Exam (elaborations) TEST BANK FOR The Mechanics Problem Solver

Exam (elaborations) TEST BANK FOR The Mechanics Problem Solver PARTICLE VIBRATIONS .................................................. 851 23 RIGID BODY VIBRATIONS .............................................. 896 24 SYSTEMS HAVING MULTI-DEGREES OF FREEDOM ... 954 Coupled Harmonie Oscillators .............................................. 954 FOI'Ced Vibration .................................................................... 964. Normal Modes of Vibration and Natural Frequencies .......... 968 25 CONTINUOUS AND DEFORMABLE MEDIA ................... 983 Fluid Flow Problems •...•......••... ~ •.....•....•.•.......................•...... 983 S tting Problems ...................................................................... 99() 26 VARIA TIONAL METHODS ............................................. 1012 Method ofVirtual Work ...................................................... 1012 's :Eq_uations ........ ,. ..••..••••.••.•••••••..•.•....••...••........•.•. 1015 Hamilton' s Principle ..... ,. ....................................................... 1 ()4.8 Examp1es of Systems Subject to Constraints ..................... 1054 INDEX .............................................................................. 1068 xii IUNITS CONVEIRSION FAOORS This section includes a particularly useful and cornprehensive table to aid students and teachers in converting between systems of units. The problerns and their solutions in this book use SI (lnternadonal System) as weil as English units. Both of these units are in extensive use throughout the world, and therefore students should develop a good facillty to work with both sets of units until a single Standard of units has been found acceptable intemationally. In worldng out or solving a problern in one systern of units or the other, essentially only the numbers change. Also, the conversion frorn one unit systern to another is easily achieved through the use of conversion factors that are given in the subsequent table. Accordingly, the units are one of the least important aspects of a problem. For these reasons, a student should not be concemed rnainly with which units are used in any particular problem. Instead, a student should obtain from that problern and its solution an understandlog of the underlying principles and solution techniques that are illustrated there. To convart To MuHiply by acres ............................ square feet ··················~-···..... 4.356 x 10" acres ............................ square meters .......................... 4047 ampere-hours .............. coulombs ................................. 3600 ampere-tums ............... gilberts ..................................... 1.257 ampere-turns per cm. . • ampere-turns per inch ............. 2.54 angstrom units ............. Inches ................................ ~.... 3.937 x 1 o-8 angstrom units ............. meters ...................................... 1 o-10 atmospheres ................ feet of water ............................. 33.90 atmospheres ................ inch of mercury at O"C ............. 29.92 atmospheres ................ kilogram per square meter ....... 1.033 x 10" atmospherea ................ millimeter of mercury at O"C .... 760 atmospheres ................ pascals ..................................... 1.0133 x 1()5 atmospheres ................ poundaper square Inch ........... 14.70 bars .............................. atmospherea ............................ 9.870 x 1 o-r bars .............................. dynes per square cm. ............... 1 rl bare .............................. paacals ..................................... 1 ()15 bars .............................. pounds per square inch ........... 14.504 Btu ............................... erga .......................................... 1.0548 x 1 010 Btu ............................... foot-pounds ............................. n8.3 Btu ............................... joules ....................................... 1 054.8 Btu ............................... kilogram-calories ..................... 0.252 calorles, gram .............. Btu ........................................... 3.968 x 1 Q-3 calorles, gram .............. foot-pounds ............................. 