BOBO(dd$H$H$HZHZ..(\-Z(Charles E. Miller, Jr.x HHHV,(hh hd'0F d@fEfH(l+/P"">@Zg, ''pE\fP@1H  HZ4(bDSET( eZ$Z0YZ(Z,gfff fff f Pg |f f Pf "Ig $g   3 S T          L               6 V W     5 Q R }      K V       & Q       " W a w X n | f h i       : U L    : ~            u w   d f K L P S U V W | ~     g w    / > K \          $ &       > X Y      * @ L M [ h i !| !~ ! ! ! " "& "H "I "x " " " " # # #= #S #f #g #h $H $J $ $ %/ %e %i % % % & & ' '( 'T '~ ' ' ' ( ( ($ (; (Z (o (~  23RTW\? C   " #   "  ")*, "-3 "4LMV "Welm "n " " "%$ "n&(A)(  P*( M "N +OT "U4,. "  " " " +   " " +&'( ") +*1 2F "G "# +  + k/ !. " J# K Q# R V X w x z " {      "     "   "     "   "     "     "  [ " \ + ] q# r + s #  +   0 . 1  1  1  1  ef 2g.h +i 2. "  " " "  "!' "(#):;< "=H "I#J#  &##  "111 "1111 "C1D "F~5.6 8.5.6 89..9..9.tu;:v.w;:. S;:T.U{|';} <~ " " !  " " " " #( ",= ">  " " + H fi1ju1vx "|~1 "# "1 " "1 "1 " 0 "4617? "@C1DK5L6N 8P.U5V6X 8[.e 8i. 8. 8. 8 <.$(%.&[ "\b "c " " " " " "A "Bk o1 " "1 "11 " 5 6  8 . #5 $6 & 8 ). N 8 O. \ 8 ]. j 8 k.  8 .  8 .  !{!|l=!}.!~! "!! "!"1""  """1" """" ""%"' ""-"/1"0 ""2"A ""G"x6"y 8"|." 8"."9"."6"9"."9" 8"."9"." 8"." 8"."9" 8"." 8".# 8#.#  8#.#  8#&.#, 8#2.#B 8#H.#T 8#Z.#h#n$H %$I($J$r !$s$y !$z$} !$~$ "$$ !$%1 "%2% "%% "%% "% +%% "%% "%% "%% &&1&&1& "&') "'*'-1'. "'0'U "'V'Y1'Z "'\' "''1' "''5'6' 8'.'5'6' 8'9'.'9'.'9'.'9'.( 8(.(/ 8(0.(N 8(O.(c 8(d <(e.(p 8(q <(r.(~( (( "(( (( "(( ((m%%33YPP[[  C0  eef u u  d@MA!A!oVzQ!1!1!S%| *'*'* !r!r!$   !| $H q q qnZ ZZ ZZZ ZZ Z$Y(YPHY 1150 Doug Davis Chapter 8; Static Equilibrium 8.3, 10, 22, 29, 52, 55, 56, 74 8.3 A 2-kg ball is held in position by a horizontal string and a string that makes an angle of 30 with the vertical, as shown in the figure. Find the tension T in the horizontal string. S F = 0 (Since I am writing this all in bold to begin with, it is difficult or impossible to explicitly tell that this is a vector equation--but it is! And that's very important!) S F = 0 But that reall means S Fx = 0 and S Fy = 0 S Fx = T + Tr cos 60 = 0 S Fy = Tr sin 60 20 N = 0 T + Tr (0.500) = 0 Tr (0.866) 20 N = 0 T = Tr (0.500) Tr (0.866) = 20 N Tr =  Tr = 23.1 N T = (23.1 N)(0.500) T = 11.5 N 8.10 The pulleys shown in the figure below are frictionless but have a mass of 4 kg each. What is the tension in the cord and in the pulley supports?   Apply the first condition of equilibrium to each pulley. First to the one one the left, S F = 0 S F = S T 40 N T = 0 S 2 T = 40 N That gives two unknowns in one equation so we must seek additional information. Of course, that will come from looking at the pulley on the right, S F = 0 S F = 2 T 40 N 200 N = 0 2 T = 240 N T = 120 N S = 40 N + 2 T = 40 N + 2 (120 N) S = 280 N 8.22 a cable and pulley arrangement shown there. The coefficient of friction between the two blocks is 0.25. The coefficient of friction between the block and the floor is 0.4. What is the maximum external horizontal force F that can be exerted on the lower block before it will move? What is the tension in the cable?  