The Tension in a Mass-Suspended String with Pulleys

In a physics problem, two masses of 6 kg and 4 kg are attached to either end of a string that is suspended over a pulley. This system is a fundamental example of how pulleys and forces interact. Here, we explore the tension in the string and the acceleration of the masses.

Understanding the Physical Situation

Let's analyze this system. When two masses are connected by a string over a pulley, the forces acting on the system include the gravitational force (weight) pulling each mass downward and the tension in the string pulling each mass upward. Since the string is inextensible and massless, the tension is the same throughout.

Method 1: Simplified Approach

The first approach simplifies the problem by finding the downward acceleration of the 6 kg mass and the upward acceleration of the 4 kg mass. The net force on the 6 kg mass is:

Net Force on 6 kg mass 6g - T

Using Newton's Second Law, F ma, the equation becomes:

6g - T 6a

For the 4 kg mass, the net force is:

Net Force on 4 kg mass T - 4g

Again, using Newton's Second Law:

T - 4g -4a

Combining these equations and solving for T, we get:

69.8 - 1.96 47 N

Thus, the tension in the string is 47 N.

Method 2: System of Equations

To explore this further, let's use two equations based on the masses and gravitational acceleration (g 9.81 m/s2).

Let m1 6 kg and m2 4 kg, and let the tension in the string be F. We can write two equations:

m1g - F m1a

F - m2g m2a

By adding these equations, we can eliminate the tension T and solve for the acceleration a:

6g - 4g (m1 m2)a

2g (10 kg)a

a 0.2g

This acceleration is consistent with the expectations that the 4 kg mass will rise and the 6 kg mass will fall, maintaining the tension in the string.

Method 3: Detailed Calculation

For a more detailed approach, let's calculate the tension specifically for each mass.

For the 4 kg mass:

F - 4g 4a

4g - 4a 4g (since a 0.2g)

F 4g 4(0.2g) 4g 0.8g 4.8g 4.8 * 9.81 m/s2 47.1 N

For the 6 kg mass:

6g - F 6a

6g - F 6(0.2g) 1.2g

F 6g - 1.2g 4.8g 47.1 N

Thus, the tension in both parts of the string is 47.1 N.

Types of Pulleys

The problem specifies "from the end of a pulley," which typically implies a single fixed pulley. In a single fixed pulley, the tension in the string is the same as the gravitational force acting on the mass. Therefore, if the system involves a single pulley, the tension would be equal to the weight of the total mass, which is 10 N.

Conclusion

In conclusion, the tension in the string is 47.1 N, and the acceleration of the masses is 0.2g. The type of pulley further influences the tension, but in a single fixed pulley, the tension equals the weight of the hanging mass, which is 10 N.

Keywords

Tension in string, Pulley system, Gravitational force, Mass acceleration