3.087 calorles, gram .............. joules ....................................... 4.185 Celsius ......................... Fahrenheit ................................ (OC x 915) + 32 = °F xiü For the reverae, rnuHJply by 2.296x1~ 2.471 X 1()-4 x 1CJ-4 0.7958 0.3937 2.54 X 1rl 1010 0.02950 3.342x1~ 9.678x1~ 1.316x1Q-3 0:9869x1~ 0.06804 1.0133 1~ 1~ 6.8947x1~ 9.486 X 1D-11 1.285 X 1Q-3 9.480 X 1()-4 3.969 252 0.324 0.2389 ( 0 f- 32) X 5/9 : 0C Toconvert To Celllua ......................... kelvin ....................................... . centlmeters •••.•••••••.••.••. angstrom unlts ....................... .. centlmetera •••••••••••••.••.• feet .......................................... . centlstokea ................... aquare metera per second ...... .. circular mils ................. square centimeters ................. . circular mill ................. square mils ............................ .. cubic feet ..................... gallons (liquid U.S.) ................ . cubic feet ..................... rrtera ....................................... .. cubic inchel ................. cubic centimetn .................... . cubic Inches ................. cubic feet ................................ . cubic Inches ................. cubic maters .......................... .. cubic inches ................. gallons (liquid U.S.) ............... .. cublc meters ................ cublc feet ................................ . cubic msters ................ cubic yards ............................. .. curies ........................... coulombs per minute ............. .. cyclea per second ........ hertz ....................................... .. degrees (angle) ............ mils ........................................ .. degrees (angle) ............ radians ................................... .. dynes ........................... pounds ................................... .. eledron volts ............... joules ...................................... . erga ........................... m foot-poundl ............................ . ergs .............................. joules ...................................... . ergs per second ........... watts ...................................... .. ergs per square cm. ..... watts per square cm ............... .. Fahrenheit .................... kilvln ............................... ._ .... .. Fahrenheit .................... Rankine ................................. ... faradays ....................... ampere-hours ......................... . fall............................... centimeters ............................ .. feet ............................ ... meters .................................... .. feet ............................... mils ......................................... . fermis ..................... ;..... meters .................................... .. foot candles ................. Iux .......................................... .. foot lamberts ................ candela& per square meter ..... .. foot-pounds ................. gram-centlmaters .................... . foot-pounds ................. horsepower-hours .................. .. foot-pounds ................. kilogram-rneters ..................... .. foot-pounds ................. kilowatt-hours ......................... . foot-pounds ................. ounce-inches .......................... .. gallons (liquid U.S.) ..... cubic meters ........................... . gallons (liquid U.S.) ..... gallona (liquid British Imperial) gamnlll ....................... teala.l .................••.•...•.............•. gausaes ........................ llnes per square cm ................ .. gausaes ........................ lines per square Inch .............. .. g8UII81 ........................ ......................................... . gausses ........................ webers per square Inch .......... .. gilbarta ......................... amper• .................................. . grada ........ ..................... radians .................................... . ........................... _. ...... -.............................. . grains ........................... pounds .................................. ... grama ................•...... _, dynea ..........•............................ grams ................... ......... gllins ....................................... . xiv llultlply -c +273.1= K 1 X 101 0.03281 1 X 10""1 5.067x 1~ 0.7854 7.481 28.32 16.39 5.787x 10""4 1.639 X 1 ()-1 4.329 X 1()-3 35.31 1.308 1.1 X 1012 1 17.45 1.745x ~ 2.248 X 1()-1 1.602 X 1o-11 7.376x1D-I 1~ 1o-7 10""3 (•F + 459.67)11.8 •F + 459.67 = •R 26.8 30.48 0.3048 1.2x10" 1o-15 10.764 3.4263 1.383x 10" 5.05x 1o-7 0.1383 3.766x1o-7 192 3.785x 10""3 0.8327 1o-' 1.0 6.452 10"-4 6.452 x1D-I 0.7958 1.571 x1~ 0.06480 ,, 1000 980.7 15.43 For the rev.,.., mutuply by K- 273.1 = -c 1 X 10""1 30.479 1 X 101 1.973 X 105 1.273 0.1337 3.531 x1 ~ 6.102 x1 ~ 1728 6.102 X 10" 231 2.832x1~ 0.7646 0.91 X 1o-12 1 5.73 x1~ 57.3 4.448 X 105 0.624x1011 1.356x 107 10" 107 101 1.8K- 459.67 •R- 459.67 = •F 3.731 1~ 3.281 x1~ 3.281 8.333x to-6 1015 0.0929 0.2918 1.235 X 1 ()-1 1.98 X 1<JII 7.233 2.655 X 101 5.208 X 1()-3 264.2 1.201 HJII 1.0 0.155 10" 1.55 X 107 1.257 63.65 15.432 7000 1.02 X to-3 6.481 x1~ Toconvert To Munlply gramt........................... ouncet (avdp) .......................... 3.527 x 1«t-2 grams ........................... poundals .................................. 7.093 x 1«t-2 hectares ....................... acrn ........................................ 2.471 hol'l8p0wer .................. Btu per minute ......................... 42.418 horaapower .................. foot-pounds per minute ........... 3.3 x 104 horaapower .................. foot-pounds per second ........... 550 horaapower .................. horsepower (metric) ................ 1.014 horaapower .................. kilowatts................................... 0.746 Inches .......................... centimeters .............................. 2.54 Inches .......................... feet ....................... .................... 8.333 x 1 (t-2 Inches .......................... meters...................................... 2.54 x 1«t-2 Inches .......................... miles ............ ............................ 1.578 x 1 Q-1 Inches .......................... mill .......................................... 1 0' Inches.......................... yards ........................................ 2.n8 x 10"2 joules .. ......................... foot-pounda ............................. 0. 7376 joules ........................... watt-houra................................ 2.n8 x 1~ kilograms ..................... tons (long) ............................... 9.842 x 1~ kilograma ..................... ton1 (ahort) .............................. 1.102 x 1G-3 kllograms ..................... pounds (avdp}.......................... 2.205 kilometerB .................... feet .............................. m.......... 3281 kllometers .................... Inches ...................................... 3.937 x 1 04 kilometers per hour •..... feet per minute ......................... 54.68 kilowatt-hours .............. Btu ........................................... 3413 kilowatt-hours .............. foot-pounds ............................. 2.655 x 1rß kilowatt-hours .............. horaepower-hours.................... 1.341 kilowatt-hours .............. jaulet ....................................... 3.6 x 1rß knots ......................... ._ feet per second ........................ 1.688 knots ............................ miles per hour .......................... 1.1508 lamberts ....................... candlel per square cm. ............ 0.3183 lamberts .............. ._...... candles per square Inch ........... 2.054 Iiters ............................. cubic centimetera .......... ........... 1 (J1 Iiters ............................. cubic Inches............................. 61.02 Iiters ............................. gallona (liquid U.S.) ................. 0.2642 Iiters ............................. pints (liquid U.S.) ..................... 2.113 Iumens per square foot foot-candles ............................. 1 Iumens per square meter foot-candles ............................. 0.0929 Iux ................................ foot-candles ............................. 0.0929 maxweils ...................... kilolines .................................... 1 G-3 maxweils ...................... webers ..................................... 1 ~ meters .......................... feet ....... .................................... 