The maximum external force F that can be exerted will be the force exerted when the friction forces are at their maxima, Ff = FN Make careful free body diagrams of the forces on each block. First, for the 2-kg block, we have  S F = 0 But that really means S Fx = 0 and S Fy = 0 S Fx = Ff T = 0 S Fy = FN 20 N = 0 Ff = T FN = 20 N With the normal force on the 2 kg block known, we can readily calculate the friction force, Ff = (0.4) (20 N) Ff = 8 N And that also means T = 8 N Now, look at all the forces on the 7-kg block,  The downward 20 N-force on top of the block is the normal force exerted by the 2-kg block. The 8-N force on the top is the friction force exerted by the 2-kg block. The tension is still 8 N just as it was at the other end of the string which is attached to the 2-kg block. The forces labeled FN and Ff are the normal force and friction force exerted at the bottom on this 7-kg block, of course. We used the same symbols to represent different forces earlier when we looked at the 2-kg block. Now we are ready to again apply the first condition of equilibrium, S F = 0 But that really means S Fx = 0 and S Fy = 0 S Fx = F Ff 8 N 8 N = 0; S Fy = FN 20 N 70 N = 0 F = Ff + 16 N FN = 90 N With the normal force on the 7-kg block known, we can readily calculate the friction force, Ff = (0.4) (90 N) Ff = 36 N F = Ff + 16 N F = 36 N + 16 N F = 52 N 8.29 A piece of pipe may sometimes be used as a "cheater" to effectively lengthen the moment arm of a wrench. If a pipe is used as shown in the figure to allow a 150 N force to be exerted 30 cm from the bolt, what torque is exerted on the bolt?  t = r F sin q t = (0.30 m) (150 N) (1) t = 45 m-N 8.35 A uniform pole 6 m long weighs 300 N and is attached by a pivot at one end to a wall. The pole is held at an angle of 30 above the horizontal by a horizontal guy wire attached to the pole 4.0 m from the end attached to the wall. A load of 600 N hangs from the upper end of the pole. Find the tension in the guy wire and the components of the force exerted on the pole by the wall.  As always, a free body diagram is essential.  From the first condition of equilibrium, we have S F = 0 But that really means S Fx = 0 and S Fy = 0 S Fx = Fx T = 0 S Fy = Fy 600 N = 0 Fx = T Fy = 600 N More information is needed to solve for T and Fx. We can get that information from the second condition of equilibrium. Before we start to calculate torques, we must decide on the reference point about which we will calculate those torques. If we choose the lower, left end of the pole, there are two forces, Fx and Fy, that will have zero torque. That will reduce the number of terms in all of our equations. Therefore, that is a good choice for the origin or reference point or axis of rotation. List each force and the torque caused by that force: Fx: t = 0 (since r = 0 in t = r F sin q ) Fy: t = 0 (since r = 0 in t = r F sin q ) T: tccw = (4 m)(T)(sin 30) = (4 m) T (0.866) = (3.46 m) T 600 N: tcw = (6 m)(600 N)(sin 60) = (6 m)(600 N)(0.5) = 1800 m-N S tccw = S tcw (3.46 m) T = 1800 m-N T = 520 N Therefore, Fx = 520 N And we already knew Fy = 600 N 8.41 A carpenter's square has the dimensions shown in the figure. It is made of uniformly thick metal. Locate its center of gravity.  