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Two masses m(1) and m(2) are connected b...

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.

Text Solution

Verified by Experts

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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