## Answers

WE The acceleration of Systems must be acuberation & any one component of the system therefore, Newton's second lowo i should be applied to each of the three bodies mpa 25.13kg 70.250m For ML T = I Mag Ty - Ma => [T=Mg-ma TR ² 22 For Me me + 19.7 kg MR - 11.7 kg E-mg - Mea => T2 = 1 = Matme For Mp TR-TR = I a I a → The above er is certarily Complicated. The first team is toene od Me & Decind item is themes from Me → The right hand and side teomis found by using relation blow angular and tangetical a cederation, y The moment & necta for a dok is given by I MR2/2 Substituile en; in enuation 6 » (Mg-Ma) R - (M9+ma] R2 mp RPå R[m, g-m,a-meg-ma] - mp Ra » (m - m2) a (m + mp] 2 1 / mpa - (m. -me) g ((mame ) + Am] ER | .2

2 mg az em, mop) (m + m₂) at 1 mp (19.70 -41,7) (9:8) (19.7+ 11.7) +1 (5.3) 78.4 31.4 +2.565 a = 78.4 22.308 m 1st 33.9965 a = 2.308 m 182 Tension in lesght mass IT beft Now acceleration his benown, the faceebution) tension for Steing attached to block Me is nothing more than =T, a M. (g-a) mgomea [m - malg um amp) + mp mg m-mo - m + mp + mp 2 - 19.7(9.8 19.7. 11.7 19.7+11.7 +5:13 2 193.06 33.965 193.0611-0.237] 193.06 (0.762] 147,2N

LT, = 147.2 Tension ful light mass To Tz © => me (gtal - mpg + ma (cm - MR) magama - meg mea [ + M + M2 + 1 mp m - MR m+MR+ Мр > 117 (98) 19.7-117 o ) 1 + 19.7*11.746-13 2 14.66[140-23 TR 2 141.83 N 0 a 22.308 alone T= 147.2 N ③ TR 2 141.83N

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