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Unformatted text preview: ESM 2104
Final Exam Study Guide
Professor Hendricks Prepared by Andrew Roberson ESM 2104 Final Exam Study
Guide Force Law of Sines: bsin0=AsinB=Csina Law of Cosines 9
b2=A2+C22ACcosG A C a must be between A and at B C in Force is a vector: Made up of a magnitude and direction Use the parallelogram method to find the components of a vector Rsin0=Bsino<=AsinB
x" f R! {513 Exam x the magnitude of the resultant is 2000 lb. 2/21 Findn’ma 5011114 11 inldeofresul
2d>:3c,o'2.9
(1): lac—e
Lami at.th
HO” Zm‘: 3002+ lﬁwl—Zm'lHOOCcSJJ
"\e 4106=Q‘i'iOStlﬁé'lOG‘ZZ‘f'lOCcas'Ii
_ l . Cos¢=(é‘1+€6"¥)lo%,2~11m"
Cl): 128.682"
l ESQSZ”: [80 9
J, — .7 4'8” Finding xdirection and ydirection components of a force vector Right triangle method
X=Lcose Y=Lsin6 Example: The unstretched length of the spring of modulus k=1.42 kN/m is 100 mm. When pin P
is in the position 0 = 30°, determine the (a)x and (b)ycomponents of the force which the spring
exerts on the pin. X=Bosm3zf = “i0
l#80 Got: 30": (31.282
5 SL2 @Ow‘ft (60%?
$1=LD°i18f+Zoz
‘31: am: 54.23 8= Moon F: Hts—106): LL’J’ZB'tDCS)
F: [HZULLOQ'J'rLOaJ
E: F5056 F= mﬂnmﬁ 1;: 5.7.58 c.03701é 20! 1215.758»
:F t ’7 Q37 E: Fame 13:51:13 sm7ﬁiézﬁ F_i1: Vector Projection Projection of vector A onto B P=ABcosGB i3 Moments The moment around a point is made up of a Force and Radius Arm
Radius arm must be perpendicular to Force Mc=FxR R Example: A force of 150 N is applied to the end of the wrench to tighten a flange bolt which holds the wheel to the axle. Determine the moment
M produced by this force about the center 0 of the wheel for the position of the wrench shown.
Consider counterclockwise to be positive and clockwise to be negative. HON 651] mm 5; \U E ’ F50 c9570” T” E5 51.303“
413 :13 iii: 1303mm”
“ x F; H0991" X: “450 @326 (j: 0150 srnZ’O”
Xrgzz'géa El : [53. cEO‘DI Zﬂor‘Fgg F F.1 (xQZS)
E M;"5!,3oa «5301001 _ [‘10‘3’15‘1(H223él ($2.5) 2M0: *79‘15Fi‘73 4 507q471L1é5
EMU: 58661:).L158N: F552 r0: bu Couples replace two or more
equal forces with one Equivalent Force and a Moment All three pictures are
equal The sum of forces in the
direction is equal. the moment around the
top of the beam is equal 1 1m Remember to draw a
2"“ picture with the
equivalent forces,
and moments Example: For the thrust configuration shown, determine the equivalent force—couple system at
point 0. Then replace this force—couple system by a single force and specify the point on the x
axis through which the line of action of this resultant passes. US I IS"F WE..
EM; ZT + ZTwsi’O“—~ lSTleD"
2' ME azgasr 2nd [35M mtﬁ=o ZMH=E*X+E{'D= .i‘MTx —2.i;357‘: .i7H7x
XS'Zés’S/iw—Iriin/S Example: Determine the height h above the base B at which the resultant of the three forces
acts. 'F
2901b ‘9‘ ‘1
J3
53 25011: F50
E: 630 ~2‘102 to = 80
ZN; EgoaswzozaZ +2010 ~18E
Z ME=7280352 80+ Z‘fo—BO
ZMgr'ZI—Iﬂo 2m! PQJ'N‘E/ FreeBody Diagram Problems In order for an object to be in static equilibrium, the sums of all forces and moments will be zero.