Think of the square as being made of two rectangles.  The center of gravity of each rectangle is at the geometric center of that rectangle and the mass (or weight) located at that position is proportional to the area of that rectangle.  Notice that the positions are all measured from the lower left corner of the carpenter's square. Now we can use the defintions of center of gravity, Equations 8.10 and 8.11, to locate the center of gravity of the entire object,    8.52 Ladder problems are fun. Ladder problems are classic. A ladder problem will surely reappear on an hour exam or the final. A man who weighs 800 N climbs to the top of a 6 meter ladder that is leaning agains a smooth (ie, frictionless) wall at an angle of 60 with the horizontal as sketched in the figure below. The non-uniform ladder weighs 400 N and its center of gravity is 2 meters from the foot of the ladder. What must be the minimum coefficient of static friction between the ground and the foot of the ladder if it is not to slip?  Using the first condition of equilibrium, we have SFx = Fwall Ff = 0 or Fwall = Ff and SFy = FN 400 N 800 N = 0 or FN = 1200 N To solve for Fwall and, thus, for Ff, we must use the second condition of equilibrium. In calculating the torques, let us calculate torques about the foot of the ladder. That choice means two of the forces, FN and Ff , will provide zero torque and that reduces the number of terms in our equations. Be very careful of the angles! t = r F sin q Fwall: tcw = (6.0 m)(Fwall)(sin 60) = (6.0 m)(Fwall)(0.866) tcw = (5.2 m) Fwall 800 N: tccw = (6.0 m)(800 N)(sin 30) = 2400 m-N 400 N: tccw = (2.0 m)(400 N)(sin 30) = 400 m-N Fwall: t = 0 Ff: t = 0 S tcw = S tccw (5.2 m) Fwall = 2400 m-N + 400 m-N (5.2 m) Fwall = 2800 m-N Fwall = 538 N Ff = 538 N  = 0.45 8.55 A uniform, rectangular, 600 N sign 0.8 m tall and 2.0 m wide is held in a vertical plane, perpendicular to a wall, by a horizontal pin through the top inside corner and by a guy wire that runs from the outer top corner of the sign to a point on the wall 1.5 m above the pin. Calculate the tension on the wire and the force exerted by the pin.  Call the components of the force exerted by the pin Fx and Fy. The angle between the guy wire and the horizontal is  Start with the first condition of equilibrium, S Fx = Fx T cos 37 = Fx 0.80 T = 0; Fx = 0.80 T S Fy = Fy + T sin 37 600 N = 0 Fy + 0.60 T = 600 N This leaves us with two equations but three unknowns, so some additional information is required. Of course, we can get that by applying the second condition of equilibrium. We will calculate torques about the pin. This means the torque caused by the force of the pin is zero. Be careful of the torque for the 600-N weight. T: tccw = (2.0 m)(T)(sin 37) =(2.0 m)(T)(0.60) =(1.2 m)T Fx: t = 0 Fy: t = 0 W: tcw = (1.0 m)(600 N) = 600 m-N S tcw = S tccw (1.2 m) T = 600 m-N T = 500 N Fx = 0.80 T Fx = 400 N Fy = 600 N 0.60 T Fy = 600 N 0.60(500 N) Fy = 300 N 8.56 A 100 N child sits in the portable high chair shown in the figure. The child sits 20 cm from the top supports and 40 cm from the bottom supports. Determine the forces that hold the chair in place.  