Question

Chapter 2, Problem 2/190 x Incorrect The rod of the fixed hydraulic cylinder is moving to the left with a constant speed VA = 48 mm/s. Determine the corresponding acceleration of slider B (positive if downward, negative if upward) when 5a = 420 mm. The length of the cord is 1075 mm, and the effects of the radius of the small pulley A may be neglected. 245 mm Answer: ag = 35.563 mm/s2

Answer 1

h=245 mm AMA

Given: L = 1075 mm va = 48 mm/s h = 245 mm Find: ag =? OSA = 420 mm AC = V8 +2452 = 4202 + 2452 = 486.236 mm AB=L - AC = 1075 - 486.2355 = 588.7645 mm 8B = VAB2 - 52 = 588.76452 – 4202 = 412.6059 mm L= Vs+h+vs+ +52 nove teme na vita (200 + 20 = 0 Put dosar va 1 ds, ------(0) V8+2 +3 SB dd Differentiate agian despor + VASAVS+ = 0 Va+(vart) - 0454 ( cover ) (s+h2) V3+ (SB*+ (+ v ) - (spder + vas) (2884*+24) (52 +8%) (6*+) 6% + 5%) (9995 + vá + vy) – (5993 +2403=0 h22 (8 + 9) (80p + B + v) -(8848 + 4181) (+12) 3/2 eg() (6% + 123/ 24 Put values in equation (i) 48 x 420 (412.6059 X Ug +48 x 420) V 4202 +2452 412.60599 +420 112.6059 X ve + 20160 41.4614+ 588.7645 →vB = -108.0232 mm/s from eq (1) 2452 x 482 (412.6059% +420°) (412.6059a3 + 108.0232? + 48) - (-412.6059 x 108.0232+ 48 x 420) (412.60592 + 4202)1,5 (4202 + 2452157 0 = 1.203025+ 346643.629 x (412.6059a3 + 13973.0117) - 5.95897 x 108 -= 2.040914 x 108 →1.203025 +0.700803 +20.81302=0 =an = -31.4156 mm/s2