PROBLEM 13-27 (page 126, 12 th edition) The mass of block A is 100 kg. The mass of block B is 60 kg. The coefficient of kinetic fiction between block B and the inclined plane is 0.4. A and B ae eleased fom est. Detemine the acceleation of block A and the tension in the cod. Neglect the mass of the pulleys and the cod.
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13-27 (page 126, 12 th edition) COMPLETION OF PROBLEM Data m A =100 kg m B = 60 kg g = 9.81 m/s 2! = 60 o µ = 0.4 Diection of motion Because of the pulley system, A moves up and B moves down the plane. Equations of motion Block A! F y = ma y : 3T! m A g = m A (1) Block B! F x = ma x m B gsin! " T " µn = m B (2)! F y = ma y N! m B gcos" = 0 (3) Dependent motion = 3 (4) Solution = [ 3m ( sin! " µ cos! ) " m B A ]g = 0.305 m/s 2 (5) m A + 9 m B Pulley system T = m A (g + ) 3 = 337 N (6) The facto 3 in the numeato of (5) indicates the multiplie effect of the pulley system; the smalle mass is able to lift the lage mass.
PROBLEM 13-41 (page 129, 12 th edition) A hoizontal foce P = 20 lb is applied to block A. The coefficients of kinetic fiction between block A and the hoizontal suface, between the two blocks, and between block B and the vetical suface ae 0.1, 0.2, and 0.3, espectively. Detemine the acceleation of each block and all nomal foces.
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13-41 (page 129, 12 th edition) COMPLETION OF PROBLEM (page 1) Data W A = m A g = 8 lb W B = m B g = 15 lb g = 32.2 ft/s 2 P = 20 lb! = 15 o Acceleations µ A = 0.1 µ AB = 0.2 µ B = 0.3 = i = j Relative motion: = + /A /A = /A (! cos" i + sin" j) Catesian components of : x component y component 0 =! /A cos" = /A sin! Then, = tan!
13-41 (page 129, 12 th edition) COMPLETION OF PROBLEM (page 2) Equations of motion Equations of motion fo A :! F x = ma x : P! N sin"! µ AB N cos"! µ A N A = m A (1)! F y = ma y :!N cos" + µ AB N sin" + N A!W A = 0 (2) Equations of motion fo B :! F x = ma x : N sin! + µ AB N cos! " N B = 0 (3)! F y = ma y : N cos! " µ AB N sin! " µ B N B "W B = m B N cos! " µ AB N sin! " µ B N B "W B = m B tan! (4) The above 4 equations contain 4 unknowns: N A N N B The poblem can be solved.
13-41 (page 129, 12 th edition) COMPLETION OF PROBLEM (page 3) SOLVING THE EQUATIONS OF MOTION Data W A = m A g = 8 lb W B = m B g = 15 lb g = 32.2 ft/s 2 P = 20 lb! = 15 o µ A = 0.1 µ AB = 0.2 µ B = 0.3 Equations to be solved m A + (sin! " µ AB cos!)n + µ A N A = P (1) (!cos" + µ AB sin")n + N A = W A (2) (sin! + µ AB cos!)n " N B = 0 (3)!m B tan" + (cos"! µ AB sin")n! µ B N B = W B (4) Numeical esults (using ef ([M ]) whee [M ] is a 4! 5 matix) = 26.0 ft/s 2 N A = 29.4 lb N = 23.4 lb N B = 10.6 lb = 6.97 ft/s 2 /A = 26.9 ft/s 2
PROBLEM 13-46 (page 130, 13 th edition) PROBLEM 13-37 (page 128, 12 th edition) The diagam shows two tiangula blocks A and B each with mass 2 kg. B is on a hoizontal suface. The sloped suface of B makes an angle! = 40 o with the hoizontal suface. The coefficient of kinetic fiction between B and the hoizontal suface is 0.5. The coefficient of kinetic fiction between the two blocks is 0.2. A hoizontal foce P = 50 N acts to the ight on A. (Not to the left on B as shown in the diagam.) Detemine the acceleation of each block and the nomal foces.
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13-46 (page 130, 13 th edition) 13-37 (page 128, 12 th edition) COMPLETION OF PROBLEM (page 1) Data m A = 2 kg m B = 2 kg W A = m A g W B = m B g g = 9.81 m/s 2 P = 50 N! = 40 o µ B = 0.5 µ AB = 0.2 Acceleations = x i + aay j = i Relative motion: = + /B /B = /B (cos! i + sin! j) Catesian components of : x component y component x = + /B cos! y = /B sin!
