 7 Trusses EDDIb
A
4. The force In member F8 is must nearly. II E; C. D
a. EIEIOIIJT ’f
40ml: b. 80'3le —)o 0 ii I J
a. 'lDUElEE E lﬁ Fin: [3 Ir“. H In '
. lElUD ,_
e. 120mb" 00'“ b
1. 12(30le We can find the force Within a member of the truss by creating a section such that the member is an
external force in the free—body diagram. We can ignore the inside elements because we are only concerned With external forces acting on the section. Furthermore, trusses have only collinear forces thus the force in member F6 is only in the F6
direction. We assume F6 is experiencing tension; if we are right FG Will be POSItIVS, else F6 is
experiencmg compression °l00
l . Draw FBD
C CD lb
. 6 2. List and count unknowns: CD, CG, J, F6 (4)
Ur
H, l 4_, J
(7
l 3. We can only write 3 eguations (Fx, l=y, M). Because there are more unknowns than eguations, we
cannot solve for F6 With this FBD by itself. We can take the whole structure though... m i“ 4. Total list and number of unknowns:
CD, CG, J, FG, Ax, Ay (6) ——9 w W 600 J 5. We have 6 unknowns, and because we can write 3 eguations ber free body diagram, we now
have enough relationships to solve for F6 2 Me = —0iOOU13+LiooLs) —eooLLi) + Mug) :0 =7 J= 750
ZM ¢:"FG'(.?>) 0100004. 750(8) : 0 => FG= 800
FG=500b Tension Notehall Page 8 Internal Forces 900 mm The largest abeelule value 0! lhe bending moment in lhe beam is 2400 b_ﬂ '
a. 3200 Iblt A f v v v v r
b. 400!) lb—ft .
c. 43:30 Iblt ' C
d. 11200 lbll B
e. 240D lb—ft _ 4 fl 4 fl 4 ft To find internal forces, we must cut a section of the beam.
Forces are not differentially distributed , therefore we Will
need three different sections for each force distribution.  . Draw FBD
We represent the distributed force by an equwalent force located at the centr0id
V qoo‘Wri‘ Xe V oioo‘blii 'LW V vioolbIH  Lilf
xl i—{I’A ‘l/k
M T [ ——ig‘ M J/~—\ M T 1%
P 4Q— Ci P v 4; ‘FYOM O‘Ll c Rom +5 (1mm e—ra
0100\5/9 '4 9+
2. Listunknowns: P M v c (4) DJ—iOo L Ll”
a s a —‘ There are more unknowns than equations (Fx=O, Fy=O, M=O) therefore we need another FBD 5? 3. List unknowns again: F‘, M, V, C, Ax, Ay(6) J T
We have 2 FBD so we have 6 equations and are y C,
in the clear EMA=aLlOOlOLSG>OO)+l9\C—O =7 (1:wa
0—4 ZMc= M—§\L%ox>+a800x=o => M: HSOX" —;LBOOX
48 2M6 M—(XABL36003*9~BOO:O :2 M= sow—Two
8—D. zMQ= M+aLl00Lx—a)i3eoo)+ awox=o a W\= Booxqeoo 4. Find the absolute max and min of each function Within each function's given domain
(04,4654 2). The largest value is the largest absolute value of the bending moment. M =43 60 lb/ft lVl
[afl Seﬁi P 03M sec.
Remember conventions for internal forces: Forces and moments shown in (+) direction x _. / < M Notehall Page 9 Centroids The centroid is located at 2" __ __2"__ a. x = 4.001'n., y = 3.50m. b. .I; = 4.00m. _I' = 2.94m, ”
c. x = 4.001‘n., g. = 2.5%. 7
d. x : 4.00m? _': 2.38nt e. x = 4.00m” _': 2mm 8r! A centrOId Is the center of an area.
We can calculate the centrOId of a SImPlC ﬁgure WIth:
— J'di :__J'}rdA : For 2—D ﬁ ures
3: A I} A 6 — J"de — fde —_ fzdlr” For3D ﬁgures x: V J}: v ’3 v When the ﬁgure consIsts of several parts, we can ﬁnd the "average centrOId”: — 2A3: — Ay
x = — ,3: = E —
EA EA
Common centrOId coordInates can be found In the back of the book. For rectangles such as those
combrlsmg the comb05Ite ﬁgures above, the centrOId Is located In the center. —:3:— "l "— y W: Q‘QLI lﬁ+l4+8 y=2.94 In Notehall Page 10 Area Moments of Inertia Find the area moment ofinertia about a, then about b. ‘IGII‘ El _ " El
()1!
