First slide

Connected bodies

Question

Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively are in contact on a frictionless surface as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

Moderate

A

2 N
B

6 N
C

8 N
D

18 N

Solution

Given, $m_{^{A}} = 4 k g, m_{B} = 2 k g a n d m_{^{C}} = 1 k g$
So, total mass, M = 4 + 2 + 1 = 7 kg
Now, $F = M a \Rightarrow 14 = 7 a \Rightarrow a = 2 {ms}^{- 2}$
FBD of block A ,
$F - F^{'} = 4 a \Rightarrow F^{'} = 14 - 4 \times 2 \Rightarrow F^{'} = 6 N$
Hence, the contact force between A and B is 6N.

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