Fx=0
Fy=0
Mc=0 lnteractions with surfaces 430.19, support There will be a force (n) on any object resting on a surface that is perpendicular to the surface
Q % There is both a vertical and horizontal force at any pin connection 5. ley sliding guide
At built in connections there is a moment, vertical and horizontal tiff Ea Example: A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is
leaning downhill. With the aid of the winch W, what tension T in the cable will be required? The
1000lb trunk has a center of gravity at G. The felling notch at O is sufficiently large so that the
resisting moment there is negligible. Horizontal Z M: O= wig gem ea—Tﬂ laws—TX lam.9“ O: lOCD 83% 6° +11% '8?  [Ssoné LTmﬂBO l8cosé° T(.581 0,02534' loan l3, gm (5":(3
l6.‘f"f‘/T: USE—83:7 T1358.‘87/lla.‘1‘1‘1 : 82 :33? Example: To test the deflection of the uniform 200lb beam the 128Ib boy exerts a pull of 30 lb
on the rope rigged as shown. Compute the force supported by the pin at the hinge O. Tm P :N’ﬁﬁoﬂﬁTE +w~(é+2.8~tr1)+3w.+ 6+T+T+LE+233
O = 2.8P+l7.8("18)+ 200(3) + 3D (63 + 30(83)
r2513: GHH + QOO+18© +2 .94 ‘P= HEBLi /2,g= 6612.286 3D Forces
FZ=Fsin<p
FY= FcoscpsinB
FX=Fcos<pcosG
3D forces can also be represented as a F=Xi+Yj+Zk Example: The cable BC carries a tension of 820 N. Write this tension as a force T acting on
point B in terms of the unit vectors i, j, and k. The elbow at A forms a right angle. 3‘8 U 8 n .34; i: If}. ~.8Los"58°=.éé0tc.
X=O—i7=L7T Q? .7+33m383: IRIS T“ T/J ﬂan? < fhﬁhi kt; LT”: 820/2.i81<‘ 7E.+ I.l01gj+.£a?l0 ‘ 5”. ._ HUN”. LEVER? .r Vector Projections: A force can be projected onto a line; the projection of F onto AB is the dot
product between the force vector and the unit vector of A to B FAB=FoAB 3D Moments The moment around a point can also be represented as a vector. Each component is the
moment around each axis. If a force is parallel to an axis, no moment due to the force around the axis exists
Always split the Force into XY—Z components, and multiply by the proper moment arm  Don't Forget to Find Direction of Moment, Clockwise or Counterclockwise, the direction is
found by looking down towards the origin of the axis (from positive to negative) Imagine you are
placing your eyeball at the letter x, y, or Z and looking down the axis Example: The 5lb force is applied at pointA of the crank assembly. Determine the moment of this force about point 0. Mdﬂi‘ﬂ 2
m3 isms 5°52}? $3.220;
Mowrtﬂ+ 2 M0,, 5555/1 33b'.7752.0017 3D Couples and Wrenches BOON Find the sum of the forces in the X, Y, 2 directions, to find equivalent forces Rx, Ry, R2 Find the moments around the X, Y, Z axis to find equivalent moments Mx, My, M2 Remember to draw a 2"“ picture with the equivalent forces, and
moments  Rx, RY, Rz Can be written as a vector that creates a single equivalent force Wrenches Wrenches are an equivalent force
with a moment around the force To find the equivalent wrench (M) Find the R unit vector M=m*R . uo Sum the moments in the first picture and compare to the second picture
~ , . 21m: —Ho (w) =Ho(¥)
? 2 dIanaJ 440K EMT: 410 ‘8) = no“) :2: n Solve for m using algebra and use m
m P m— L—?ﬁ)m‘]lm)\.,£)  to find M
~ my, a m: o “zoolpr
Kim M ﬁlo $=Uiox *ﬁ‘ﬁ, M
O: L70X+EZO’ZOM/Lﬁblf7
O: ‘ZOxLl'OM/bfLIJ
’ZOX  8516M : Lioxi 3102‘4‘1’WVI
x (*20410) = 310 4 Mlwntsﬁs)
)(t*533  .DD7M
O r 20 S .33— .Dowm) ~ “lam/W7
Ovloem ,‘~1“lM_%OISM l, M t  1 Dad £7 M ? I 01' '63 Example: The resultant of the two forces and m t [(73 .1133 91 y _ . _ couple may be represented by a wrench. Determine
— / U1 7 < 2'01} 010“ ) the vector expression for the moment M of the wrench and find the coordinates of the point P in the xz plane through which the resultant force
of the wrench passes. R: < lBOL + Hwy;
ﬁ:I/J70,3<\30L+HOD EMJO T'Rg'E+n/i E3C%7O3 MT M'm'f’omﬁlmv
OI‘IJO%+.7¢3M ﬂ: €$§§<II~GWID~D “02:7 63m .2: ,OD7M
EM; “1% JED.3: M “‘9'?er Rx? ll =_C‘jém+ 00754431:
IS‘i'SmﬁH m=706 M = <6_HZ‘16+"1.SGIJ) 2: _OO'77 {0,5,3 {DE1'93 3D Equilibrium
1 _ Sum of forces in X,Y,Z directions must
