What’s the Easiest Way to Find Tension in Pulley Systems?

To prove the formula for the tension T in the pulley system given in the image, let's go through the process step by step.

The formula provided is:

$$T = \frac{(C_1 + C_2 + C_3 + C_4 + \dots) \cdot g}{\left(\frac{C_1^2}{M_1} + \frac{C_2^2}{M_2} + \dots\right)}$$

Where:

• $C_i$ represents the constant related to each pulley or mass.
• $M_i$ represents the mass of each block connected to the pulleys.
• $g$ is the acceleration due to gravity.

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Step-by-Step Proof

1. Identify the Setup of the System: In this pulley system, we have multiple blocks of masses $M_1,M_2,M_3,\dots \; connected\; by\; strings\; and\; pulleys.\; Each\; string\; is\; associated\; with\; a\; constant$$C_i$, which could represent a tension constant for each block due to its pulley configuration.

2. Consider Forces Acting on Each Mass: For each mass $M_i$, the tension in the string pulling it up can be represented as $C_i \cdot T$ According to Newton's second law, the net force on each block will be the difference between the upward force (tension) and the gravitational force.

   For each block, we have:

   $$C_i \cdot T - M_i \cdot g = M_i \cdot a$$
   

   where $a$ is the acceleration of the system, and $T$ is the common tension in the main string connecting all pulleys.

3. Set Up the System of Equations: Rearranging each equation, we get:

   $$C_i \cdot T = M_i \cdot (g + a)$$
   

   Summing up these equations for all masses in the system:

   $$\sum C_i \cdot T = \sum M_i \cdot (g + a)$$
   

   This gives:

   $$T \sum C_i = (g + a) \sum M_i$$

4. Solve for T in Terms of g: Using the fact that the system has an overall acceleration due to the collective effect of all masses and pulleys, we simplify the expression to isolate $T$ based on the given constants