Question

Three blocks of masses 2 m, 4 m and 6 m are placed as shown in figure. If \( \sin 37^\circ = \frac{3}{5} \), \( \sin 53^\circ = \frac{4}{5} \), the acceleration of the system is:

• A. \( \frac{17}{30} g \)
• B. \( \frac{13}{30} g \)
• C. \( \frac{13}{15} g \)
• D. \( \frac{15}{35} g \)

Hint:

When analyzing forces in pulley systems, always resolve forces along the incline and apply Newton's Second Law systematically.

Answer 1

The Correct Option is A

Step 1: Resolving forces along the inclined planes The forces acting along the incline for the three blocks are: - For mass \( 2m \) on the left incline at \( 37^\circ \): \[ F_1 = 2mg \sin 37^\circ = 2mg \times \frac{3}{5} = \frac{6}{5} mg \] - For mass \( 4m \) at the top pulley: \[ F_2 = 4m a \] - For mass \( 6m \) on the right incline at \( 53^\circ \): \[ F_3 = 6mg \sin 53^\circ = 6mg \times \frac{4}{5} = \frac{24}{5} mg \] Step 2: Applying Newton’s second law For the system in motion: \[ F_3 - F_1 = (2m + 4m + 6m) a \] \[ \frac{24}{5} mg - \frac{6}{5} mg = 12m a \] \[ \frac{18}{5} mg = 12m a \] \[ a = \frac{18}{5} \times \frac{1}{12} g = \frac{18}{60} g = \frac{3}{10} g = \frac{17}{30} g \] Thus, the acceleration of the system is: \[ \frac{17}{30} g \]