5H
1) h
The area moment of Inertia abouta l5 given by
Ix=Jy3dA
'1 EX '1 3x ‘1
1m: vholll = Wot 0M: Y3 olx: are 3
l) 0H0 Y Cl E10 [Tax otx
0
— E": X4 :lbq lﬂq
‘ mm 1: O The area moment oflnertla about another lIne can be ﬁgured wuth the parallel aXI5 theorem:
3’; = II + Ad: + 2ng ydA
We can simplify thIS equation by taking moment of Inertia from the centroud. TI’IIS Simpliﬁes to
I; : Ix + An“? Using this equation, we can compute the moment of lﬂCI‘tIa about line b as shown below IXc + 6% £31) : Ixa ‘3 IKC. =IXOL— (9": CR1)
1x0: lea—(95:3 u“): ’54
be=1xc4 m L8*)=\761 a. Notehall Page 11 Friction The force P = 120 lb is applied to the EDD—lb block when it is at
rest. The magnitude and directien of the friction force exerted by P
the eurlace en1he block is meet nearly qﬁf mg“)
a. 52.0 ll: up the slope "1—
b. 52.0 II: down the slope c. 39.? ll: up the slope _ d. 39.? lb down the slope l u. — 0.30
e. 33.DII:uplhe $009 In 1, pk — (125
i. 33.0 lb down the slope n g. 12.?6 lb up the slope 30" h. 12.?6 lb down the slope Friction is a force that opposes a net force to an extent. Friction increases With net force while the body
remains at rest until reaching a maXImum equal to the coefficient of static friction times the normal force: f : .ueN
After which the body moves and friction lowers to the normal force times the coeffICIent of kinetic friction:
f 2 #kN l . Draw free—body diagram. We don't know whether the obJect is movmg up the ramp or down the ramp at this
pomt, thus we must assume the direction. I chose to assume the force of friction was up the ramp. If I am
right, my answer for friction Will be positive; if I am wrong, my computed friction Will be negative and  Will
know friction is actually down the ramp. 4
e l’\/7‘ V 39/ \/
\ \ Q 2. Llst unknowns: N, f (2)
"’ \ \ I can write two equations (Fx and Fy) so I
’50 \ should be able to solve this problem. W 3. Solve for f ZFX=»9\oosm some C0530 +{2=o :> «0: «mg, 4. The force we calculated above is actually the force reqUIred to keep the block stationary. We
must make sure this force is not above the maXImum static friction. 7H=7\OOLOb so + l3~05\na0 n—N=o => N = 1’53
‘l‘smatx:fLSN=\3Ll59t)’3°it(94 > lame Friction is l 2.76 down the slope Notehall Page 12 NonIdeal Pulleys The cylinder is pinned Ell O and has a radius of 2 in. There is H _ n 72
a cable that goes around the Cylinder, The minimum " '
moment M ihat will allew1he cylinder ID begin TCI turn is most nearly P = 5] ll'i
a. 186.6 in.ll:= D. 2953 ir'i.—ID C. 354 in.—Ib d. EDD in.—llJ R in e. 600 in.—lh ll {11. Friction affects nonideal pulleys in such a manner that
T: = Tlell‘l'a
Where mu is the coetFICIent of friction and B is the number of radians around which the rope is wrapped l . Draw a free body diagram OY
M Ox 2. List unknowns: Ox, Oy, M, Ti , T2 (5)
’ We have more unknowns than equations so we
Ta T1 must have another FBD solb 3. List unknowns: Ax, Ay, Ox, Oy, M, Ti ,T2, (7)
We have two force equations and one moment equation per FBD, and we also have the
relationship between the two tensions, so we should be able to find a solution. Fir Z ME} = (l2’5)Ti " l QLSOD 3O :7 TI : \50 We want to find the pomt at
which the cylinder Will begin to Ta: [509)an gqq‘q turn, so we use the static
coetFICIent of friction zmo= RLM‘iH)—a(iso)—M=o :7 m: gold? M=299 in—lb Notehall Page 13 ...
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This note was uploaded on 10/02/2011 for the course AERO 1234 at Virginia Tech.
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