Question

The load L is lowered by the two pulleys which are fastened together and rotate as a single unit. For the instant represented, drum A has a counterclockwise angular velocity of 3.6 rad/sec, which is decreasing by 2.6 rad/sec each second. Simultaneously, drum B has a clockwise angular velocity of 5.0 rad/sec, which is increasing by 5.5 rad/sec each second. Calculate the accelerations of points C and D and the load L. 6' 6" O D 12/ Rau с L Answers: ac = i) ft/sec an = li) ft/sec a = li) ft/sec2 +

Answer 1

10 find: ACCELERATIONS of POINTS Ĉ, & THE LOAD L. GIVEN: from figuse and data ! The angular velocity of drum A, WA = 3.6 radls. The radius of drum AVA = 6 in 0.5 dt The angular velosing of drum B, Wp 2 5 radls. The radius of dwum B.To - 6in: 0.5dt: The radius at point D = 24 in= 2 ft A has a counter clour wise angulas velocity of 3.6 rods, which is decreasing by 26 radlls each second. B has a cloumwibe angulas velouing of 5 rodes, which is increasing by 8-6 redke ANSWER The relation for veloury is given by, Va r. W & where v-radius w- angular velocing velocity of the drum B, VB = VB WB : 0.5 5 = 2.5 fts Velour of the drum A, VA "A.WA : 0.5 * 3.6 - 1.8 ft)s Then, The acceleration of the drum A can be calculated as aA = VAXKA = 0.5% 2.6 = 1.3 ft/32 the acceleration of the drum B can be calculated as, ag = P x dog 20.5X 562 2.75 ft) 62 Now, we can easily find out the angular velocity of fastened pulley by using thie following relation: W-VA-VB NA-VB 1.8-2.5 Votre 2 -0.23 radla & + 2) Hint No 24 in: aft re: 12 in 1st

NOW find out the angular andleeation of pulleys using the following relation ! 1.3 +2.75 1+2 x= OLA+ AB AA+ AB 1.35 rad 13² rg + 12 Notre The acceleration of Point d' can be calculated as: ab=a0+ +9010 = + @ojo) + (y) = (0-ag)j + (0 Code + w x)) +("JXKk) = (as - ap)| +(?my T})+ (* 7) =(1-3-2.25)$+- (60.23% 29) — (2 x 1.35) -1.556 9 – 2.7 d'uus, aceleration of point, D, ão - (1.66631 – 2-71) Pt 162 The acceleraton of point wc can be calculated as: 9.2 ag taco ao.+ acro+ (ayolt = (x - ap)h + (WR ( 281 xWK)) + (roltak) = (A - ap): -(W27) - (***) 3) -(1.3 -2.75)3 – (60-23)2 21) – (2*1-35) =(1-3 - 2-45)9 - 231-36): - (0.233* 21) - 4.159 - 0.105 1 Thus, acceleration of point, C,(-4.169 -0.05 m) 44/02 Now, consides de acceleration ratio, AB 2.75 YA (12424-0) 1.3 36- n

2-35 1.3 2.75 (36-W=1.30 = 99-2.7502 1.3n M 36-0 99-4.05n n99 2 24.63 in 4.05 Now, the acceleration of the load can be calculated as: AL . 2 MA 24 AL MAX12 24 1-3 2 0.65 86/32 12 Thus Acceleration of point caca -0.105 1 - 4.15 h ft/5? Acceleration of point D, ap: -2.11 - 15569 ft) 52. Acceleration of load, L, AL = 0.65 St / 32