Find the force of interaction between the bodies as shown in $F i g . 6.36$ . Blocks are in contact.

Open in App

Solution

All the bodies move together along horizontal direction with an acceleration,
$a = \frac{{[F_{h o r i z o n t a l}]}_{n e t}}{m_{t o t a l}} = \frac{F cos θ}{m_{1} + m_{2} + m_{3}}$
By Newton's second law for block $m_{1}$ ,
$N_{1} = m_{1} a = m_{1} (\frac{F cos θ}{m_{1} + m_{2} + m_{3}})$
and for block $m_{2}$ , we have $N_{1} - N_{2} = m_{2} a$
or $N_{2} = N_{1} + m_{2} a = N_{1} + m_{2} (\frac{F cos θ}{m_{1} + m_{2} + m_{3}})$
Here the force of interaction is of compressive nature.
Hence, $N_{2} = \frac{(m_{1} + m_{2}) F cos θ}{(m_{1} + m_{2} + m_{3})}$
We can calculate the value of $N_{2}$ by taking $m_{2}$ and $m_{2}$ together as system. From FBD of $^{'} m_{1} + m_{2}^{'}$ as $F i g . 6.37$ (b)
or $N_{2} = (m_{1} + m_{2}) a = \frac{(m_{1} + m_{2}) F cos θ}{(m_{1} + m_{2} + m_{3})}$

Suggest Corrections

Similar questions

1 k g

2 k g

F i g s . 6.34

6.35

1 k g

2 k g

3 k g

4 k g

Join BYJU'S Learning Program