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Unformatted text preview: homework 09 – BAUTISTA, ALDO – Due: Feb 10 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. The pulley is massless and frictionless. A massless inextensible string is attached to these masses: 1 kg, 2 kg, and 9 kg. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 1 m 21 . 1 cm ω 2 kg 1 kg 9 kg T 2 T 1 T 3 What is the tension T 1 in the string between the block with mass 1 kg and the block with mass 2 kg (on the lefthand side of the pulley)? Correct answer: 14 . 7 N (tolerance ± 1 %). Explanation: Let : R = 21 . 1 cm , m 1 = 1 kg , m 2 = 2 kg , m 3 = 9 kg , h = 2 . 1 m , v = ω R , I = 1 2 M R 2 , and K disk = 1 2 I ω 2 = 1 4 M v 2 . Consider the free body diagrams 1 kg 2 kg 9 kg T 1 T 2 T 3 m 1 g T 1 m 2 g m 3 g a a Basic Concept : For each mass in the system ~ F net = m~a . Solution : Since the string changes direc tion around the pulley, the forces due to the tensions T 2 and T 3 are in the same direction (up). The acceleration of the system will be down to the right ( m 3 > m 1 + m 2 ), and each mass in the system accelerates at the same rate (the string does not stretch). Let this ac celeration rate be a and the tension over the pulley be T ≡ T 2 = T 3 . In freebody diagram for the lower lefthand mass m 1 the acceleration is up and T 1 m 1 g = m 1 a . (1) In freebody diagram for the upper lefthand mass m 2 the acceleration is up and T T 1 m 2 g = m 2 a . (2) In freebody diagram for the righthand mass m 3 the acceleration is down and T + m 3 g = m 3 a . (3) Adding Eqs. (1), (2), and (3), we have ( m 3 m 1 m 2 ) g = ( m 1 + m 2 + m 3 ) a . (4) Therefore a = m 3 m 1 m 2 m 1 + m 2 + m 3 g (5) = 9 kg 1 kg 2 kg 1 kg + 2 kg + 9 kg g = 6 kg 12 kg (9 . 8 m / s 2 ) = 1 2 (9 . 8 m / s 2 ) = 4 . 9 m / s 2 . The tension in the string between block m 1 and block m 2 (on the lefthand side of the homework 09 – BAUTISTA, ALDO – Due: Feb 10 2006, 4:00 am 2 pulley) can be determined from Eq. (1). T 1 = m 1 1 2 + 1 g (6) = (1 kg) 3 2 (9 . 8 m / s 2 ) = 3 2 kg (9 . 8 m / s 2 ) = 14 . 7 N . Question 2 Part 2 of 2. 10 points....
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 Spring '08
 Turner
 Acceleration, Force, Friction, Mass, Work, kg, – BAUTISTA

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