Three blocks , of masses `m_(1) = 2.0 ,m_(2) = 4.0 ` and `m_(3) =6.0 kg` are connected by strings on a frictionless inclined plane of `60^(@)` , as shown in the figure. A force `F = 120N` is applied upward along the incline to the uppermost block, causing an upward movement of the blocks. The connecting cords are light. The values of tensions `T_(1)` and `T_(2)` in the cords are.

Options

A. `T_(1) = 20N, T_(2) = 60 N`
B. `T_(1) = 60 N, T_(2) = 60 N`
C. `T_(1) = 30N, T_(2) = 50N`
D. `T_(1) = 20 N, T_(2) = 100 N`

Answer

A

Explanation

Here
`m_(1) = 2.0 kg , m_(2) = 4.0kg , m _(3) = 6.0 kg F = 120 N`
Let a be common acceleration of the system.
The equation of motion of block 1 is
`T_(1) - m_(1) g sin 60^(@) = m_(1) a" "…….(i)`
The equation of motion of block 2 is
`T_(2) - T_(1) - m_(2)g sin 60^(@) = m_(2)a " "............(ii)`
The equation of motion of block 3 is
`F - T_(2) - m_(3)g sin 60^(@) = m_(3)a" "..........(iii)`
Adding (i), (ii) and (iii) , we get
`F - (m_(1) + m_(2) + m_(3)) g sin60^(@) = (m_(1) + m_(2) + m_(3))a`
`a = (F - (m_(1) + m_(2) + m_(3))g sin 60^(@))/(m_(1) + m_(2) +m_(3))`
` = (120N - (2 kg + 4kg + 6kg)(9.8m//s^(@))(0.866))/(2kg + 4kg + 6kg)`
`=(120 - (12kg)(9.8m//s^(2))(0.866))/(12 kg) = 1.51m//s^(2)`
From equation (i) ,
`T_(1) = m_(1)g sin 60^(@) + m_(2) a= (2kg) (9.8m//s^(2))(0.86) + (2kg) (1.51m//s^(2))`
` = 20 N`
From equation (iii) ,
`T_(2) = F - m_(3) g sin 60^(@) - m_(3)a = 120 - (6kg)(9.8m//s^(2))(0.866) - (6kg)(1.51 m//s^(2))`
= 60 N.