From the first condition of equilibrium, we have Ftop = Fbottom + 100 N Calculate torques about the position of the child, W: t = 0 Ftop: tccw = (18 cm)Ftop Fbottom: tcw = (45 cm) Fbottom Now apply the second condition of equilibrium, tccw = (18 cm)Ftop = (45 cm) Fvottom = tcw (18 cm)Fttop = (45 cm) Fvottom Ftop = (45/18) Fbottom Fttop = 2.5 Fbottom Ftop = Fbottom + 100 N 2.5 Fbottom = Fbottom + 100 N 1.5 Fbottom = 100 N Fbottom = 66.7 N 8.74 Figure 8.81, on page 310, shows a person lifting a 30 kg (or 294 newton) mass. When his back is horizontal, as in the sketch, what is the tension T in his back muscles and the compressive force C in his spinal disks?  Ooops, I forgot to give you values for w1 and w2; w3 is the weight of 294 N (the 30-kg mass that is being lifted). Read problem 8.73 for those values; w1 = 480 N and ww = 60 N. From the first condition of equilibrium we have Cx = T cos 12 = T (cos 12 = 0.9999978 1.00) and T sin 12 = 0.208 T = Cy + w1 + w2 + w3 0.208 T = Cy + 480 N + 60 N + 294 N 0.208 T = Cy + 834 N We now have two equations with three unknowns so we must get additional information from the second condition of equilibrium. We will calculate torques about the hip; that means the torque exerted by the compressive force C will be zero. C: t = 0 T: tccw = (48 cm) T (sin 12) = (48 cm) T (0.208) = (9.98 cm) T w1: tcw = (36 cm)(480 N) = 17 280 cm-N w2: tcw = (72 cm)(60 N) = 4 320 cm-N w3: tcw = (48 cm)(294 N) = 14 112 cm-N S tccw = S tcw (9.98 cm) T = (17 280 + 4 320 + 14 112) cm-N (9.98 cm) T = 35 712 cm-N T = 3 580 N Cx = 3 580 N 0.208 T = Cy + 834 N 0.208 (3 580 N) = Cy + 834 N 745 N = Cy + 834 N Cy = 89 N The negative sign here simply means that while we assumed that Cy pointed down, Cy actually points up.ZNDSET.HY |YB6* |D  "m  5 dNISIdNISI  wwww 1 dNISIdNISI  UUUU - dNISIdNISI  """" ) dNISIdNISI  "K "? >M #dNISI 1>M,"New Century Schlbk" .+J30dNISI   """"T X u>T "YdNISI 0u>T(?TdNISI  ?N"IdNISI 0?N(KTdNISI  DS"KdNISI 0DS+ rdNISI dNISIdNISI  "PdNISI 0( m = 2 kgdNISI  "dNISI 0(w = 20 NdNISI   aq0 "{H a~MaB0 \"{ a>R0 M"|  "}S UUUUa>qR0 M"{ a~0 "{" N] #dNISI 0N](ZTdNISI  Sb"ZdNISI 0Sb+ rdNISI dNISIdNISI  G]"_dNISI 0G](HTdNISI  X"RdNISI 0X+Pw = 20 NdNISI  "YdNISI 0(TdNISI  "dNISI 0+ rdNISI dNISIdNISI  "dNISI 0(cos 60dNISI  :ZIp"dNISI 0:ZIp(F[TdNISI  ?eNs"FedNISI 0?eNs+ rdNISI dNISIdNISI  9oH"KmdNISI 09oH(Epsin 60dNISI DSET.HY |%YB6* |D""' ' '"m %"xprWEwdxpr "WEWE" %WEWEp"New Century SchlbkWE q"2WE q"0WE q" WE q"N,"New Century Schlbk" .%%+ 20 NWEWE q"0WE q".WE q"8WE q"6WE q"6(0.866DSET.HY |3gB6* |D 5 5 5"md SPNT 3"3% P2-v15 - Copyright 1988 Silicon Beach Software, Inc. userdict/md known{currentdict md eq}{false}ifelse{bu}if currentdict/P2_d known not{/P2_b{P2_d begin}bind def/P2_d 27 dict def userdict/md known{currentdict md eq}{false}ifelse P2_b dup dup /mk exch def{md/pat known md/sg known md/gr known and and}{false}ifelse/pk exch def{md /setTxMode known}{false}ifelse/sk exch def/b{bind def}bind def/sa{matrix