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Two masses m_(1) and m_(2) are connected by means of a light string, that passes over a light pulley as shown in the figure. If m_(1)=2kg and m_(2)=5kg and a vertical force F is applied on the pulley, then find the acceleration of the masses and that of the pulley when (a) f = 35N, (b) F = 70N, (c) F = 140N. |
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Answer» SOLUTION :Since string is massless and friction is ABSENT, hence tension in the string is same every where. (a) Let acceleration of the pulley be `a_(p)`. For `a_(p)` to be non-zero, F.B.D. of the pulley `T gt m_(1)g ""...(1)` `rArr T gt 2g "" ...(2)` From (1) and (2), we get `F gt 2 xx (2g) rArr F gt 40N` Therefore, when `F=35N` `a_(p)=0` and hence `a_(1)=a_(2)=0` As mass of the pulley is negligible `F-2T=0` `rArr T=F//2 rArr T=35N` To lift `m_(2)`, `T ge m_(2)g rArr T ge 50N` Therefore, block `m_(2)` will not move `rArr T-m_(1)g=m_(1)a_(1) rArr 15=2a_(1)` `rArr a_(1)=(15)/(2)m//s^(2)` CONSTRAINT equation `y_(p)+y_(p)-y_(1)=` constant `rArr 2y_(p)-y_(1)=c rArr 2(d^(2)y_(p))/(dt^(2))-(d^(2)y_(1))/(dt^(2))=0` `rArr a_(p)=(a_(1))/(2)=(15)/(4) m//s^(2)` (c) When F = 140N T = 70N `rArr T-m_(1)g=m_(1)a_(1)"" ...(1)` `rArr 70N-20N=2xxa_(1) rArr a_(1)=25m//s^(2)` `T-m_(2)g=m_(2)a_(2)"" ...(2)` `rArr 70N-50N=5a_(2) rArr a_(2)=4m//s^(2)` Constraint equation `y_(p)-y_(2)+y_(p)-y_(1)=c` `rArr 2y_(p)-y_(1)-y_(2)=c` `rArr 2(d^(2)y_(p))/(dt^(2))-(d^(2)y_(1))/(dt^(2))-(d^(2)y_(2))/(dt^(2))=0` `rArr a_(p)=(a_(1)+a_(2))/(2)=(29)/(2)m//s^(2)` 
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