13-46 (page 130, 13 th edition) 13-37 (page 128, 12 th edition) COMPLETION OF PROBLEM (page 2) Equations of motion Equations of motion fo A :! F x = ma x : P! N sin"! µ AB N cos" = m A x P! N sin"! µ AB N cos" = m A ( + /B cos") (1)! F y = ma y : N cos! " µ AB N sin! " W A = m A y N cos! " µ AB N sin! " W A = m A /B sin! (2) Equations of motion fo B:! F x = ma x : N sin! + µ AB N cos! " µ B N B = m B (3)! F y = ma y :!N cos" + µ AB N sin" + N B! W B = 0 (4) The above 4 equations contain 4 unknowns: /B N N B The poblem can be solved.
13-46 (page 130, 13 th edition) 13-37 (page 128, 12 th edition) COMPLETION OF PROBLEM (page 3) SOLVING THE EQUATIONS OF MOTION Data m A = 2 kg m B = 2 kg W A = m A g W B = m B g g = 9.81 m/s 2 P = 50 N! = 40 o µ B = 0.5 µ AB = 0.2 Equations to be solved m A + m A /B cos! + (sin! + µ A/B cos!)n = P (1)!m A /B sin" + (cos"! µ A/B sin")n = W A (2)!m B + (sin" + µ AB cos")n! µ B N B = 0 (3) (!cos" + µ AB sin")n + N B = W B (4) Numeical esults (using ef ([M ]) whee [M ] is a 4! 5 matix) = 4.86 m/s 2 /B = 5.03 m/s 2 N = 40.9 N N B = 45.7 N x = 8.71 m/s 2 y = 3.23 m/s 2 = 9.29(cos20.4 o i +sin20.4 o j) m/s 2
PROBLEMS 13-53 and 13-54 (page 138, 12 th edition) A 1700 kg spots ca tavels aound a 100 m adius hoizontal cicle on a 20 o banked tack. The coefficient of static fiction between the ties and the oad is 0.20. 13-53: Detemine the maximum constant speed the ca can tavel without sliding up the slope. 13-54: Detemine the minimum constant speed the ca can tavel without sliding down the slope.
13-53 and 13-54 (page 138, 12 th edition) COMPLETION OF PROBLEMS (page 1) Data m = 1700 kg g = 9.81 m/s 2! = 20 o! = 100 m µ = 0.2 Kinematics Foces a = v2! u n F g =! mg k speed v is constant N = N (sin! u n + cos! k) F f = µn (cos! u n " sin! k) 13-53 F f = µn (! cos" u n + sin" k) 13-54 Equations of motion (13-53) Then! F n = ma n : N sin! + µn cos! = mv2 "! F z = ma z : N cos! " µn sin! " mg = 0 13-53 13-54 v 2 g! v 2 g! = sin" + µ cos" cos" # µ sin" = sin" # µ cos" cos" + µ sin" v = 24.4 m/s v = 12.2 m/s The two speeds do not geneally diffe by a facto of 2.
13-53 and 13-54 (page 138, 12 th edition) COMPLETION OF PROBLEMS (page 2) SPEED IN TERMS OF THE ANGLE OF FRICTION Data m = 1700 kg g = 9.81 m/s 2! = 20 o! = 100 m µ = 0.2 µ = tan!! = tan "1 µ! = 11.31 o Fom page 1 13-53 v 2 sin" + µ cos" = g! cos" # µ sin" = tan! + tan" 1 # tan" tan! = tan(! + ") 13-53 v = g! tan(" + #) v = 24.4 m/s 13-54 v = g! tan(" # $) v = 12.2 m/s
13-53 and 13-54 (page 138, 12 th edition) COMPLETION OF PROBLEMS (page 3) USING THE RESULTANT FORCE OF FRICTION (Solution 2) Data m = 1700 kg g = 9.81 m/s 2! = 20 o! = 100 m µ = 0.2 µ = tan!! = tan "1 µ! = 11.31 o Foces F g =! mg k R f = R f [sin(! + ") u n + cos(! + ") k] 13-53 R f = R f [sin(! " #) u n + cos(! " #) k] 13-54 Equations of motion (13-53)! F n = ma n : R f sin(! + ") = mv2 #! F z = ma z : R f cos(! + ") # mg = 0 Then v 2 = tan(" + #) g! 13-53 v = g! tan(" + #) v = 24.4 m/s 13-54 v = g! tan(" # $) v = 12.2 m/s
13-53 and 13-54 (page 138, 12 th edition) COMPLETION OF PROBLEMS (page 4) USING THE VECTOR EQUATION OF MOTION (Solution 3) Data m = 1700 kg g = 9.81 m/s 2! = 20 o! = 100 m µ = 0.2 µ = tan!! = tan "1 µ! = 11.31 o Foces F g =! mg k R f = R f [sin(! + ") u n + cos(! + ") k] 13-53 R f = R f [sin(! " #) u n + cos(! " #) k] 13-54 Vecto equation of motion R f + F g = m a Fom the vecto diagam, 13-53 v 2 = tan(" + #) g! 13-53 v = g! tan(" + #) v = 24.4 m/s 13-54 v = g! tan(" # $) v = 12.2 m/s