Four blocks of the same mass m connected by cords are pulled by a force F on a smooth horizontal surface as shown in Determine the tensions `T_1 ,T_2` and `T_3` in the cords.

Options

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

Answer

A

Explanation

Here, `m_(1)=2.0kg, m_(2)=4.0 kg, m_(3)=6.0kg`.
F=120
Let a be common acceleration of the system.
The equation of motion of block 1 is
`T_(1)-m_(1) g sin60^(@)=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 sin 60^(@)=(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-(2kg+4kg+6kg)(9.8m//s^(2))(0.866))/(2kg+4kg+6kg)`
`(120-(12kg)(9.8m//s^(2))(0.866))/(12kg)`
`=1.51 m//s^(2)`
From equation (i)
`T_(1)=m_(1)g sin 60^(@)+m_(1)a`
`=(2kg)(9.8m//s^(2))(0.866)+(2kg)(1.51) m//s^(2))`
=20N
From equation (iii)
`T_(2)=F-m_(3)g sin 60^(@)-m_(3)a`
`=120N-(6kg)(9.8 m//s^(2))(0.866)-(6kg)(1.5m//s^(2))`
=60N