Animated Solution for Physics - Laws of Motion: Three identical blocks of masses $m=2\text{ kg}$ are drawn by a force $F=10.2\text{ N}$ with an acceleration of $0.6{\text{ ms}}^{-2}$ on a frictionless surface, then what is the tension (in N) in the string between the blocks $B$ and $C$?

Imagine you are pulling a train of three identical blocks across a surface. The force you apply at the front has to drag not just the first block, but the entire chain behind it. This means the tension in the connecting strings will vary depending on how much mass is trailing behind each point. Let's break down this classic mechanics problem step-by-step.

We are given a system of three identical blocks, let's call them A, B, and C, from right to left. Each block has a mass of $m=2\text{ kg}$. A force $F=10.2\text{ N}$ is applied to the leading block (Block A), causing the entire system to accelerate at $a=0.6{\text{ m/s}}^2$.

Our mission is to find the tension in the string connecting the last two blocks, B and C. To do this systematically, we will define the tension between A and B as $T_1$, and the tension between B and C as $T_2$. The secret to solving connected body problems is to isolate each block and draw its Free Body Diagram (FBD).

Block A

Let's isolate the leading block, Block A. It is being pulled forward by the external force $F$, but it is also being held back by the tension $T_1$ from the string connecting it to Block B.

According to Newton's Second Law, the net force on an object equals its mass times its acceleration ($F_{net}=ma$). For Block A, this gives us our first master equation:

We know all the values except $T_1$. Let's substitute them in:

Solving for $T_1$, we get:

So, the string between A and B is experiencing a tension of $9\text{ N}$.

Block B

Now, let's move down the chain to Block B. This block is being pulled forward by the tension $T_1$ (which we just found), and it is being held back by the tension $T_2$ from the string connecting it to Block C.

Applying Newton's Second Law to Block B, we get:

Substitute the value of $T_1$ we just calculated, along with the mass and acceleration:

Solving for $T_2$, we find our final answer:

You might have noticed something strange if you have a keen eye for physics. If the surface were truly frictionless, the acceleration of the $6\text{ kg}$ system under a $10.2\text{ N}$ force would be $1.7{\text{ m/s}}^2$, not $0.6{\text{ m/s}}^2$. The given acceleration implies there is some hidden retarding force (like friction) acting on the system. However, in exam scenarios like this, strictly following the sequential force equations block-by-block from the point of applied force ensures you arrive at the examiner's intended mathematical solution.