currentmatrix P2_tp concat aload pop}b/sb{matrix currentmatrix exch concat P2_tp matrix invertmatrix concat aload pop}b/se{matrix astore setmatrix}b/bb{gsave P2_tp concat newpath moveto}b/bc{curveto}b/bl {lineto}b/bx{closepath}b/bp{gsave eofill grestore}b/bf{scale 1 setlinewidth stroke}b/be {grestore}b/p{/gf false def}b/g{/gf true def}b g pk{/_pat/pat load def/_gr/gr load def}{/_gr {64.0 div setgray}b}ifelse sk{/_sTM/setTxMode load def}if/gx{/tg exch def}b 0 gx/x6{av 68 gt {false}if}b end P2_b pk end{/pat{P2_b gf{end pop sg av 68 gt{pop}if}{/_pat load end exec} ifelse}bind def}{/pat{P2_b pop _gr end}bind def}ifelse P2_b sk end{/setTxMode{P2_b/_sTM load end exec P2_b tg/_gr load end exec}bind def}{/setTxMode{pop P2_b tg/_gr load end exec}bind def}ifelse}if "g106 48 1 index neg 1 index neg matrix translate 3 1 roll currentpoint 2 copy matrix translate 6 1 roll "2344 355 currentpoint 1 index 6 index sub 4 index 9 index sub div 1 index 6 index sub 4 index 9 index sub div matrix scale 11 1 roll o[ 9 1 roll cleartomark 3 2 roll matrix concatmatrix exch matrix concatmatrix /P2_tp exch def P2_b mk end{bn}if "3d SPNT d SPNT  P2_b p end "D"D1$8d SPNT  P2_b g end  $$d SPNT d SPNT d SPNT BPRdd SPNT C 1$sn 8d SPNT d SPNT B0a3d SPNT B"ic?d SPNT B"gYd SPNT @dSPNT &>Z(d SPNT P2_b 0 gx x6 end 1 setTxMode ,! Avant Garde! .33(!20 kgd SPNT d&SPNT LA*} "THo#d SPNT #d&SPNT }&2@@ UUUU 7 P2_b sa end 3P2_b [0.99928 0 0 1.0062 0.63873 -0.45241 ] sb end P2_b p end "D"DAqH P2_b se end P2_b se end @@@qqqqpqqq@d SPNT dSPNT idSPNT P2_b g end  "pd SPNT d SPNT d&SPNT LA*|d SPNT BPd SPNT C 1 8d SPNT d SPNT B $dSPNTDSET.Hg |gB6* |D  "m PAd5a -0 "O# aq-0 v"V#% a0 "TF a*>0 "Q4K Pwa]q0 l" a`mt0 o"w a~0 "& a0 "c *@ #dNISI 0 *@,"New Century Schlbk" .++SdNISI  |2"4dNISI 0|2(40NdNISI  9O"dNISI 09O+# TdNISI  w"CdNISI 0w(TdNISI  wi" dNISI 0wi)hTdNISI  u"sdNISI 0u(TdNISI  c"dNISI 0c(d40 NdNISI  "ddNISI 0+4:200 NdNISI DSET.Hg |PgB6* |D R R R"md SPNT P UUUU"%% P2-v15 - Copyright 1988 Silicon Beach Software, Inc. userdict/md known{currentdict md eq}{false}ifelse{bu}if currentdict/P2_d known not{/P2_b{P2_d begin}bind def/P2_d 27 dict def userdict/md known{currentdict md eq}{false}ifelse P2_b dup dup /mk exch def{md/pat known md/sg known md/gr known and and}{false}ifelse/pk exch def{md /setTxMode known}{false}ifelse/sk exch def/b{bind def}bind def/sa{matrix currentmatrix P2_tp concat aload pop}b/sb{matrix currentmatrix exch concat P2_tp matrix invertmatrix concat aload pop}b/se{matrix astore setmatrix}b/bb{gsave P2_tp concat newpath moveto}b/bc{curveto}b/bl {lineto}b/bx{closepath}b/bp{gsave eofill grestore}b/bf{scale 1 setlinewidth stroke}b/be {grestore}b/p{/gf false def}b/g{/gf true def}b g pk{/_pat/pat load def/_gr/gr load def}{/_gr {64.0 div setgray}b}ifelse sk{/_sTM/setTxMode load def}if/gx{/tg exch def}b 0 gx/x6{av 68 gt {false}if}b end P2_b pk end{/pat{P2_b gf{end pop sg av 68 gt{pop}if}{/_pat load end exec} ifelse}bind def}{/pat{P2_b pop _gr end}bind def}ifelse P2_b sk end{/setTxMode{P2_b/_sTM load end exec P2_b tg/_gr load end exec}bind def}{/setTxMode{pop P2_b tg/_gr load end exec}bind