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Q4.16:
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Solution :

Step 1: Given Variables & System Parameters

We are given two masses connected by a light inextensible string passing over a massless, frictionless pulley. The parameters are defined as follows:

  • Mass $m_1 = 8\text{ kg}$
  • Mass $m_2 = 12\text{ kg}$
  • Acceleration due to gravity $g \approx 9.8\text{ m/s}^2$
  • String: Light and inextensible (Tension $T$ is uniform throughout the string)
  • Pulley: Frictionless and massless
m₁ (8kg) m₂ (12kg) T

Step 2: Free Body Diagram (FBD) and Equations of Motion

Since $m_2 > m_1$, the system will accelerate with $m_2$ moving downward with acceleration $a$ and $m_1$ moving upward with the same acceleration $a$ [Per Newton’s Second Law].

For mass $m_1$ (upward motion):

$T - m_1g = m_1a$ --- (Equation 1)

For mass $m_2$ (downward motion):

$m_2g - T = m_2a$ --- (Equation 2)

Step 3: Calculating Acceleration ($a$)

Adding Equation 1 and Equation 2 to eliminate $T$:

$(T - m_1g) + (m_2g - T) = m_1a + m_2a$ $(m_2 - m_1)g = (m_1 + m_2)a$ $a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right)g$

Substituting the values:

$a = \left( \frac{12 - 8}{12 + 8} \right) \times 9.8$ $a = \left( \frac{4}{20} \right) \times 9.8 = 0.2 \times 9.8$ $a = 1.96\text{ m/s}^2$

Step 4: Calculating Tension ($T$)

Substitute $a$ into Equation 1:

$T = m_1(g + a)$ $T = 8 \times (9.8 + 1.96)$ $T = 8 \times 11.76$ $T = 94.08\text{ N}$

Step 5: Tabular Summary of Results

Parameter Result
Acceleration ($a$) $1.96\text{ m/s}^2$
Tension ($T$) $94.08\text{ N}$

Final Solution: The acceleration of the masses is $1.96\text{ m/s}^2$ and the tension in the string is $94.08\text{ N}$.


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