:Mt‘ [1021:34an ‘. 33 #123"): be zero
: 10. _ . ~_— .. Sum of moments around X, Y, Z axis
“'0 X 1 38 ED ' 53 WZ' 6" must be zero
XGG'fLS' “faﬁbﬂlo : '. D 'i Draw Free Body Diagram, and list
Unknowns Use moment axis that go through forces to eliminate forces that are used in calculation (see
example) Example: The industrial door is a uniform rectangular panel weighing 910 lb and rolls along the
fixed rail D on its hangermounted wheels A and B. The door is maintained in a vertical plane by
the floormounted guide roller C, which bears against the bottom edge. For the position shown
compute the horizontal side thrust on each of the wheels A and B, which must be accounted for
in the design of the brackets. Detail of door hanger Z Vigortiw+ Marx
l32CE"!"ll0 0.527.519 EMAE=D= Cy(7“i)ll+iSI113,4
56353422757]; BHFGBLH
EMBJOECgﬂlﬂiﬂiz — 1341A):
fang27.964140 Ashlin? Example: A rectangular sign over a store has a mass of 148 kg, with the center of mass in the
center of the rectangle. The support against the wall at point C may be treated as a balland
socket joint. At corner D support is provided in the ydirection only. Calculate (a) the tensions T1
and (b)T2 in the supporting wires, (c) the total force supported at C, and (d) the lateral force R supported at D. i T Isl—z. R “Cy GD
1.; :0
ll I
s l Wﬁ‘bﬁtLsyszh) =1L ('Z.SL+I.SJ‘+3,2K) Z ‘9'. 31a JLJ' XML; or Law/a —TI {391 41125 T.,. I “in
TtBWZEﬂéz 41.3sz = Haas3% Wme '3>1+2.5.3“2 H.szq*%£%+%%ﬁ=2%ue
2 .0: i2 T +2334 T1: 246648 XML; :Z‘STZS _3HT':‘ 25' I'STVlfszq: 3.“1' I'STﬂriﬁlof
8643:1041? T1: lion, TL=LZOIMZSI3Q=
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—— _ 5.52L1T,=2%5é8 7]: 9233a
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Riff "5/1.qu ‘Tz'%alq
1? = ims I’m: = “#612
E ETD: CwTlerzx
CK=TI . 3.91A1_q0_1 Cf 2013.53+ 293.S3=587.oc; 21EE:O:TZ.EETIf+ Czﬂ LU. ‘ C%=375.723~27é.167'MSDH Cf magnum
C1=C§+CE C: #172. Trusses Truss has to be made up entirely of two force members
Tension forces pull out away from the member (positive sign)
Compression forces push in towards the center of the member (negative sign)
Use sectioning method to find unknown forces (see Example)
Cut through the members that you need to find the force of Use moments to eliminate extraneous forces (if a force goes through a point, the moment
around that point due to the force is zero) Example: Determine the forces in members (a)CH and (b)CF. (Note: compression forces must
be negative.) 21n—‘<—2 111T?! 113 2111 56.421113!) 1
r T A" n D '2; E CF, 2.
A B T E ' u '
t c a: 23 GE I T:
g F
231;}; £17113
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in
Machines E Fig: D : —ZE 2LF23C'H L3
Machines are made up of both truss 3. ' CH {355% 5—3 =— £3 members and nontruss members 13 __ _.
Draw free body diagram of each part of the CHCDEH‘S machine (split the pieces at the joints) CH: H All internal forces must be equal and
opposite, if a force goes to left on one FBD, on the other FBD with the same point the force will go to the right (see example) Example: A lifting device for transporting 135kg steel drums is shown. Calculate the
magnitude of the force exerted on the drum at E and F. EB 1‘10 CM GB,GH@
“an ,2"? mo 3?le ZE=U=GﬂiJQBX 3‘”
6A=Gg
12m SEED: 13+ GBHJHSHE
136 ‘l.8‘’ Zc:1 A girl 3%.05 290 290
mm mm
\ C A = News
1 mEXEﬁCQ
an _ __ c EgoGigi;
llltl‘i U, "— E3: ‘HngSSmSUl.OSH=*QGlH£
EE1 EMgo:Gill,log—.lZEgEﬂrza—zs)
“El—“430:3 3%.00'..r_uc_—,1ZEK+.D‘1 Eﬂ : o
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E¥E3+E§'E4573l%3 Centroids The centroid is the center of the area of an object. X=X dAdA Centroid X=X dmdm Center of mass Remember to use the right bounds, and proper use of dx and dy for the double integral Example: Determine the xcoordinate of the mass center of the tapered steel rod of length L
where the diameter at the large end is 6.30 and the diameter at the small end is 1.30. Din. = 6.3D 3.lSﬂ=.é5‘D+L‘k
2.50 = KL Jazz—f” '2.
T’flx «:5 +145 0'“ L J 1
a { g 1.. ‘XS y 'fflzgﬂlx +3150 “Kll 612.512??? 0L3:
: IE! —— 1
K: C __ __ __ _ L. L.— \  7— L.