def}ifelse}if "h120 120 1 index neg 1 index neg matrix translate 3 1 roll currentpoint 2 copy matrix translate 6 1 roll "O456 261 currentpoint 1 index 6 index sub 4 index 9 index sub div 1 index 6 index sub 4 index 9 index sub div matrix scale 11 1 roll o[ 9 1 roll cleartomark 3 2 roll matrix concatmatrix exch matrix concatmatrix /P2_tp exch def P2_b mk end{bn}if "%d SPNT  DD1y8d SPNT 1yP8d SPNT  "wd SPNT !xId SPNT d SPNT B0Zud SPNT B0=Xd SPNT d SPNT "Id"SPNT D"pad SPNT "gY#"gMdSPNT d SPNT dSPNT &'Td SPNT P2_b 0 gx x6 end 1 setTxMode ,! Avant Garde! .PP+O2 kgd SPNT dSPNT &+Qd SPNT P2_b 0 gx x6 end 1 setTxMode +7 kgd SPNT dSPNT &d SPNT P2_b 0 gx x6 end 1 setTxMode  (Y'Fd SPNT PJc8d SPNT "bd SPNT  UUUU"R!dSPNTDSET.Hg |[gB6* |DBB] ] ]"mdNISI  KOZ """""HdNISI 0KOZ,"New Century Schlbk" .[[+PW w = mg 20 NdNISI dNISIdNISI  [ 0$8fa8 L0 ["!B [aK9_M0 P"3C" [a 4 B0 ["*) [a|30 [")_'dNISIdNISI  Mc """"#dNISI 0Mc( NFdNISI  Ve" XdNISI 0 Ve + NdNISI dNISIdNISI  &"_dNISI 0& (#TdNISI dNISIdNISI  *5"# dNISI 0*5 +fdNISI  "1 "3dNISI 0"1 (.FdNISI dNISIDSET.Hg |gB6* |D  "m 02IMa)j=~0 ."t+ } #dNISI 0},"New Century Schlbk" .+~ 20 NdNISI  a%B9VB0 Q"/l M*f #dNISI 0M*f('N8 NdNISI  a7K,B0 '"AI />5 #dNISI 0/>5(;T = 8 NdNISI  ak0 "BuN x #dNISI 0x+I w = mg 70 NdNISI  a7`Kt0 F"j a5I0 "?O aH$\8B0 3"Rq Z%i; #dNISI 0Z%i;(f&FdNISI  c/n<"f0dNISI 0c/n< + fdNISI dNISIdNISI  |Qg"l4dNISI 0|Qg +"FdNISI  [h"\dNISI 0[h + NdNISI dNISIdNISI  ,;"ddNISI 1,; (8FdNISI DSET.H g |egB6* |Dg g g"md SPNT e UUUU"u% P2-v15 - Copyright 1988 Silicon Beach Software, Inc. userdict/md known{currentdict md eq}{false}ifelse{bu}if currentdict/P2_d known not{/P2_b{P2_d begin}bind def/P2_d 27 dict def userdict/md known{currentdict md eq}{false}ifelse P2_b dup dup /mk exch def{md/pat known md/sg known md/gr known and and}{false}ifelse/pk exch def{md /setTxMode known}{false}ifelse/sk exch def/b{bind def}bind def/sa{matrix currentmatrix P2_tp concat aload pop}b/sb{matrix currentmatrix exch concat P2_tp matrix invertmatrix concat aload pop}b/se{matrix astore setmatrix}b/bb{gsave P2_tp concat newpath moveto}b/bc{curveto}b/bl {lineto}b/bx{closepath}b/bp{gsave eofill grestore}b/bf{scale 1 setlinewidth stroke}b/be {grestore}b/p{/gf false def}b/g{/gf true def}b g pk{/_pat/pat load def/_gr/gr load def}{/_gr {64.0 div setgray}b}ifelse sk{/_sTM/setTxMode load def}if/gx{/tg exch def}b 0 gx/x6{av 68 gt {false}if}b end P2_b pk end{/pat{P2_b gf{end pop sg av 68 gt{pop}if}{/_pat load end exec} ifelse}bind def}{/pat{P2_b pop _gr end}bind def}ifelse P2_b sk end{/setTxMode{P2_b/_sTM load end exec P2_b tg/_gr load end exec}bind def}{/setTxMode{pop P2_b tg/_gr load end exec}bind def}ifelse}if "h149 140 1 index neg 1 index neg matrix translate 3 1 roll currentpoint 2 copy matrix translate 6 1 roll "d420 241 currentpoint 1 index 6 index sub 4 index 9 index sub div 1 index 6 index sub 4 index 9 index sub div matrix scale 11 1 roll o[ 9 1 roll cleartomark 3 2 roll matrix concatmatrix exch matrix concatmatrix /P2_tp exch def P2_b mk end{bn}if "ud SPNT dSPNT 0HHd SPNT cam"e     0` `0          {w<K @~>|  pH  Gic8dY+ ?  0 | pᅠa`p1` 0€0@c @b`` c`1@`!