Atwood Machines: Finding Accelerations and Tension in Different Scenarios, Study notes of English Language

Download Atwood Machines: Finding Accelerations and Tension in Different Scenarios and more Study notes English Language in PDF only on Docsity! Atwood Machines Kevin S. Huang 1 Fixed Pulley Let’s begin our analysis by studying the fundamental Atwood machine - a fixed pulley with two masses. Question Find the accelerations of the masses and the tension in the string. We solve problems involving Atwood machines by using F=ma equations and an equation of conservation of string. F=ma equations T −m1g = m1a1 T −m2g = m2a2 Conservation of String a1 + a2 = 0 Accelerations a1 = m2 −m1 m1 + m2 g 1 a2 = m1 −m2 m1 + m2 g Tension T = 2m1m2g m1 + m2 We can use the idea of effective mass of a fixed pulley to simplify our analysis in fixed pulley combinations. Effective Mass of a Fixed Pulley M = 2T g = 4m1m2 m1 + m2 1.1 Fixed pulley combination Question Find the accelerations of the masses and the tension in the string. F=ma equations T −m1g = m1a1 T/2 −m2g = m2a2 T/2 −m3g = m3a3 We can use the idea of effective acceleration of a fixed pulley to simplify finding the equation for conservation of string in fixed pulley combinations. Effective Acceleration of Fixed Pulley A = a2 + a3 2 2 a2 = 3m1 −m2 9m1 + m2 g Tension T = 4m1m2 9m1 + m2 g 2.2 n Strings We can generalize this to n strings. Question Find the accelerations of the masses and the tension in the string. F=ma equations T −m1g = m1a1 nT −m2g = m2a2 Conservation of String a1 + na2 = 0 Accelerations a1 = nm2 − n2m1 n2m1 + m2 g a2 = nm1 −m2 n2m1 + m2 g Tension T = (n + 1)m1m2 n2m1 + m2 g 5 2.3 Free Pulley In Between Question Find the accelerations of the masses and the tension in the string. F=ma equations T −m1g = m1a1 2T −m2g = m2a2 T −m3g = m3a3 Conservation of String a1 + 2a2 + a3 = 0 Accelerations a1 = 3m2m3 −m1(m2 + 4m3) m2m3 + m1(m2 + 4m3) g a2 = −m2m3 + m1(4m3 −m2) m2m3 + m1(m2 + 4m3) g a3 = −m2m3 + m1(3m2 − 4m3) m2m3 + m1(m2 + 4m3) g Tension T = 4m1m2m3 m2m3 + m1(m2 + 4m3) g 6 2.4 Massless Free Pulley Equivalence of Free Pulleys The subject of massless free pulleys is an interesting topic in Atwood machines. We can use a key idea derived here to solve more interesting problems. Question Find the acceleration of the free pulley. Procedure Assume the free pulley has mass m2. This is equivalent to the case with a free pulley derived above. T −m1g = m1a1 2T −m2g = m2a2 a1 + 2a2 = 0 We have m2 = 0. The force on the pulley must be zero so T = 0. We can use the first equation and third equation to solve for a2. Note that we cannot use the second equation. a2 = g 2 7