Find the minimum value of `F` and the tension in the string so that the blocks move with constant velocity. The coefficient of friction at all surfaces is `mu`.
(a)
(b)
( c )
(d)

Explanation

(a) `N_(0) = mg , f_(1) = mu N_(0) = mu mg`
`F - T - f_(1) = 0` (i)
`N = (m + 4m)g = 5 mg , f_(2) = mu N = 5 mu mg`
`T - f_(1) - f_(2) =0` (ii)
`T = f_(1) + f_(2) = 6 mu mg`
`F = T + f_(1) = 7 mu mg`
(b) `N_(0) = mg , f_(1) = mu N_(0) = mu mg`
`T - f_(1) = 0 rArr T = mu mg`
`N = (m + 4 m)g = 5 mg , f_(2) = mu N = 5 mu mg`
`F - T - f_(1) - f_(2) = 0`
`F = T + f_(1) + f_(2) = 7 mu mg`
(c ) `uarr : N_(1) + (T)/(sqrt(2)) - 100 = 0 rArr N_(1) = 100 - (T)/(sqrt(2))`
`f_(1) = mu N_(1) = (1)/(4) (100 - (T)/(sqrt(2))) = 25 - (T)/(4 sqrt(2))`
`rarr : (T)/(sqrt(2)) - f_(1) = 0 rArr (T)/(sqrt(2)) - 25 + (T)/(4 sqrt(2)) rArr (5 T)/(4 sqrt(2)) = 25`
`T = 20 sqrt(2) N`
`N_(1) = 100 - (T)/(sqrt(2)) = 80 N , f_(1) = 25 - (T)/(4 sqrt(2)) = 20 N`
`uarr : N_(2) - N_(1) - 200 = 0 rArr N_(2) = N_(1) + 200 = 280 N`
`f_(2) = mu N_(2) = (1)/(4) xx 280 = 70 N`
`larr : F - f_(1) - f_(2) = 0 rArr F - 20 - 70 = 0`
`F = 90 N`
Applied force `F = 90 N` , Tension `T = 20 sqrt(2) N`
(d) `m : darr /, mg - N_(1) = ma_(0)`
`N_(1) = m(g - a_(0)) = (mg)/(2)`
`f_(1) = mu N_(1) = (mu mg)/(2)`
`rarr : T - f_(1) = 0 rArr T = f_(1) = (mu mg)/(2)`
`4 m: darr : 4 mg + N_(1) - N_(2) = 4 ma_(0) = (4 mg)/(2)`
`N_(2) = 2 mg + (mg)/(2) = (5 mu mg)/(2)`
`larr : F - T - f_(1) - f_(2) = 0`
`F = T + f_(1) + f_(2) = (7 mu mg)/(2)` ltbr5gt Note : When a block is moving in a lift with acceleration `a_(0) (a_(0) lt g)` in downward direction , we can use `(g - a_(0))` in place of `g`. In part `(b)` , for calculating normal reactions `N_(1)` and `N_(2)` , replace `g by (g - a_(0))`.