Ropes and Pulleys

Few basic tips for solving problems on ropes(strings) and pulleys...

$\mathbf{1.}$ Tension in a string is a very useful concept in problem solving. It always acts towards the direction away from the connected mass.

$\mathbf{2.}$ Whenever strings are to be considered of negligible mass, then the tension throughout the string is to be kept constant.

$\mathbf{3.}$ For simplicity in many problems, pulleys are considered massless (negligible mass) and frictionless.

$\mathbf{4.}$ The effect of acceleration due to gravity is always downward, thus it's effect along with the tension in string determines the direction of acceleration of masses connected.

$\mathbf{5.} \quad \bullet$ If $T > mg$ , acceleration is towards the direction of tension, with the magnitude $\dfrac{T}{m}-g$

$\quad\quad\bullet$ If $T<mg$ then acceleration is towards gravity with the magnitude of $g-\dfrac{T}{m}$

In above note, $T$ is tension, measured in the SI unit $\text{newton (N)}$ , $m$ is mass of the connected body in $\text{kilogram (kg)}$ , and $g$ is acceleration due to gravity, which is taken as $9.8 \text{ms}^{-2}$, and in problem solving , for simplicity taken as $10\text{ms}^{-2}$.

Practising ropes-pulleys, Tension, a worked example.

$\textbf{Question}$- In this question, we're given that $2$ bodies are connected to each other via a string passing over a pulley. Masses of the bodies are $2kg$ and $4kg$ as shown. Find the tension $T$ is the string, and acceleration of both the bodies.

($g=10\text{ms}^{-2}$, string and pulley are massless, frictionless, length of the string doesn't change due to the masses.)

$\textbf{Solution}$-

First of all, we see by logic that the acceleration of $4kg$ body will be downwards (towards gravity force) and of $2kg$ body will be upwards (opposite to gravity)

Because the length of string is the same, we conclude that acceleration of both the masses is same by magnitude.

Let this acceleration be $a$.

Then, by using $F=ma$, we have the force equations for the 2 bodies as

for $4kg$ body - $(4g-T=4a)\implies 4a+T=40$

for $2kg$ body - $(T-2g=2a)\implies T-2a=20$

Solving the 2 equations for $T$ and $a$, we have $T=\frac{40}{3} \text{N}$ and $a= \frac{10}{3} \text{ms}^{-2}$

This gives tension in string is $\frac{20}{3} \text{N}$ and acceleration of both the masses, which is equal, as $\frac{10}{3} \text{ms}^{-2}$