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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' = x{ + *, x = x + x2) are mäf = m¡(*{ + #) = mig - T = mog - T so mig - T - mäí = m,(g – a) – T m2 = mag – T - mgöz = m2(g – æ) – TJ (2.70) where = * = a. We have #, ing T: - *1, so we solve for , as before by eliminat- = (т, — т,) m, + mg * = - *, = (g – a)- (2.71) and 2m, my(g – a) T = (2.72) m, + m2

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' = x{ + *, x = x + x2) are mäf = m¡(*{ + #) = mig - T = mog - T so mig - T - mäí = m,(g – a) – T m2 = mag – T - mgöz = m2(g – æ) – TJ (2.70) where = * = a. We have #, ing T: - *1, so we solve for , as before by eliminat- = (т, — т,) m, + mg * = - *, = (g – a)- (2.71) and 2m, my(g – a) T = (2.72) m, + m2

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.4: Plane Curves And Parametric Equations
Problem 39E
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Show the step by step in the equation of 2.71 and 2.72
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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