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# ch05 - Chapter 5 Multiparticle Systems 5.1 Force Balance...

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Chapter 5 Multiparticle Systems 5.1 Force Balance and Linear Momentum 131

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5.1.1 GOAL: Find center of mass of two particles. GIVEN: Position and mass of particles. DRAW: FORMULATE EQUATIONS: r G = 1 m 2 i =1 m i r i SOLVE: r G = 1 (5 kg) [(2 kg)(4 m) + (3 kg)(4 ı m)] = 1 . 6 m + 2 . 4 ı m r G = (2 . 4 ı + 1 . 6 )m 132
5.1.2 GOAL: Find mass center of four particles. GIVEN: Particle mass and position. DRAW: FORMULATE EQUATIONS: r G = 1 m 4 i =1 m i r i / O SOLVE: r G = 1 (45 slug) [(5 slug)(2 ft) + (10 slug)[(2 ı + 2 )ft +(20 slug)(2 ı ft) + (10 slug)(3 ı ft)] r G = (2 ı + 2 3 )ft 133

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5.1.3 GOAL: Find system’s center of mass GIVEN: Position and mass of four particles. FORMULATE EQUATIONS: r G = n i =1 m i r i m (1) SOLVE: (1) r G = (2 kg)(0 . 5 ı m) + (2 kg)(1 . 2 m) + (2 kg)( - 0 . 5 k m) + (2 kg)( - 1 . 2 + 1 . 5 k m) 8 kg r G = . 125 ı + 0 . 25 k m 134
5.1.4 GOAL: Find system’s center of mass GIVEN: Position and mass of three particles. DRAW: FORMULATE EQUATIONS: r G = n i =1 m i r i m (1) SOLVE: (1) r G = (5 kg)( - 0 . 2 m) + (2 kg)( - 0 . 2 + 0 . 25 k m) + (6 kg)(0 . 3 ı - 0 . 2 k m) (5 + 2 + 6) kg r G = (0 . 138 ı - 0 . 108 - 0 . 0538 k ) m 135

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5.1.5 GOAL: Find system’s linear momentum. GIVEN: Position, mass and velocity of four particles. DRAW: FORMULATE EQUATIONS: v G = n i =1 m i v i m (1) SOLVE: (1) v G = (2 slug)(5 ft/s) + (3 slug)(6 ı ft/s) + (4 slug)( - 5 ı ft/s) + (4 slug)( - 6 ft/s) (2 + 3 + 4 + 4) slug = ( - 0 . 154 ı - 1 . 08 ) slg · ft/s L = mv G = ( - 2 ı - 14 ) slg · ft/s 136
5.1.6 GOAL: Find system’s linear momentum. GIVEN: Position, mass and velocity of three particles. DRAW: FORMULATE EQUATIONS: v G = n i =1 m i v i m (1) SOLVE: (1) v G = (2 kg)(10 m/s) + (3 kg)(20 m/s) + (10 kg)( - 10 m/s) (2 + 3 + 10) kg = - 1 . 3 m/s L = mv G = - 20 kg · m/s 137

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5.1.7 GOAL: Find the total force acting on the floor of a dance platform. GIVEN: Dancers A and B weigh 120 lbs, dancer C weighs 110 lbs and dancer D weighs 150 lbs. v A = a A = v B = a B = 0 v C = v D = 5 ı ft/s a C = a D = - 6 ı ft 2 / s DRAW: SOLVE: Force balances: Restricting ourselves to horizontal motions: m D ¨ x D = - F D (1) m C ¨ x C = - F C (2) m B ¨ x B = - F B (3) m A ¨ x A = - F A (4) Using the given accelerations yields F D = 110 lb 32 . 2 ft/s 2 (6 ft/s 2 ) = 20 . 50 lb F C = 150 lb 32 . 2 ft/s 2 (6 ft/s 2 ) = 27 . 95 lb F B = 110 lb 32 . 2 ft/s 2 (0) = 0 lb F A = 110 lb 32 . 2 ft/s 2 (0) = 0 lb Total force acting on the floor: F TOTAL = F A + F B + F C + F D = 20 . 50 lb + 27 . 95 lb = 48 . 45 lb 138
5.1.8 GOAL: Determine the velocity of a four car pileup. GIVEN: v A = 40 m/s, m A = 1900 kg v B = 0 m/s, m B = 1200 kg v C = 10 m/s, m C = 1400 kg v D = 5 m/s, m D = 2000 kg DRAW: FORMULATE EQUATIONS: The linear momentum of the system of particles is conserved. Σ L i = CONSTANT SOLVE: Σ L t 1 = Σ L t 2 All motion is in the ı direction and we have: (2000 kg)(5 m/s) + (1400 kg)(10 m/s) + (1200 kg)(0 m/s) + (1900 kg)(40 m/s) = (2000 kg + 1400 kg + 1200 kg + 1900 kg)¯ v ¯ v = 15 . 4 m / s 139

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5.1.9 GOAL: Find the distance m 2 has moved when θ = 90 and its speed at that time. GIVEN: m 1 = 2 kg, m 2 = 4 kg L = 3 m, ˙ x (0) = 0, θ = 45 The attachments are frictionless.
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