Tension in an Inextensible String Attached to Equal Masses over a Frictionless Pulley: Principles and Calculations

In this article, we will explore the concept of tension in an inextensible string that is attached to equal masses and passes over a frictionless pulley. Understanding the principles and calculations involved in such a system is crucial for grasping basic mechanics and statics.

Introduction to the System

The scenario typically involves an inextensible string that passes over a frictionless pulley with equal masses ( m ) attached to each end. Such a system is often analyzed in introductory physics courses to understand stable equilibrium and force interactions. The key to solving these problems lies in applying Newton's laws of motion and the principles of force equilibrium.

Forces Acting on the Masses

Each mass ( m ) in the system experiences two forces:

The gravitational force ( F_g mg ) acting downwards. The tension ( T ) in the string acting upwards.

Where ( m ) is the mass and ( g ) is the acceleration due to gravity, approximately ( 9.81 , text{m/s}^2 ).

Equilibrium Condition

When the system is at rest or moving with a constant velocity, the net force acting on each mass is zero. Hence, we can write the equation for one of the masses as:

( T - mg 0 )

Rearranging this, we get:

( T mg )

Analysis of Forces Using Free Body Diagrams

Consider a free body diagram (FBD) of the system. For each mass, draw the force vectors. On the left side, the forces are:

Tension ( T_1 ) acting upwards. Gravitational force ( W_1 mg ) acting downwards.

Similarly, on the right side, the forces are:

Tension ( T_2 ) acting upwards. Gravitational force ( W_2 mg ) acting downwards.

Since the string is inextensible and the pulley is frictionless, the tensions on both sides of the pulley are equal, i.e., ( T_1 T_2 ).

According to Newton's second law of motion, the net force on each mass must be zero for the system to be in equilibrium:

( T_1 - mg 0 ) and ( T_2 - mg 0 )

This gives us:

( T_1 T_2 mg )

Equilibrium in a Bearing-Supported Pulley System

Consider a bearing-supported pulley system. If the supported mass on each side is the same and there is no externally applied torque, the system is in static equilibrium. Therefore, the angular acceleration is zero:

( alpha 0 )

For moment equilibrium about point A:

( T_1R - T_2R I_Aalpha )

Since ( T_1 T_2 ) and ( alpha 0 ):

( T_1R - T_2R 0 )

This simplifies to:

( T_1 T_2 )

Thus, the tension in the string is equal to the weight of one of the masses:

( T mg )

Conclusion

The tension in the inextensible string attached to equal masses over a frictionless pulley is constant and equal to the weight of the masses attached to it. This principle is fundamental to understanding statics and the behavior of systems under such constraints.

Many students mistakenly believe that equilibrium means no motion. In fact, a system can be in static equilibrium when the masses move at a constant velocity, with no acceleration.

Understanding and applying these concepts correctly is crucial for solving more complex problems in mechanics and engineering.