Q.Two bodies of masses $10\ \text{kg}$ and $20\ \text{kg}$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $F = 600\ \text{N}$ is applied to

The key idea is that the two bodies, being tied by a string and moving together, will have a common acceleration. We apply Newton's Second Law ($F_{net} = ma$) to the system as a whole to find this acceleration, and then to individual bodies to find the tension.

Given: $m_A = 10\ \text{kg}$, $m_B = 20\ \text{kg}$, $F = 600\ \text{N}$. The surface is smooth, so there is no friction.

1. Calculate common acceleration:

   The total mass of the system is $M = m_A + m_B = 10\ \text{kg} + 20\ \text{kg} = 30\ \text{kg}$.

   The net external force on the system is $F = 600\ \text{N}$.

   The common acceleration $a = \frac{F}{M} = \frac{600\ \text{N}}{30\ \text{kg}} = 20\ \text{m/s}^2$.

2. Case (i): Force $F$ applied to A

   Consider the free-body diagram of mass B. The only horizontal force acting on B is the tension $T$ in the string.

   Applying Newton's Second Law to mass B: $T = m_B a$.

   $T = (20\ \text{kg})(20\ \text{m/s}^2) = 400\ \text{N}$. …

When dealing with connected bodies, the fundamental principle is Newton's Second Law of Motion. The problem describes two masses connected by a "light string" on a "smooth, horizontal surface." These are crucial details:

• Light string: This means the string's mass is negligible. Consequently, the tension is uniform throughout the string, and we don't need to account for its inertia.
• Smooth surface: This implies there is no friction acting on the bodies, simplifying our force analysis.
• Tied to the ends of a string: This means the two bodies move together as a single unit. They will have the same acceleration.

Our strategy will be to first find the common acceleration of the system by considering both masses as a single unit. Then, we will isolate one of the masses and apply Newton's Second Law to it to find the tension in the string.

Let $m_1 = 10\ \text{kg}$ and $m_2 = 20\ \text{kg}$. The applied force is $F = 600\ \text{N}$.

Case (i): Force $F$ applied to mass A ($m_1$)

1. Determine the common acceleration of the system.

   Since the string is inextensible, both masses $m_1$ and $m_2$ will move together with the same acceleration, let's call it $a$. We can treat the two masses as a single system with a total mass $M = m_1 + m_2$. The net external force acting on this system in the horizontal direction is $F$.

   According to Newton's Second Law, $F_{\text{net}} = M a$.

   So, $F = (m_1 + m_2) a$.

   Substituting the given values:

   $600\ \text{N} = (10\ \text{kg} + 20\ \text{kg}) a$

   $600\ \text{N} = (30\ \text{kg}) a$

   $a = \frac{600\ \text{N}}{30\ \text{kg}} = 20\ \text{m/s}^2$

   This is the acceleration of both $m_1$ and $m_2$.

2. Find the tension in the string by analyzing mass B ($m_2$).

   Now, consider mass $m_2$ in isolation. The only horizontal force acting on $m_2$ is the tension $T$ from the string, pulling it in the direction of acceleration.

   Drawing a Free Body Diagram (FBD) for $m_2$:

   • Horizontal forces: Tension $T$ (to the right, assuming $F$ is applied to the right).
   • Vertical forces: Normal force $N_2$ (upwards) and gravitational force $m_2 g$ (downwards). These balance each other as there is no vertical acceleration.

   Applying Newton's Second Law to $m_2$ in the horizontal direction:

   $T = m_2 a$

   Substitute the values for $m_2$ and $a$:

   $T = (20\ \text{kg}) (20\ \text{m/s}^2)$

   $T = 400\ \text{N}$

   Tip

   When finding tension in a connected system, it's often easier to analyze the block not directly experiencing the external pulling force. This way, the tension is the only horizontal force on that block, simplifying the equation.

Case (ii): Force $F$ applied to mass B ($m_2$)

1. Determine the common acceleration of the system. …

Concept: Newton's Second Law for Two Bodies Connected by a Light String

Step 1: Find the common acceleration of the system

The string is light and inextensible and the surface smooth, so $m_A=10\ \text{kg}$ and

$m_B=20\ \text{kg}$ move together with one acceleration $a$. Treating both as a single

system of total mass $M=m_A+m_B$:

This acceleration is the same in both cases (i) and (ii), since $F$ and $M$ don't change.

Step 2: Case (i) — force applied to A

The tension is the only horizontal force on B, so isolate B:

Step 3: Case (ii) — force applied to B

Now the tension is the only horizontal force on A, so isolate A: …