Question

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible strings that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.

Answer 1

$\bf \: {SOLUTION : - }$

The given system of two masses and a pulley can be represented as shown in the above attached figure ↑↑.

$\rm \: Smaller \: mass \: m_1 = 8kg$

$\rm \: Larger \: mass \: m_2 = 12kg$

Tension in the string = T

$\rm Mass \: m_2 \: owing \: to \: its \: weight \: moves \: downward \\\rm \: with \: acceleration \: a \: and \: mass \: m_1 \: moves \: upward. \:$

Applying Newton's second law of motion to the system of each mass

$\rm \: For \: mass \: m_1 \: the \: equation \: of \: motion \: can \: be \: written \: as :$

$\rm \: T - m_1g = ma \: \: \: \: \: \: \: ...(i)$

$\rm \: For \: mass \: m_2 \: the \: equation \: of \: mass \: can \: be \: written \: as :$

$\rm \: m_2g - T = m_2a \: \: \: \: \: \: \: ...(ii)$

Adding both equations (i) and (ii), we get:

$\rm \: (m_2 - m_1)g = (m_1 + m_2)a$

$\therefore \rm \: a = \bigg( \frac{m_2 - m_1}{m_1 + m_2} \bigg)g \: \: \: \: \: \: \: ...(iii)$

$\rm \: = \bigg( \frac{12 - 8}{12 + 8} \bigg) \times \frac{4}{20} = 2m s {}^{2}$

Therefore, the acceleration of the masses is 2m/s².

Substituting the value of a in equation (ii), we get:

$\rm \: m_2g - T = m_2 \bigg( \frac{m_2 - m_1}{m_1 + m_2} \bigg) \: g$

$\rm \: T = \bigg( m_2 - \frac{m_2 {}^{2} - m_1m_2}{m_1 + m_2} \bigg) \: g$

$\rm \: = \bigg( \frac{2m_1m_2}{m_1 + m_2} \bigg) \: g$

$\rm \: = \bigg( \frac{2 \times 12 \times 8}{12 + 8} \bigg) \times 10$

$\rm \: = \frac{2 \times 12 \times 8}{20} \times 10 = 96N$

Therefore, the tension in the string is 96 N.

Answer 2

If two masses $\sf{m_1}$ and $\sf{m_2}$ (in kg) where $\sf{m_1>m_2}$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley, then the acceleration of the system and the tension in the string are respectively,

• $\sf{a=\left(\dfrac{m_1-m_2}{m_1+m_2}\right)g}$

and,

• $\sf{T=\dfrac{2m_1m_2g}{m_1+m_2}}$

Let $\sf{g=10\ m\,s^{-2}.}$

In the question,

• $\sf{m_1=12\ kg}$

• $\sf{m_2=8\ kg}$

Then, acceleration of the system,

$\sf{\longrightarrow a=\left(\dfrac{m_1-m_2}{m_1+m_2}\right)g}$

$\sf{\longrightarrow a=\left(\dfrac{12-8}{12+8}\right)10}$

$\sf{\longrightarrow\underline{\underline{a=2\ m\,s^{-2}}}}$

And, tension in the string,

$\sf{\longrightarrow T=\dfrac{2m_1m_2g}{m_1+m_2}}$

$\sf{\longrightarrow T=\dfrac{2\times12\times8\times10}{12+8}}$

$\sf{\longrightarrow\underline{\underline{T=96\ N}}}$