In the arrangement shown in figure mass of blocks A,B and C is `18.5 kg .8 kg` and `1.5 kg` respectively . The bottom surface of A is A is smooth while coefficient of friction that between block A and C is `mu_(1) = 1//3` between B and floor is `mu_(2) = 1//5` System is relased from rest `r = 0` and pulley are light and friction Calculate acceleration of block A ,B and C

Answer

`a = 1 ms^(-2),b= 0.5ms^(-2)c = 3 ms^(-2)`

Explanation

Let the acceleration of A , B, and C are a,b, and c respectively
` `
Net sum of charge of segment length in string should be zero
`2(-x_(A)) + 2(-x_(B)) + x_(c) = 0 rArr c = 2(a+ b)`

For A in vertical direction
`N' = m_(A)g + T+ f_(1)`...(i)
where `f_(1) = mu_(s) N_(1) = (N_(1))/(3)`
In horizontal direction
`2T - N_(1) = m_(A)a`...(ii)
For C orizontal direction
`N_(1) = m_(c)a`...(iii)
In vertical direction `m_(c)g - f_(1) - T = m_(c) 2 (a + b)`...(iv)
`m_(c)g - (N_(1))/(3) - T = m_(c) 2(a + b)`...(v)
For B: in vertical direction `N'' = m_(B) g`....(vi)
In horizontal direction
`2T - f_(2) = m_(B)b`
where `f_(2) = mu_(2)N' = (N')/(3)`....(vii)
After solving above equation we get
`a = 1 ms^(-2) , b= 0.5ms^(-2)`
`c = 2 (a+b) = 2 (1+0.5) = 3 ms^(-2)`
Net acceleration of `c = sqrt(a^(2) + c^(2)) = sqrt(1^(2) + 3^(2)) = sqrt(10) ms^(-2)`