Calculating Tension in a Pulley System with Two Masses

Understanding the mechanics involved in a pulley system with two different masses is crucial for various engineering and physics applications. In this article, we will analyze a specific system, where a 5kg mass is resting on a frictionless table connected to a 3kg hanging mass via a light string passing over a frictionless pulley. We will calculate the tension in the string when the masses are released. Let’s break down the problem step-by-step.

Introduction to the System

We have two masses in our system:

A 5kg mass resting on a perfectly level, frictionless table (denoted as (m_1)).A 3kg mass hanging off the side, subject to gravity (denoted as (m_2)).

Given:

(m_1 5 , text{kg})(m_2 3 , text{kg})Gravitational acceleration, (g 9.81 , text{m/s}^2)

Forces Acting on the Masses

Let’s analyze the forces acting on each mass:

Hanging Mass ((m_2))

Weight of the mass: (W_2 m_2 cdot g) (3 , text{kg} cdot 9.81 , text{m/s}^2 29.43 , text{N})Tension in the string: (T)

The net force acting on (m_2) when released is:

(F_{text{net}2} m_2 cdot g - T)

Mass on the Table ((m_1))

The only force causing it to accelerate is the tension in the string: (T)

The net force acting on (m_1) is:

(F_{text{net}1} T)

Applying Newton's Second Law

According to Newton’s Second Law, for (m_2):

(m_2 cdot a m_2 cdot g - T) -- equation 1

And for (m_1):

(m_1 cdot a T) -- equation 2

Solving equation 2 for (T) in terms of acceleration (a):

(T m_1 cdot a) -- equation 3

Substitute equation 3 into equation 1:

(m_2 cdot a m_2 cdot g - m_1 cdot a)

Combine the terms:

(m_2 cdot a m_1 cdot a m_2 cdot g)

(a (m_1 m_2) m_2 cdot g)

Solving for (a) :

(a frac{m_2 cdot g}{m_1 m_2})

Substituting the values:

(a frac{3 , text{kg} cdot 9.81 , text{m/s}^2}{5 , text{kg} 3 , text{kg}} frac{29.43 , text{N}}{8 , text{kg}} 3.67875 , text{m/s}^2)

Calculating Tension

Now, substitute (a) back into equation 3 to find (T) :

(T m_1 cdot a 5 , text{kg} cdot 3.67875 , text{m/s}^2 18.39375 , text{N})

Conclusion: The tension in the string when the masses are released is approximately (18.39 , text{N}). If your calculated (T) was 14.7 N, there might have been a miscalculation in the acceleration or the forces involved.

My solution involves a similar approach:

Free-body diagrams for each Newton’s Second Law to each mass:Hanging Mass: (Sigma F_y m_2 cdot a - T m_2 cdot g) -- Equation 1Mass on the Table: (Sigma F_x T m_1 cdot a) -- Equation 2Solving the system of equations, we find:(T 18.4 , text{N}) and (a 3.68 , text{m/s}^2)