@!@ 1A  1A  A A `@  D  0 0` p<`@ @`0## #"u#"k #"*$#"% #"/ #"* #"*   *",! Avant Garde! .* 3.5 cm  MB#)3.0 cm  }H '#)45 cm  #)60 cm  " $02u#0Ns   '")42 cm b  Q" $ X Qd]me# X   " *y  > #)x  " `Z#`I]Z   KQ")M+ 1(  = 180(%X+ 1(% = 30 cm(>Y+1(> = 1.5 cm  G Kd#( M+ 2(  = 147(%X+ 2(% = 58.25 cm(>Y+2(> = 24 cm  ">Jq/737/3 33#q"#q@HD@HDq@*HD*H*@D !DDDSET.Hgx |!AgtB6* |D!A A! !A"D ?"xprWEwdxpr"WEWEWEWEPWEWEp"New Century SchlbkWE q"x,"New Century Schlbk" .??*xWEWE q"cWE q"g + cgWEWE q" WE q"=WE q" WE %WE ( = WE q"(WE q"1WE q"8WE q"0WE q")WE q"(WE q"3WE q"0WE q" WE q"cWE q"mWE q")WE q" WE q"+WE q" WE q"(WE q"1WE q"4WE q"7WE q")WE q"(WE q"5WE q"8WE q".WE q"2WE q"5WE q" WE q"cWE q"mWE q")( %(180)(30 cm) + (147)(58.25 cm)WE WE q"1WE q"8WE q"0WE q" WE q"+WE q" WE q"1WE q"4WE q"7+I 180 + 147WE q" WE q"=WE q" WE q"4WE q"2WE q".WE q"7WE q" WE q"cWE q"m( = 42.7 cmWEDSET.Hgp |!1glB6* |D!1 1! !1"D /"xprWEwdxpr"WEWEWEWEPWEWEp"New Century SchlbkWE q"y,"New Century Schlbk" .//*yWEWE q"cWE q"g + cgWEWE q" WE q"=WE q" WE %WE ( = WE q"(WE q"1WE q"8WE q"0WE q")WE q"(WE q"1WE q".WE q"5WE q" WE q"cWE q"mWE q")WE q" WE q"+WE q" WE q"(WE q"1WE q"4WE q"7WE q")WE q"(WE q"2WE q"4WE q" WE q"cWE q"mWE q")( %(180)(1.5 cm) + (147)(24 cm)WE WE q"1WE q"8WE q"0WE q" WE q"+WE q" WE q"1WE q"4WE q"7+A 180 + 147WE q" WE q"=WE q" WE q"1WE q"1WE q".WE q"6WE q" WE q"cWE q"m( = 11.6 cmWEDSET.Hgh |gdB6* |D  "D  ""t&(9{9{({(WW99p&(9{9{({(WW99"W""W "W "y"y ("( "( "" 9}"9 "9 "}"}"0"0"0"0 "0  T| #dNISI 0 T|,! Avant Garde! .j(U3.5 cmdNISI  -"!UdNISI 0-q(3.0 cmdNISI  "dNISI 0(45 cmdNISI  "dNISI 0(60 cmdNISI dNISI   a0   a+=H0 6 8-8 3GB] #dNISI 13GB] x(?HydNISI  "?PdNISI 1ɑ(xdNISI  R  X ~ "dNISI 0~ (x = 42.7 cmdNISI  "dNISI 0  +cgdNISI dNISIdNISI  "dNISI 1  (y = 11.6 cmdNISI  "dNISI 0  +cgdNISI dNISIDSET.Hg` |*'g\B6* |D,) ), ,)"DdNISI  *' EE E% wwww D a :!N0 *?"@ *'aYCmW0 ^'"MD *'a0 '"+ *'a0 *'") *'aB0 *'"  """"#dNISI 0,"New Century Schlbk" .*'*'+2.0 mdNISI  U$|"dNISI 0U$|(!V6.0 mdNISI  /6"0VdNISI 0 /6(,F(wall)dNISI  oM~v";dNISI 0oM~v+LO800 NdNISI  k"NdNISI 0k+M400 NdNISI  "ldNISI 0+(60dNISI dNISIdNISI  "dNISI 0+eFdNISI   "dNISI 0 +fdNISI dNISIdNISI   "dNISI 0  (FdNISI  "dNISI 0 + NdNISI dNISIDSET.HgX |!rgTB6* |D**!r r! !r"D p"xprWEwdxpr "WE WEp"New Century SchlbkWE q"WE q" WE q"=WE q" WE"WE,"New Century Schlbk" .pp* = WEPWEWE q"F( FWEWE q"f + fWEWEWEPWEWE q"F (FWEWE q"N + NWEWE q" WE q"=WE q" WE q" WE"?1WE (+ = WE q"5WE q"3WE q"8WE q" WE q"N(C538 NWEWE q"1WE q"2WE q"0WE q"0WE q" WE q"N(?1200 NDSET.HgP |gLB6* |DTT  ##0"4 19# 8# =6#qq"*#q7B7BB7q 9#q(0,(0,q<(G0<,G0G(<,">,L  )Y /","New Century Schlbk" .* 0.8 m  G#)1.5) m  #( 2.0) m \ " qyy"~J  {& ("( 600 N  " "qMYYMMY"T#qz8Az=A8z="|<   ")F+x  $#( F+y 9e " qs~su~s"t0  S "( TDSET.HgH | gDB6* |DLL   "  "xprWEwdxpr"WE!WEp"New Century SchlbkWE q"tWE q"aWE q"nWE q" WE pSymbolWE qq,"New Century Schlbk" .