“[035” L3: “3 dx ] H113 01+ may $2331;de
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.2111: :42 L “353 m 15:25 wi
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.LTZZEX +.Q15x1+2 oggugl" f’TZLELl LEZEL +2,DEF5L.
L, ' L} U 24. "I 2.657 E/f—Mm Lrt.C"lH;l_
Example: Calculate the coordinates of the mass center of the metal die casting shown ((a)x
coordinate; (b)z—coordinate). Dilnensicna In 111 illilDIE‘T em Distributed Loads
Find the equivalent force by integrating W between the boundaries.
Use centroid equation to find where equivalent force is applied. Example: A cantilever beam supports the variable load shown. Calculate (a) the magnitude of
the supporting force RA and (b) the magnitude of the moment MA at A. i 'J
Ei' = it?” + ELI." ..
I ‘5‘
I
l 1 26Ib9'ﬂ $0.5M: ’ : J‘ H d):
a? 3 A FEW3.5 +.Ol?..x1
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w 0 k): 1 ‘ii 1r—:QOZS.L’I_Z
lZéTﬁ—h  L I 31% I Shear SH ~
KEOIZIl F Forces :rﬂho:KqF
w3705+~0l21x1 F‘ lZIq: F; (20342 6
a ?..3
_— F XII Kudx 0 quto: Mq“ Mgcomz (mad 3 7H2)
MA: ‘8320101 Shear forces are internal forces within a loaded structure such as a beam. There will be a shear (V) and moment (M) in
every beam that is cut w whim!
When W = The equation for the distributed force .1. w
wax l a 5;: V=de M=de ‘7
Don't forget to solve for constants after N integrating with initial conditions, V(0)=R, M<°>=° V Ax V+AV Example: What are the values of the (a) shear and (b) moment at x = 6.1 m? (c) Determine
the maximum bending moment Mmax. rad m N/m
A . B
L2m+lx—3m4>L2m 21nd 1500 N Z MEEOHH (241+Z+3)+ “:00 "2+ F~LUFZTLSB U: A’ 830(Y‘ZX
QA: 3100+ 53'2'4‘710 V: Ignza'aww mm
A: U: 22*930x
ZZZ—£5“ F: 850'3‘2‘1'510 Mthdx
1‘9 lav“ M13537. 22 x — Luzmac.
if t 0 : A + U _ F {Ma‘Orwmm ‘Z'lS)‘C°"A Liana =6. (Ks—7.2a5‘frsléIfﬁ,
"j 222q90(2_ﬁ) C!:"ll57,85 l877.ZZZ“?”IOV‘O *
L u..78
MW: 35 3112012 £2341 IS (“I 26232—157. as x, M=3337.22x_=115x1— I15785 0: 3537.2? 830x
830x=363122
I W*ogr'r9b th'ZdZ—bb’lw Friction Friction is equal to the coefficient of friction (static or kinetic) multiplied by the normal force of
the surface on the object F=u*n Example: The P: 110 lb wheel rolls on its hub up the circular incline under the action of the
p=26 b weight attached to a cord around the rim. (a) Determine the angle 6 at which the
wheel comes to rest, assuming that friction is sufficient to prevent slippage. (b) What is the
minimum coefficient of friction which will permit this position to be reached with no slipping? . . g “  g. ER=OZILHOSMB awe
{ﬂame lsasmnmmss
mam33%;; @= 3065” 2611: 2‘ 153:0: {3— HO maQchme
ml 3W3 30.9511“?
WP: ME“ Friction on Pulleys As a rope passes around a cylinder there is friction that causes two different tensions in the
rope. The relationship between the larger T and T1 Tl'1=eb*u b = the angle in radians of contact between the rope and pulley Example: A counterclockwise moment M = 1600 Ibin. is applied to the flywheel. If the
coefficient of friction 11 between the band and the wheel is 0.23, compute the minimum force P
necessary to prevent the wheel from rotating. T,T.feo@
1‘19 0' if" mpgM1
arm TI: T, ﬁzsrr
i Ti<.jxlT, TﬂogT
J EMM:M+CIT—01‘rl:o
 LIEOOZTMGIZBQ
FWD;
EMU: O: ~ST+I3T, aw
ZGP= BLDG1L1? LrSmn
P=Ha53 Moments of Inertia
Similar to centroid but deal with mass (area) as well x=y2da
y=x2da Poar moment of inertia is the sum of IX and y Example: Determine the moments of inertia of the rectangular area about the x and yaxes and
find the polar moment of inertia about point 0. h I'll—>— Parallel Axis When finding the moment of inertia around a point that is not the center of mass of an object
(lp). the parallel axis formula is used p=c+Ad2 The original moment of inertia (lc), Area of the object (A), and distance between the center of
mass and new point (d) ...
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