Question

Two weights ${\text{w}}_1$ and ${\text{w}}_2$ are suspended on the ends of a string on a frictionless pulley. When pulled up with an acceleration g, the tension in the string is?

Answer 1

Step1: Given

Weigh of the First body is ${\text{w}}_1$

Weight of the second body is ${\text{w}}_2$

Mass of first body : ${\text{m}}_1$

Mass of the second body : ${\text{m}}_2$

Step 2: Calculating of the tension

FBD of the system is shown below.

Equilibrium of the forces for mass ${\text{m}}_2$ is as follows.

${\text{T-w}}_2-{\text{m}}_2{\text{a}}_0={\text{m}}_2\text{a}$

Where tension in the string is $\text{T}$, acceleration of the body in moving upward is ${\text{a}}_0$ and it is equal to acceleration due to gravity, acceleration of the mass is $\text{a}$.

Equilibrium of the forces for mass ${\text{m}}_1$ is as follows.

$$\begin{gathered} {\text{m}}_1{\text{a}}_0+{\text{w}}_1-\text{T}={\text{m}}_1\text{a} \\ -{\text{T+m}}_1{\text{g+w}}_1{\text{=m}}_1\text{a} \\ -{\text{T+w}}_1+{\text{w}}_1{\text{=m}}_1\text{a} \\ \text{a}=\frac{1}{{\text{m}}_1}\left(\text{-T}+2{\text{w}}_1\right) \end{gathered}$$

Substitute value of $\text{a}$ in above equation.

$$\begin{gathered} \text{T}-{\text{w}}_2-{\text{m}}_2{\text{g=m}}_2\text{×}\frac{1}{{\text{m}}_1}\left(-\text{T}+2{\text{w}}_1\right) \\ \text{T}-{\text{w}}_2-{\text{w}}_2{\text{=m}}_2\text{g×}\frac{1}{{\text{m}}_1\text{g}}\left(-\text{T}+2{\text{w}}_1\right) \\ \text{T}-2{\text{w}}_2{\text{=w}}_2\text{×}\frac{1}{{\text{w}}_1}\left(-\text{T}+2{\text{w}}_1\right) \\ \text{T=}\frac{4{\text{w}}_1{\text{w}}_2}{{\text{w}}_1{\text{w}}_2} \end{gathered}$$

Hence, the required tension is $\frac{4{\text{w}}_1{\text{w}}_2}{{\text{w}}_1{\text{w}}_2}$.