*tan WE q" , Symbol)qWE q"=WE q" WE" 4WE") = WE q"oWE q"pWE q"p( 4oppWEWE q"aWE q"dWE q"j+adjWE q" WE q"=WE q" WE" `(WE(P = WE q"1WE q".WE q"5WE q" WE q"m( `1.5 mWEWE q"2WE q".WE q"0WE q" WE q"m*2.0 mWE q" WE q"=WE q" WE q"0WE q".WE q"7WE q"5WE q";WE q" WE q" WE q" WE q" WE q" WE qq( = 0.75; WE q" )DqWE q"=WE q" WE q"3WE q"7WE q"") = 37DSET.Hg@ |g<B6* |D   " d SPNTdSPNT ]d SPNT d SPNT dSPNT 0d SPNT cam  ` ?o@000000888 08 008 0 0 0 0080x?0?sppppp000<00p Fp0 <0 <08's08g0~0900@0 00 0080 8898;8s8 a q y8? {x0 < , > ` {` ` ` 8` 0` 0` ` ` x         8 8 8 8 8 8 8 8 8 0 0 0 0 p 0 p p p p p ` ` ` ` ` ` ` ` ` ` ` ` `s{?d SPNT dSPNT &-Zd SPNT ,! Avant Garde! .+845 cmd SPNT dSPNT &DYd SPNT (W = mgd SPNT dSPNT &"*<d SPNT ({Fd SPNT dSPNT &)4Md SPNT + topd SPNT  1%<0P8d SPNT dSPNT &d SPNT (.?18d SPNT dSPNTDSET.Hg8 |q g4B6* |D;;s s s" 09 q  HHw`ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33ff33̙ff33̙̙̙̙ff̙33̙ffffffffffff33ff33333333ff333333ff33ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33ffffffffffff33ffffffff̙ffffff33ffffffffffffff33ffffffffffffffffffffffff33ffffff33ff33ff33ff33ffff3333ff33ffffffffffff33ff33333333ff333333333333̙33ff33333333333333ff33333333ff33ff33ff33ffff33ff3333ff3333333333333333ff333333333333333333ff333333ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33wwUUDD""wwUUDD""wwUUDD""wwwwwwUUUUUUDDDDDD""""""  ! %"  "!%%"            9HHw`ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33ff33̙ff33̙̙̙̙ff̙33̙ffffffffffff33ff33333333ff333333ff33ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33ffffffffffff33ffffffff̙ffffff33ffffffffffffff33ffffffffffffffffffffffff33ffffff33ff33ff33ff33ffff3333ff33ffffffffffff33ff33333333ff333333333333̙33ff33333333333333ff33333333ff33ff33ff33ffff33ff3333ff3333333333333333ff333333333333333333ff333333ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33wwUUDD""wwUUDD""wwUUDD""wwwwwwUUUUUUDDDDDD""""""90    %* 5= :=J< 5(%$!$'*2.## !$"#-&$!(!%)#"" !#$%!9zHHw`ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33ff33̙ff33̙̙̙̙ff̙33̙ffffffffffff33ff33333333ff333333ff33ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33ffffffffffff33ffffffff̙ffffff33ffffffffffffff33ffffffffffffffffffffffff33ffffff33ff33ff33ff33ffff3333ff33ffffffffffff33ff33333333ff333333333333̙33ff33333333333333ff33333333ff33ff33ff33ffff33ff3333ff3333333333333333ff333333333333333333ff333333ff33̙ff33ff33ffffffffffff33ff33333333ff333333ff33wwUUDD""wwUUDD""wwUUDD""wwwwwwUUUUUUDDDDDD""""""9z0q "%$               DSET.HY]l'`6*]lDSET\.HYSSUSp %S @ SSSeS5SP S  S! S!$u S$'E SVp & @ f6P    !! $v$ 'F'`6*]lDSET.Hg`0*`0DSETT (YYYYY(`>(> A( A & )Y)PHY 1150, Homework, Chapter 8, page  ZNFNTMCUTSDSUM/Charles E. Miller, Jr.HDNISTYL2@STYL((l((((|(t(pj        i                )   ! d"  # $ %  &  (  )  *  + !,  9".  #/  $0  %1 &2  '5  (6  #)8  *9  +: ! ,; " -<  .= # />  0A> 1B <2HASHd*d* d* d*d*d( $"- "-"-"-"-""/"/"/#-2#-i$-%$-"Ģ* %x1%p& Af*0B"--B"-B"-"B"/*B"/)B#-&B#-B#-$B#-4,B#-A!B#-H+B#-NB#-]B#-x#B#-y.B#-{B$-(B$-"'C(/D* L:Qhjazo}Bod$ CHAR(   "     S W \  W@ @ @%nAP 4k ;'lHASH          $&"3:!@O[jk#mR CELL(,0"HASH  GRPH(,.fn.HASHog& l RULR(x@..}$.l...D..}@$.l...D.... $.l...D..@$.l...D..}@..@..X HASH\^ Ü@@^ A  B CC LKUP  '- !3"#$%&47'(?)*+,-./@01 $NAMEDefault Default SSHeaderBodyFooterFootnoteFootnote IndexdDFNTM HelveticaGeneva"New Century SchlbkNew YorkSymbolETBL@FNTME?CUTSEGDSUMEOHDNIESTYLEETBLZ8