3.28 meters .......................... inches ...................................... 39.37 meters .......................... mllea ........................................ 6.214 x 1~ meters .......................... yarda ........................................ 1.094 miles (nautical) ............ feet ........................................... 6076.1 milea (nautical) ............ meters ...................................... 1852 miles (statute) .............. feet ........................................... 5280 milea (statute) .............. kilometers ................................ 1.609 miles (statute) .............. miles (nautlcal) ........................ 0.869 miles per hour .............. feet per second ........................ 1.467 miles per hour .............. knots ........................................ 0.8684 millimeters ................... microns.................................... 101 XV For the reveru, muHiplyby 28.35 14.1 0.4047 2.357 X 1(t-2 3.03x 1()-1 1.182x1~ 0.9863 1.341 0.3937 12 39.37 6.336 X 104 1G-3 36 1.356 3600 1016 907.2 0.4536 3.408x1~ 2.54x 1()-1 1.829 X 1 (t-2 2.93x1~ 3.766x10'"7 0.7457 2.n8x10'"7 0.5925 0.869 3.142 0.4869 1G-3 1.639 X 1(t-2 3.785 0.4732 1 10.764 10.764 101 101 30.48 X 1(t-2 2.54 X 1()-2 1609.35 0.9144 1.646x1~ 5.4x1~ 1.894x1~ 0.6214 1.1508 0.6818 1.152 1G-3 CHAPTER1 STATICS REPRESENTING FORCES BY VECTORS • PROBLFM 1-1 + Express the force f as shown in the diagram in terms of " " the basis l, j and also in terms of e1, e2. 2 1 I ·~ l-- ~ + Solution: The vector f can be written (1) where f 1 and f 2 are components of f along the coordinate axes. ,. ,.. Assuming that i and j are orthogonal unit vectors + fl = l . i = f cos a (2) + + f2 = f . j = f sin a (3) + where f - magnitude of f and e is the angle between f ,.. and i. (8 = 30° in this example.) + ,.. ,. f = (1:3 i + j) lbs. (4) + f can also be written in terms of the basis vectors e1 and e2. This is not an orthogonal set of unit vectors, so that simple result used above to evaluate the components cann<t be used in the general case. However, since flies alon· one of the basis vectors, the result follows by inspection. Writing 1 f ,... ,... = m1 e 1+ ffi2 e2 , (5) we see that ffil = 0 and that mz = f -... le2l (6) Since the length of e2 is not specified, we cannot evaluate m2 further. • PROBLEM 1-2 + A force f is expressed with respect to the basis e1, e2 by the equation f = 2(3;1 + ,... 4e2) lb. If the directions of ,... er and f are known as shown in the Figure, find the vector ... ff=lO lb Solution: The vector e 2 is shown in Figure 1. The construction of Fig. 1 follows from the formula and the figure given in the problem. Point C is constructed so sides BC and AC of the triangle ABC intersect at right angles. In the smaller triangle (OBC) the side DC is given by (all units are lbs), OC = AC - 6 = 10 cos 30° - 6 = 2.660. ~ S ~ The side BC is given by, BC = 10 sin 30° = 5. ~-$--' --f~ ct> I '~,~ e ' The side OB is given by, 2 '6" OB = I(BC) 2 + (oc} 2 = 5.664 1 el 8 66 10 I • ... 1~~~ Since OB = 8 !e2l we have, ,... 5.664 Fig. le2l = = o. 708 8 We can write e2 on the (i, j) basis (defined in Figure 2) as 2 " j .7 unit length length Fig. 2 = - " =- {0.708) {0.883) i + {0.708) {0.470)j " e 2 =- 0.625 i + 0 . 333 j. • PROBLEM 1-3 A telephone pole in a rural area supports a wire which carries a force of 50 lbs directed along the wire. The wire configuration is indicated in the figure shown below. Find the force on the pole from the wire. 3 k 2 F 1=50 lb i Solution: Forces Ft and F2 must be put into vector form us~ng x, y. and z coordinates before they can be added to + find the total force on the pole. We can write ~directly as + " Ft = 30j - 40k. (1) F2 has three components which can be found from vector laws as follows: + + + + F2 = (x)i- (3x)j - {2x)k {2) + IF2 I =50= {x 2 + 9x 2 + 4x2 X = 13.361 + + + + giving F2 = 13.36i - 40.08j - 26.72k {3) + + Adding F1 and F2 gives the force on the pole as + + + + F = (13.36i - 10.08j - 66.72k)lb. (4) J • PROBLEM 1-4 Find the resultant of the three forces acting at 0 by computing the magnitude and direction of the resultant. P•50 N ·Y P = <5o c o s 2 5•) t y P•SO~,---- ~~~-~~~~--~-i-nlt0°)l ' : Q•160N Q~160 N I I N t S X., ( 1 1 0 CO S 5 0°)1 t Flg. ---- S=11 ON • SY={110 sln50°)1 F•g. 2 Solution: Each of the three force vectors is resolved into its x and y-components using the method shown in Figure 2. The x and y-components of force P is determined as follows: Px =-(50 sin 25°)i = (-21.1N)i Py = (50 cos 25°)j = (45.3N)j For force Q; Qx = (160 cos 40°)i = (122.6N)i and Qy = (160 sin 40°)j = (102. SN) j For force S; sx = (110 cos 50°)i = (70. 