Ch03 Kinetics of Particles.docx

KINETICS OF PARTICLES 365760-95250B A 5kg 10kg 125N 00B A 5kg 10kg 125N Knowing that the coefficient of friction is 0.30 at all surfaces of contact, determine (a) the acceleration of plate A, (b) the tension in the cable. (Neglect bearing friction in the pulley) Kinematics 23050543815B A xB yt l xA xt 00B A xB yt l xA xt - Assume cable is inextensible and, therefore of constant length xA + xB = constant aA = -aB ……………….……. (i) Motion of Plate B Free-body diagram 44577041910T WB = 5(9.81)N = 49.05N T NB FB 00T WB = 5(9.81)N = 49.05N T NB FB Fy = 0: NB – 49.05 = 0 NB = 49.05 N Friction: FB = 0.3NB = 0.3(49.05) = 14.72 N Fx = MB aB: FB-T = 5 aB T = 14.72 – 5aB ………………(ii) Motion of Plate A Free-body diagram 365760112395125N NB = 49.05N FB = 14.72N NA FA T WA = 10(9.81)N = 98.1N 00125N NB = 49.05N FB = 14.72N NA FA T WA = 10(9.81)N = 98.1N Fy = 0 : NA – 49.05 – 98.0 = 0 NA = 147.15 N Friction: FA = 0.3 NA = 0.3(147.15) N FA = 44.15 N Fx = mAaA: 125 – T – 14.72 – 44.15 = 10aA………….(iii) With (ii), (iii) becomes 125 – (14.72 – 5aB) – 14.72 – 44.15 = 10aA 51.41 + 5aB = 10A aB = -aA from (i) From (ii) (a) Disk about to slide to the right 18288060960Y 60o X a 00Y 60o X a 2806065152400mg F N 00mg F N Free-body diagram Fx = max: 0.2 N = m a cos 60 ………………… (i) Fy = may: mg – N = m a sin 60 ………………..(ii) Solve (i) and (ii) simultaneously (b) Disk about to slide to the left 3099435143510N mg F 00N mg F 144780124460Y 60o X a 00Y 60o X a Free-body diagram Fx = max: 0.2 N = m a cos 60 ………………… (iii) Fy = may: N - mg = m a sin 60 ………………..(iv) Solve (iii) and (iv) simultaneously 42062409652000 In a manufacturing process, disks are moved from one elevation to another by the lifting arm shown; the coefficient of friction between a disk and the arm is 0.20. Determine the magnitude of the acceleration for which the disks slide on the arm, assuming the acceleration is directed (a) downward as shown, (b) upward. 347472014859000For the pulley arrangement shown below, with s = 0.25 and k = 0.20, calculate the acceleration of each block and the tension in the cable. Neglect the small mass and friction of the pulleys. Kinematics Assume cable is inextensible xA – 2yB = constant aA = 2aB …….. (i) 376809013779520g T T B 0020g T T B Free-body diagrams 18288050165T N 60g F 30o A 00T N 60g F 30o A Check if block A slips For no slipping, Fx = 0 : 60 g sin 30 – T – F = 0 F = 30(9.81) – T = 294.3 – T Also Fy = 0 : 2T – 20 g = 0 T = 10(9.81) = 98.1 N F = 294.3 – 98.1 = 196.2 N Friction: Fmax = sN = 0.25 (60 g cos 30) = 127.4 N F > Fmax A slips down inclined plane. Motion of A Fx = mAaA : 60 g sin 30 - T – F = 60 aA 294.3 – T – F = 60 aA …………(ii) Fy = 0 (A does not move in the y direction): N – 60 g cos 30 = 0 N = 509.7 N…(iii) Friction: F = kN = 0.2(509.7) = 101.9 N ………(iv) with (iv), (ii) becomes 294.3 – T – 101.9 = 60 aA 60 aA + T = 192.4 ……..(v) Motion of B Fy = mBaB: 2 T – 20 g = 20 aB T – 10 aB = 98.1 ………..(vi) Rearrange (i), (v) and (vi) aA – 2aB + (0)T = 0 60 aA + (0)aB + T = 192.4 (0)aA – 10 aB + T = 98.1 - solve the above using Cramer’s rule A 30-kg crate rests on a 20-kg cart. The coefficient of static friction between the crate and cart is 0.25. If the crate is not to slip with respect to the cart, determine (a) the maximum allowable magnitude of P, (b) the corresponding acceleration of the cart. 06477030 kg 20kg P – F = 0 P = F P x y 0030 kg 20kg P – F = 0 P = F P x y Crate: (Assume slipping impending) 8572553340N W = 30-kg F = 0.25N 00N W = 30-kg F = 0.25N 219456013335030 a 0030 a Fy = 0 N = 30g = F = 0.25(30g) = 7.5g 1828800160020a = 2.45 m/s2 00a = 2.45 m/s2 FA = max 7.5g = 30 a ax = 0.25g -2857515240050g P 0050g P 21031207620050 a 0050 a = Crate & Cart Fx = ma P = 50(0.25g) = 12.5g 36576045720P = 122.6 N 00P = 122.6 N Check on assumption of slipping Fmax = SN = 0.25 x 30g = 7.5g P – F = 0 F = 12.5g F > Fmax crate slips 4480560-9144000A 2 kg ball revolves in a horizontal circle as shown at a constant speed of 1.5 m/s. Knowing that L = 600 mm, determine (a) the angle that the cord forms with the vertical, (b) the tension in the cord. Free-body diagram of ball 114490526670x y T 2g 00x y T 2g - ball does not move in the y-direction Fy = 0: Tcos = 2(98.1) = 19.62 ………… (i) - Motion of ball in the x-direction Fx = max: or Tsin2 = 7.5 ……(ii) T = 7.5/sin2 substitute for T in (i) sin2 = 1 – cos2 7.5 cos = 19.62 – 19.62cos2 Block A slides along a plane inclined at an angle to the horizontal. The block is initially at a height, h, above a point B. Determine the velocity of the block at B if it starts from rest. 94615-361950 W A h B 00 W A h B Solution (i) (Case of no friction) - Free-body diagram of block 27432041910NP F = NP Y X 00NP F = NP Y X Principle of Work and Energy UiB =TB - Ti UiB (initial position) Ti = 0 (Block starts from rest) UiB = Solution (ii) (Friction considered) Block A does not move in direction inclined plane Fy = 0: NP – W cos = 0 UiB = 317817518161000 3632200196850as in solution (i) 00as in solution (i) For = 90 cot = 0 V = For = 0 , V = For real values of V, 2gh 2gh cot 411480011493500 A 20 kg package is projected up a 20 incline with an initial velocity of 12 m/s. The coefficient of friction between the incline and the package is 0.15. Determine (a) the maximum distance the package will move up the inclined plane; (b) the velocity of the package when it returns to its original position. (a) Free-body diagram of block 36576091440NA F = NA 196.2N 20o Y X 00NA F = NA 196.2N 20o Y X Principle of Work and Energy UAB = TB - TA TA = At maximum distance up the incline VB = 0 TB = 0 104775016637000 UAB = -(196.2 sin 20 + 0.15 NA)d (Force and displacement in opposite directions) Package does not move to inclined plane Fy = 0 NA – 196.2 cos 20 = 0 NA = 196.2 cos 20 -(196.2) (sin 20 + 0.15 cos 20)d = 0 – 1440 (b) Free-body diagram 83248596520NB F = 0.15NB 196.2N 20o 00NB F = 0.15NB 196.2N 20o Principle of Work and Energy UBA = TA - TB TA = UBA