Find the acceleration and tension in the string in the following cases . The pulley and string are massless.
(a)
(b)

Explanation

(a) `50 - T = 5a` (i)
`uarr : N - 100 = 0 rArr N = 100`
`f = mu N = 0.6 xx 100 = 60 N`
`T - f = 10 a`
`T - 60 = 10 a` (ii)
Solving `a = -(2)/(3) m//s^(2)` , but the acceleration cannot be negative . The blocks are at rest, `T = 50 N`.
In this problem , we committed a mistake , we applied the maximum value of friction force but friction is a self - adjusting force which varies from zero to a certain maximum value. Here the value of friction `50 N` is sufficient to prevent relative motion. When there is relative motion , friction is fully developed but when the blocks are at rest , friction or acceleration force `= 5g = 50 N` , opposing force (maximum value) `= 60 N`.
(b) For block of mass `2m`,
`N_(1) = 2mg cos 45^(@) = sqrt(2) mg`
`f_(1) = mu_(1) N_(1) = (sqrt(2))/(3) mg`
Along the plane : `2 mg sin 45^(@) - T - F_(1) = 2 ma`
`sqrt(2) mg - T - (sqrt(2))/(3) mg = 2 ma` (i)
For block of mass `m`:
`N_(2) = mg cos 45^(@) = (mg)/(sqrt(2))`.
`f_(2) = mu_(2) N_(2) = ( 2mg)/( 3 sqrt(2))`
Along the plane : `T - mg sin 45^(@) - f_(2) = ma`
`T - (mg)/(sqrt(2)) - (2mg)/(3 sqrt(2)) = ma` (ii)
Adding (i) and (ii) , we get
`3 ma = - (2mg)/(3 sqrt(2)) rArr a = -(g)/(9 sqrt(2)) = -ve`
Let the block of mass `m` move downward by solving , acceleration comes out to be `- ve` , i.e. system will not move , the blocks are at rest.
OR
Let the block of mass `2 m` moves downward , the force which tries to accelerate the system `= 2 mg sin 45^(@) - mg sin 45^(@) = (mg)/(sqrt(2))`.
Oppsing force `= f_(1) + f_(2) = ((sqrt(2))/(3) + (2)/(3 sqrt(2))) mg = (4 mg)/( 3sqrt(2))`
`(mg)/(sqrt(2)) lt ( 4mg)/(3 sqrt(2))` , so the system is at rest
Tension : Block of mass `2 m`
`2 mg sin 45^(@) = T + f_(1)`
`sqrt(2) mg = T + (sqrt(2))/(3) mg`
`T = (sqrt(2) - (sqrt(2))/(3))mg = ( 2sqrt(2) mg)/( 3)`
Block of mass `m`
`T = mg sin 45^(@) + f_(2) = (mg)/(sqrt(2)) + f_(2)`
`f_(2) = ((2 sqrt(2))/(3) - (1)/(sqrt(2))) = (mg)/(3 sqrt(2))`
The friction force is fully developed , where the friction coefficient is smaller , hence we have to use the block of `2 m` to find tension.