Question

In the arrangement shown tension in the string connecting 4 kg and 6 kg masses is:

A. 8N
B. 2N
C. 6N
D. 4N

Answer 1

We will first find the maximum friction force exerted by the system. Then we will use it to check the type of motion of the system. Then we will solve by writing the equation of motion for any one of the two given blocks.

Complete step by step answer:
Since the motion is in horizontal direction, therefore, we will only consider the forces that are acting horizontally,
Now, let us find the total (maximum) friction force,
${f_1} = 0.2 \times 60 = 12$N…………………………..(i.)
${f_2} = 0.2 \times 40 = 8$N…………………………….(ii.)
${f_3} = 0.2 \times 20 = 4$N…………………………..(iii.)
On adding eq., (i.), (ii.) and (iii.), we get,
$f = 24$
Since the total friction force is greater than the force applied to the system. Therefore, the system will not be moving.

${T_1} + 8 = 16N$
$\Rightarrow {T_1} = 8N$
Here,${T_1}$ is the tension between the 4 kg and 6 kg masses.
Hence, option (A) is the correct answer.

Note: In such questions, always begin with making the FBD. It simplifies the problem. Also, be careful while calculating the total friction force exerted by all three blocks. A diagram that shows a unit of a particular system representing all the external forces acting on it is called a free body diagram,