7N) i and sy = -(110 sin 50°)j = -(B4.3N)j The total resultant is a summation of all the x and ycomponents of the three forces acting on A Thus R = R + R X y = (Px + Qx + Sx)i + (Py + Qy + Sy)j = (-21.1 + 122.6 + 70.7)i + (45.3 + 102.8- 84.3)j = ( 1 72 • 2N) i + ( 6 3. SN) j Notice that both the x and y-components of the resultants are represented by positive numbers. Therefore, Rx will act to the right and R will act upwards as shown in Figure 3. y Ry•(6J,8N)I~ Rx•(172.2N)I Fig. 3 4 The magnitude and direction of the resu1tant R can be determined from Figure 3. or tan 9 sin9 R =-:L R R = 184N 63. SN = 172. 2N 0. 371 and 9 = and R = l= 63.8N sin9 sin 20.4 • PROBLEM 1-5 One end of a rod is fixed to a point s, whi1e the other end P, carries a 200 1b. weight. Find (a) the moment of the force about S due to the weight; (b) the sma1lest force at P which when app1ied resu1ts in the same moment about point S; (c) the magnitude of the horizontal force which when applied produces the same moment about S. Fig. Solution: (a) The moment of the force about s, M8 is given as M8 = Fd, where F is the force and d is the perpendicular distance from s to the line of action of the 200 lb-force. From Figure 1, d is found from trigonometry COS 30° d = 30in and d = 30cos30° = 26.0in Thus M8 = (200 lb) (26.0in) = 5200 1b·in acting i~ the clockwise direction (b) The smallest force to produce the same moment as in part(a)occurs when the perpendicular distance is a maximum. F Flg. 2 5 Referring to Fig. 2, d = 30in From the relation M5 = Fd (c) To determine the magnitude of the horizontal force refer to figure 3. In this case M5 = 5200 lb·in 5200 lb·in = Fd From figure 3, sin60° = d 30 or d = 15in. Thus, ,----•F Fig. 3 and d = 30sin60° F 5200 lb in = 347 lb lSin = • PROBLEM 1-6 Find the moment M about P of a lOOON force acting on 0 point A as shown in Figure 1, using vector algebra. F " j=1000 y M 0 I I t-0.20m-44 I cos20° I Fig. 1 Solution: The moment of the lOOON force about P is given as the vector product of Kr and F. Thus (1) 6 The position and force vectors are now resolved into reetangular x, y, and z components. Thus lr = 6x1 + 6yj = (0.20m)1 + (O.lSm)j and ,.. = (939.7N)1 + (342N}j From equation 1 M = [(0.20m)i + (O.lSm)j] x [(939.7N)1 + (342N}j] 0 = (68.4N.m)k- (140.9N.m)k = -(72.SN.m)k According to convention the moment R is pointing into the 0 plane of the paper since it has a negative value. • PROBLEM 1-7 A square concrete slab acts as an anchor for two cables. These cables exert the 100 lb and SO lb forces indicated in the Figure. Determine the moment about the corner 0 for these forces. lOOlb ini 1 ,i2 plane Solution: Since moments are vectors, the sum of several moment vectors acting at the same point is obtained by a vector sum. In this case, .... .... .... mo = m& + m~ where m& is the moment of the 100 lb force, and m~ is the moment of the SO lb force (conveniently determined with the definition). ? ... The SO lb force acts in the i1 direction. Its moment about 0 is ... ... ... li 3 x 50i 1 • SOis. The other moment,m0,can be determined usinq the ... components of the 100 lb force which are in i 1 direction: 100 lb (cos 45°) = 70.7 lb ... in i2 direction: 100 lb (ein 45°) = 70.7 lb. ... il i2 i3 + mo = 2 0 0 + 50i2 70.7 70.7 0 = (141.4!, + so12)in.-lb. • PROBLEM 1-8 A model iceboat is beinq studied to determine the effect of various riqqinq chanqes. When the boat sails with the wind cominq over the left side, only the riqht runner and the steerinq runner touch the ice. Durinq one test four of the forces actinq on the boat were found to have the averaqe values indicated on the sketch. Find the coup1e vector which represents their turninq effect. Fiq. 1 Fiq. 2 Solution: Choose the riqht-handed orthonormal basis i1, !2, is. We select the position vectors beeween the 20 1b forces and between the 10 1b forces so that the po- sition vectors will have simple component representations. ~ = (t X f)lOlb + (t x f)20lb For the 10 lb