InClassLecture24

# InClassLecture24 - 9~l* Determine the horizontal force P...

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Unformatted text preview: 9~l* Determine the horizontal force P required to start moving the 250-lb block shown in Fig. PQ-l up the inclined surface. The coefficient of friction between the inclined surface and the block is # = 0.30. S SOLUTION 3 A free-body diagram for the block is shown at the right. For impending motion: Fr = HF“ = 0.30Fn + T ZFY = F cos 30o - Fr sin 300 — W n F cos 30° — 0.305n sin 30° — 250 = 0 n 349.15 lb '1‘?) II + —+ IF = p — Fn sin 30° - Ff cos 30° F - 349.15 sin 30° - 0.30(349.15) cos 30° = 0 P = 265.29 lb E 265 lb Ans. 9-3 Workers are pulling a 400-lb crate up an incline as shown in Fig. P9-3. The coefficient of friction between the crate and the surface is 0.20, and the rope on which the workers are pulling is horizontal. (a) Determine the force P that the workers must exert to start sliding the crate up the incline. (b) If one of the workers lets go of the rope for a moment, determine the minimum force the other workers must exert to keep the crate from sliding back down the incline. SOLUTION A free—body diagram for the block is shown at the right. (a) For impending motion of the blo k th ' l‘ : 0 up e inc me O 2A” ﬁn + x” XFX = p cos 15° — 400 sin 15° - 0.2An = 0 + "\ ZFY = An — P sin 15° - 400 cos 15° = 0 Solving yields: A“ = 437.6 lb P = 197.78 lb a 197.8 lb Ans. . . . 400 it (b) For impending motion of the ,5. block down the incline: 3"..’7L u” oann Fin + x” SE = p cos 15° — 400 sin 15° + 0.2An = o + \ ZFy = An — P sin 15° — 400 cos 15° = 0 Solving yields: A = 393.0 lb P = 25.80 lb n 25.8 lb Ans. 9-21* A 120—lb girl is walking up a 48-lb uniform beam as shown in Fig. P9-21. Determine how far up the beam the girl can walk before the beam starts to slip 1% (a) The coefficient of friction is 0.20 at all surfaces. (b) The coefficient of friction at the bottom end of the beam is increased to 0.40 by placing a piece of rubber between the beam and the floor. SOLUTION (a) From a free-body diagram for the beam when motion is impending: A = MA = 0.2A f n n B = MB = 0.28 f n n ¢ = tan‘1 2 = 36.870 + l M '11 II B sin 36.870 — 0.23n cos 36.870 — 0.2An = 0 n + T 2F = B cos 36.870 + 0.28n sin 36.870 + An - 120 — 48 = 0 V n Solving yields: A = 118.46 lb B = 53.85 lb n n + C EMA = 48(6) cos 36.870 + 120(x) cos 36.870 - 53.85(10) = 0 X = 3.209 ft g 3.21 ft , Ans. 0'2 Bn 320 “a (b) From a free-body diagram for the beam when motion is impending: ‘3 Ar = HA“ = 0.4An OvQFI B = MB = 0.2B ‘n1f""' n f n n . o o R + -4 53Fh = B Sln 36.87 — 0.2B cos 36.87 — 0.4A = 0 y‘ . n n n + T EFV = Bn cos 36.870 + 0.213n sin 36.87° + An — 120 — 48 = 0 Solving yields: A = 91.49 lb B = 83.17 lb n n + C EMA = 48(6) cos 36.870 + 120(x) cos 36.870 - 83.17(10) = 0 x = 6.264 ft 2 6.26 ft Ans. 9-36 The masses of blocks A and B of Fig. P9—36 are mA = 40 kg and mB = 85 kg. If the coefficient of ffiction is 0.25 for both surfaces, determine the force p required to cause impending motion of block B. SOLUTION From a free—body diagram for block A when motion is impending: -T- _ _ ----x WA = mAg = 40(9.80/) = 392.28 N Ar = Af(max) = HAAn = 0.25An / Rn + T ZFy = A cos 45° + Af sin 450 — W n A - 392.28 = 0 -0 A cos 45° + 0.25An sin 4o n 32 ll 443.81 N f 0.25An = 0.25(443.81) = 110.95 N > II From a free-body diagram for block B when motion is impending: I? II B mBg = 85(9 807) = 833.60 N Bf = Bf(max) = MB“ = 0.25Bn .17 : 0— ' 0— — . / _Fx P cos 20 WE Sln 45 Ar Bf = P cos 20o - 833.60 sin 45° — 110.95 - 0.25Bn= 0 + K\ :F = p sin 20° — w cos 45° — A + B y B n n 0 = 9 sin 20 — 833.60 cos 45° - 443.81 + Bn = 0 Solving yields: P = 935.14 N E 935 N Ans. 9-120* The masses of blocks A and B of Fig. P9-120 are mA = 50 kg and mB = 25 kg. If the coefficient'of static friction is 0.15 for both surfaces, determine the force ﬁ required to cause impending motion of block B. SOLUTION From a free-body diagram for block A when motion is impending: WA = mAg = 50(9.807) 490.4 N A cos 45° — B x f A cos 45° - 0.1513n = o + —+ 2F A = 0.21213n + T 2F = B — A sin 45° — w = B — 0.21213 sin 45° — 490.4 = 0 y n A n n Bn = 576.93 N Br = ya“ = 0.15(576.93) = 86.54 N From a free—body diagram for block B when motion is impending: W = mBg = 25(9.807) B 245.2 N + T 2F = C — B — W n Y n B = Cn — 576.93 - 245.2 = 0 Cn = 822.13 N Cr = uCn = O.15(822.13) = 123.32 N + -* EFX = P - Br - Cf = P - 86.54 - 123.32 = 0 P = 209.86 N 3 210 N Ans. 9-13* The ZOO-lb block of Fig. P9—13 is sitting on a 300 inclined surface. The coefficient of friction between the block and the surface is 0.50. A light, inextensible cord is attached to the block, passes around a frictionless pulley. and is attached to a second block of mass M. Determine the minimum and maximum masses, M . and M , such min max that the system is in equilibrium. Is impending motion by slipping or by tipping? SOLUTION From a free—body diagram for the ZOO-lb block: + “\ ZFY = A — 200 cos 30° n O A = 173.21 lb n MA = 0.50(173.21) = 86.61 lb A,(max) 1 n For slipping down the incline: + /” 2px = p + Af(max) — 200 sin 30° = p + 86.61 — 200 sin 30° = 0 P3 = 13.39 lb For tipping: + c 2MB = P(0.5) - 200 sin 30° (1) + 200 cos 30° (0.5) = 0 Pt = 26.79 lb > PS (Block tips before it slips as P is reduced) Therefore: Pmin = 26.79 lb For slipping up the incline: + /” ZFX = P — Af(max) — 200 sin 30° = P - 86.61 - 200 sin 30° = 0 P = 186.61 lb U) P _ min _ 26.79 _ . . Mmin — -—E- — EETI7— — 0.833 slug (Tipping) Pmax 186.61 _ M = —-—- = -—-—-— = 0.80 slug (Slipping) Ans. max g 32.17 ...
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## This note was uploaded on 07/05/2010 for the course ENGR 120 taught by Professor Jones during the Spring '10 term at Florida State College.

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InClassLecture24 - 9~l* Determine the horizontal force P...

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