PHY302K-Sai-Midterm 1-solutions

PHY302K-Sai-Midterm 1-solutions - Version 130 Midterm 1...

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Version 130 – Midterm 1 – sai – (58015) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points The distance between two telephone poles is 58 m. When a 1 . 36 kg bird lands on the telephone wire midway between the poles, the wire sags 0 . 111 m. The acceleration oF gravity is 9 . 8m / s 2 . How much tension in the wire does the bird produce? Ignore the weight oF the wire. 1. 1308.58 2. 925.697 3. 1242.66 4. 959.181 5. 549.983 6. 1741.06 7. 978.017 8. 889.274 9. 1123.86 10. 1197.95 Correct answer: 1741 . 06 N. Explanation: 0.11 m 58 m θ Given : m =1 . 36 kg , Δ x = 58 m , and Δ y =0 . 111 m . T T mg tan θ = Δ y 1 2 Δ x = 2 Δ y Δ x θ = tan - 1 ± 2 Δ y Δ x ² = tan - 1 ± 2 × 0 . 111 m 58 m ² =0 . 219303 . ³ F y =2 T y - mg =2 T sin θ - mg =0 since it is in equilibrium, so T = mg 2 sin θ = (1 . 36 kg)(9 . 8m / s 2 ) 2(sin0 . 219303 ) = 1741 . 06 N . 002 (part 1 oF 2) 10.0 points A pulley is massless and Frictionless. 1 kg, 3 kg, and 8 kg masses are suspended as in the fgure. 3 . 1m 20 . 6m ω 3 kg 1 kg 8 kg T 2 T 1 T 3 What is the tension T 1 in the string be- tween the two blocks on the leFt-hand side oF the pulley? The acceleration oF gravity is 9 . 8m / s 2 . 1. T 1 = ± 17 12 kg ² (9 . 8m / s 2 ) 2. T 1 = ± 7 6 kg ² (9 . 8m / s 2 ) 3. T 1 = ± 7 4 kg ² (9 . 8m / s 2 )

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Version 130 – Midterm 1 – sai – (58015) 2 4. T 1 = ± 13 12 kg ² (9 . 8m / s 2 ) 5. T 1 = ± 11 6 kg ² (9 . 8m / s 2 ) 6. T 1 = ± 19 12 kg ² (9 . 8m / s 2 ) 7. T 1 = ± 5 3 kg ² (9 . 8m / s 2 ) 8. T 1 = ± 4 3 kg ² (9 . 8m / s 2 ) correct 9. T 1 = ± 5 4 kg ² (9 . 8m / s 2 ) 10. T 1 = ± 3 2 kg ² (9 . 8m / s 2 ) Explanation: Let : R = 20 . 6m , m 1 = 1 kg , m 2 = 3 kg , m 3 = 8 kg , and h =3 . 1m . Consider the free body diagrams 1 kg 3 kg 8 kg T 1 2 3 m g a For each mass in the system " F net = m"a. Since the string changes direction 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 acceleration rate be a and the tension over the pulley be T T 2 = T 3 . For the lower left-hand mass m 1 the accel- eration is up and T 1 - m 1 g = m 1 a. (1) For the upper left-hand mass m 2 the acceler- ation is up and T - T 1 - m 2 g = m 2 a. (2) For the right-hand 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) a = m 3 - m 1 - m 2 m 1 + m 2 + m 3 g (5) = 8 kg - 1 kg - 3 kg 1 kg + 3 kg + 8 kg g = 4 kg 12 kg (9 . 8m / s 2 ) = 1 3 (9 . 8m / s 2 ) .
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PHY302K-Sai-Midterm 1-solutions - Version 130 Midterm 1...

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