forces(from Fiqure ~' 8 r2 = 6 ft, f2 = 0 r, = 0, f, = 10 lb For the 20 lb forces, r1 = - 3 ft, fl = 0 ao r2 = 0, f2 = 20 lb r, = 5 ft, f.3 = 0. ,.. " ,.. " ,.. t> il i2 i3 il i2 1.3 So e = 0 6 0 + - 3 0 5 0 0 10 0 20 0 " " " = 60it - lOOit - 60i3 + " ,.. c = (- 40il - 60i3)ft-lb. • PROBLEM 1-9 A winch-equipped bulldozer is trying to lift a pile of scrap meta! from a scrap pile. The bulldozer is lifting one part of the meta! with the lip of its shovel and is pulling another with the winch. If the loads from the bulldozer can be idealized as drawn in Figure 2, determine the moment of these loads about the point o. Fig. 1 ~ B Solution: + + + + + m0 = (a x f) +(b x G) Fig. 2 G=lOOO lb from blade F=450 lb ~--------~~Tfrom winch + + where a and b are the vectors from 0 to points A and B respectively. We choose the right-handed orthonormal basis shown for the mornent vector space at o. Any position vectors can be selected to the forces, but we select ones which give convenient components with respect to a basis parallel to il, i2, i3. 9 So From Figure, 2 for the 1000 lb force: bl = 10 ft, Gl = 0 b2 = 0, G2 = 1000 lb b, = 0, G, = o. For the 450 lb force: a1 = 0, fl = 300 lb a2 = 0, f2 = 300 lb a, = 5 ft, f, = - 150 lb. il i2 i, il i2 i, mo = 10 0 0 + 0 0 5 - 150 = il (0) - i2 (0) + i3 (10,000) + il (- 1500) - i2(- 1500) + i3(0) mo = (- 1500il + 1500i2 + 10,000i 3)ft-lb. EQUILffiRIUM OF A PARTICLE • PROBLEM 1-10 Determine the magnitude and direction o~ the resultant of forces F 1 and F2 acting at point S in Figure 1, using a trigonometric solution. s s Fig. 1 Flg. 2 Solution: The resultant of the two forces F 1 and F 2 is sketched as shown in Fig. 2 using the law of cosines. It's magnitude is determined using the relation 10 From inspection angle R2 = (BON) 2 ~ (50N) 2 - 2 (BON) (50N) cos 150° R = 125.8N The law of sines is now used to determine the direction of the resultant Thus, and sinS sinA ~=tr sinS sin 150° BON = 125.BN sin S (SON) sin 150° = = 125.8N s = 1B.5°. Angle 41 is thus 15° + S Thus, 41 = 33.5°. 0.3179 The resultant is thus 125.8N ~ 33.5° • PROBLEM 1-11 The resultant of two forces, A and B, acts vertically downard with a magnitude of 2700N. Calculate the magnitude and direction of force B if A is a 500N force. A "' SOON B Flg. 1 11 Solution: In Fig. 2, forces A and B are resolved into their x and y components. Bx • 433N R • 27DON Bx By • 2950N Flg. 2 Flg. 3 - Bx - (1) -A + = Rx X - - -r (2) A.x - B = y Using trigonometry Ax = F cos 30° = (SOON) (0.8660) = 433N A.x = F sin 3QO = (500N) (0. 50) = 250N But Rx = 0 since it acts vertically downward, thus substituting for Ax' equation (1) becomes -433N + Bx = 0 or Bx = 433N Also equation (2) becomes, 250N - By = -2700N or Using Figure 3, B = 2950N y 12 tane Bx 433N =- = = 0.1468 By 2950N and 9 = 8.4° The magnitude of s is IBI=/Bx'- + B 2 = /(433N) 2 + 2950N) 2 = 2982N y • PROBLEM 1-12 In Fig. 1, a 50- N tension is required to maintain the box B in equi1ibrium with force F. Ca1cu1ate the maqnitude of F given that d = lOcm and r = Sem. 50 N TensIon I I I I ,d I I r-----.,--.J ..----~ B Fig. 1 Solution: The free body diagram of the box is shown in Fig. 2. It accounts for all forces acting on the box. Since only F is required, it is sufficient to co~sider only the xdirection. It is given that the box is in equilibrium, thus the summation of all the forces in the x-direction must be zero. I I d = 10cm I F..-----l B -F + (SON) (sin9) = 0 Fig. 2 (1) lJ From trigonometry, sin6 = r l!oz + rz substituting for sin6 in equation (1) gives F = (SON) r But r = Sem F = (SON)( Sem ) /10 2 + 5 2 = (SON)( S ) 1100 + 25 or F = 22.4N • PROBLEM 1-13 An 80 1b 1id to a sewage tank (Fig. 1) is he1d by three cab1es attached to a fork 1ift at point P. Each cab1e is 5 ft 1ong, and the cab1es form ang1es of 6 1 = 160°, 6 2 = 110° and 6 3 = 90° with one another. Ca1cu1ate the tension in each cab1e, given the radius of the 1id as 3 ft. Q. I ' ' ·z ' X Fig. 1 14 Solution: This kind of 3-dimensional problem demands systematic treatment. Consider a top view, place the x, y-axes through P parallel to OR and OS respectively, since their projections are already perpendicular. s 0 0" .... R X ao 3 N 1.03 Flg. 2 Figure 2 is a free body diagram of the "knob" where the three cables join at the top. Actually it is a 2-D