Show the step by step of equations of 2.68, 2.69 ,2.71 and 2.72 EXAMPLE 2.9Atwood's machine consists of a smooth pulley with two masses suspended froma light string at each end (Figure 2-11). Find the acceleration of the masses andthe tension of the string (a) when the pulley center is at rest and (b) when thepulley is descending in an elevator with constant acceleration a.FixedElevatorT(a)(b)FIGURE 2-11 Example 2.9; Atwood's machine.CASE ISolution. We neglect the mass of the string and assume that the pulley issmooth-that is, no friction on the string. The tension Tmust be the samethroughout the string. The equations of motion become, for each mass, forcase (a),m,k, = mg - Tm2ä = mag – T(2.66)(2.67)Notice again the advantage of the force concept: We need only identify theforces acting on each mass. The tension T is the same in both equations. Ifthe string is inextensible, then * = - , and Equations 2.66 and 2.67 may becombinedm,ä = mig – (mog – mgž2)= mig - (mog + m2ä)Rearranging,g(m, - m2)*, == -*,(2.68)т + т,If m, > mg, then ä, > 0, and #, < 0. The tension can be obtained fromEquations 2.68 and 2.66:T = mig - m(m, - mg)T= mg - migm, + m22m, mogT =m, + m2(2.69)------------ CASE I.For case (b), in which the pulley is in an elevator, the coordinate systemwith origins at the pulley center is no longer an inertial system. We need an in-ertial system with the origin at the top of the elevator shaft (Figure 2-11b). Theequations of motion in the inertial system (x = xi + x, xg = x + x2) aremä¡ = m, (* + #1)%3Dmig - TM2g - Tsomig - T- mží = m1(g – a) – Tma = mag – T- mgäž = m2(g – æ) – TJ(2.70)where * = = a. We have ä,ing T:- #1, so we solve for ä, as before by eliminat-(m,* = - * = (g – a)-m2)(2.71)m, + mgand2m, my(g – a)T =(2.72)m1 + m2

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Physics

Show the step by step of equations of 2.68, 2.69 ,2.71 and 2.72

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter4: Dynamics: Force And Newton's Laws Of Motion

Section: Chapter Questions

Problem 2CQ: What properties do forces have that allow us to classify them as vectors?

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