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Atwood's machine consists of a smooth pulley with two masses suspended from a light string at each end (Figure 2-11). Find the acceleration of the masses and the tension of the string (a) when the pulley center is at rest and (b) when the pulley is descending in an elevator with constant acceleration a. Fixed Elevator mig m2 (b)

Atwood's machine consists of a smooth pulley with two masses suspended from a light string at each end (Figure 2-11). Find the acceleration of the masses and the tension of the string (a) when the pulley center is at rest and (b) when the pulley is descending in an elevator with constant acceleration a. Fixed Elevator mig m2 (b)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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EXAMPLE 2.9 Atwood's machine consists of a smooth pulley with two masses suspended from a light string at each end (Figure 2-11). Find the acceleration of the masses and the tension of the string (a) when the pulley center is at rest and (b) when the pulley is descending in an elevator with constant acceleration a. Fixed Elevator T (a) (b) FIGURE 2-11 Example 2.9; Atwood's machine. CASE I Solution. We neglect the mass of the string and assume that the pulley is smooth-that is, no friction on the string. The tension Tmust be the same throughout the string. The equations of motion become, for each mass, for case (a), m,k, = mg - T m2ä = mag – T (2.66) (2.67) Notice again the advantage of the force concept: We need only identify the forces acting on each mass. The tension T is the same in both equations. If the string is inextensible, then * = - , and Equations 2.66 and 2.67 may be combined m,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 from Equations 2.68 and 2.66: T = mig - m (m, - mg) T= mg - mig m, + m2 2m, mog T = m, + m2 (2.69) --- ---------
Transcribed Image Text:EXAMPLE 2.9 Atwood's machine consists of a smooth pulley with two masses suspended from a light string at each end (Figure 2-11). Find the acceleration of the masses and the tension of the string (a) when the pulley center is at rest and (b) when the pulley is descending in an elevator with constant acceleration a. Fixed Elevator T (a) (b) FIGURE 2-11 Example 2.9; Atwood's machine. CASE I Solution. We neglect the mass of the string and assume that the pulley is smooth-that is, no friction on the string. The tension Tmust be the same throughout the string. The equations of motion become, for each mass, for case (a), m,k, = mg - T m2ä = mag – T (2.66) (2.67) Notice again the advantage of the force concept: We need only identify the forces acting on each mass. The tension T is the same in both equations. If the string is inextensible, then * = - , and Equations 2.66 and 2.67 may be combined m,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 from Equations 2.68 and 2.66: T = mig - m (m, - mg) T= mg - mig m, + m2 2m, mog T = m, + m2 (2.69) --- ---------
CASE I. For case (b), in which the pulley is in an elevator, the coordinate system with 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). The equations of motion in the inertial system (x = xi + x, xg = x + x2) are mä¡ = m, (* + #1) %3D mig - T M2g - T so mig - T- mží = m1(g – a) – T ma = 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, + mg and 2m, my(g – a) T = (2.72) m1 + m2
Transcribed Image Text:CASE I. For case (b), in which the pulley is in an elevator, the coordinate system with 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). The equations of motion in the inertial system (x = xi + x, xg = x + x2) are mä¡ = m, (* + #1) %3D mig - T M2g - T so mig - T- mží = m1(g – a) – T ma = 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, + mg and 2m, my(g – a) T = (2.72) m1 + m2
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