projection of the free-body diagram. The vertical configuration of each of the wires is the same, and is shown in figure 3. Flg. 3 Direction cosines for vectors Q, S, R are Vector 1 = x/5 m = y/5 n = z/5 0 1.03 2.819 4 -s- 5 - 5 s 0 + ~ 4 5 - 5 R +~ 0 4 5 - 5 These direction cosines are obtained from the coordinates of the ring ends of the wires relative to the overhead 15 "knob" (origin) . The same direction cosines apply to the desired tensions. Equilibrium requires IFx = 0 IFy = 0 IF z = 0 since We have, 010 + s15 + Rl = R 0 QmQ + sm5 + ~= 0 QnQ + sn5 + Rn = R 80lb putting in numerical values and multiplying through by the denominator, 5ft, given -1.030 - 0 + 3R = 0 -2.8190 + 3S + 0 = 0 -40 - 4S - 4R = -(80) (5) from the last of these we have 0 + S + R = 100 and, from the first two 3 0 = I:öJ R = 2.9R S = 2.8i90 = 2.;19 (2.9R) = 2.725R substituting these last two into the immediately preceding ~uation gives (2.9R) + (2.725R) + R = 100. Therefore R = lOO = 15.09 lb 1 + 2.725 + 2.9 16 and Q = 2.9R = 2.9(15.09) = 43.76 lb S = 2.819R = 42.54 lb • PROBLEM 1-14 A 4000-lb force is kept in equilibrium by two cables PS and RS as shown in Fig.l. Cable PS makes a 10° angle with the vertical, while cable RS makes a 40° angle with the horizontal. Calculate the tension in the cable RS. p PS A R Fig. 1 RS Solution: The free body diagram is shown in Fig. 2. From !J. BCD, we observe that 41 = 50 °. Since a. is the supplement of 41, it must equal 130 o. e = 40° From the law of sines PS RS sin 130° sin 10° = 4000 lb sin 40° PS= 4000 lb(sin 130) 0 = 4000 lb(0.7660)_ 4766 lb sin 40° 0.6428 - RS = 4000(sin 10°) = sin 40° 1080 lb Fig. 2 • PROBLEM 1-15 In Fig. 1 a 20 ft-frame PQ is supported at two points L and M, 6 ~t and 4 ft respectively from the edges. If a 300 lb load ~s attached to edge Q, determine the range of load W that must be placed at P to keep the frame in equilibrium. 17 I i' 1: I •• rlt '3--1 J L " I 2S 2S b 30! lb I I Q3 Fig. 2 Fig. 1 Solution: The conditions for equilibriurn of the beam are: Net force acting on beam = 0 Net torque acting on beam about any axis = 0 First, identify the forces acting on the beam. (See Figure 1.) We are told to ignore the mass of the beam, therefore, there is no gravitational force acting on it. Recall that the torque exerted by a given force lying in the plane of the paper, about a rotation axis perpendicular to the paper is qiven by tan = rF sine, where r is the vector fram the rotation axis to the line of action of the force, and 9 is the angle between the force F and r. Let's arbitrarily define torquesthat cause a counterclockwise rotation to be positive. Consider two cases: a) W is large enough to cause the beam to rotate counterclockwise; b) W is small enough so the beam will rotate clockwise. a) Suppose that W is large enough to just turn the beam counter-clockwise about the axis through support point L. When W has the required magnitude the force at support M will go to zero, i.e., FM+ 0. Now, using quantities defined in Figure 1 and equation 2) we have 18 TL - Net torque about axis throuqh L (Note that the force FL produces no torque since it acts alonq a line throuqh L.) Since e1 = e2 = 90° we have If TL~ 0, the beam will not rotate counter-clockwise. This allows an upper limit to be placed on W: w < 300 lb = 300 lb 14 ft 6 ft = 700 lb. b) Think of w decreasinq in value until the beam just beqins to rotate clockwise. The force exerted on the beam by the support at L will 90 to zero at this value of W. The net torque about an axis throuqh support M will be: (refer to Figure 2) The beam will not rotate clockwise, provided TM~ o. We can now place a lower limit on W. TM = r W - r (300 lb) > 0 r3 w ~ (300 lb) - = r .. 4ft (300 lb) 16 ft = 75 lb. Therefore, the ranqe of values for W that will leave the beam in equilibrium is 75 lb < w < 700 lb. EQUIV ALENT FORCE SYSTEMS e PROBLEM 1-16 The cube shown in Fiq. 1 has applied to it two couples. Represent these couples by a sinqle equivalent couple. 19 10 1 b. 8 lb. ~ , ....4-- 5 In ---a .. ~;p' Flg. 1 Solution: A couple vector which is perpendicular to the plane of the couple is used to represent each of the two couples. The two couple vectors are shown in Fig. 2 with respect to a set of coordinate axes. 60 z Flg. 2 The magnitude of the resultant of the two cou