(a) Explanation Show the free body diagram as in Figure (1). Express the equation for length of cord in terms of s A and s B . L = L A C + L A B = s A 2 + 250 2 + s A 2 + ( s B − 250 ) 2 = ( s A 2 + 250 2 ) 1 / 2 + [ s A 2 + ( s B − 250 ) 2 ] 1 / 2 (I) Obtain derivative of Equation (I). d d t { ( s A 2 + 250 2 ) 1 / 2 + [ s A 2 + ( s B − 250 ) 2 ] 1 / 2 − L } = 0 { 1 2 ( s A 2 + 250 2 ) − 1 / 2 ( 2 s A v A ) + 1 2 [ s A 2 + ( s B − 250 ) − 1 / 2 ] [ 2 s A v A + 2 ( s B − 250 ) v B ] } = 0 (II) Here, velocity at A and B is v A and v B respectively. Obtain derivative of Equation (II). 1 2 ( s A 2 + 250 2 ) − 1 / 2 ( 2 s A s ˙ A ) + 1 2 [ s A 2 + ( s B − 250 ) 2 ] − 1 / 2 [ 2 s A s ˙ A + 2 ( s B − 250 ) s ˙ B ] { v A 2 ( s A 2 + 250 2 ) − 1 / 2 + s A a A ( s A 2 + 250 2 ) − 1 / 2 − s A 2 v A 2 ( s A 2 + 250 2 ) − 3 / 2 + [ s A a A + v A 2 + v B 2 + ( s B − 250 ) a B ] × [ s A 2 + ( s B − 250 ) 2 ] − 1 / 2 − [ s A v A + ( s B − 250 ) v B ] × [ s A 2 + ( s B − 250 ) 2 ] − 3 / 2 [ s A v A + ( s B − 250 ) v B ] } = 0 (III) Show the free body of slider B as in Figure (2). Here, tension in the cord is T , frictional force acting on the slider is F and normal reaction force is N . Express the angle made by tension T with x axis. α = tan − 1 [ s B − 250 s A ] (IV) Express equation of motion in vertical direction. ∑ F y = m a y T sin α − m g − F = m a B (V) Conclusion: Substitute 425 mm for s A and 1050 mm for L in Equation (I). 1050 = ( 425 2 + 250 2 ) 1 / 2 + [ 425 2 + ( s B − 250 ) 2 ] 1 / 2 [ 180625 + ( s B 2 + 62500 + 500 s B ) ] 1 / 2 = 556 (b) To determine Determine the tension in the cable.

8th Edition

ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 3.4, Problem 45P

(a)

To determine

Determine the tension in the cable.

(b)

To determine

Determine the tension in the cable.

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Chapter 3.4, Problem 44P

Chapter 3.4, Problem 46P

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