Two blocks, A and B, of masses 2m and 3m respectively, are attached to the ends of a light string - Edexcel - A-Level Maths Mechanics - Question 3 - 2019 - Paper 1

To find the tension T in the string, we need to analyze the forces acting on both blocks A and B.

For Block A:

• The forces acting on A are:
  • Tension upward (T)
  • Weight component down the slope: $2mg \sin \alpha$
  • Frictional force: $F = \frac{2}{3} R$, where R is the normal reaction force.

Calculating R: Since the normal reaction R is perpendicular to the slope: $R = 2mg \cos \alpha$ Thus, $F = \frac{2}{3} R = \frac{2}{3} (2mg \cos \alpha) = \frac{4mg \cos \alpha}{3}$

Now writing the equation of motion for A: $T - \frac{4mg \cos \alpha}{3} - 2mg \sin \alpha = 2ma \quad (1)$

For Block B:

• The forces acting on B are:
  • Weight acting downward: $3mg$
  • Tension acting upward:

Writing the equation of motion for B: $3mg - T = 3ma \quad (2)$

Using the angle given by ( \tan \alpha = \frac{5}{12} ), we can find ( \sin \alpha ) and ( \cos \alpha ) using the triangle ratios: $\sin \alpha = \frac{5}{13}, \quad \cos \alpha = \frac{12}{13}$

Substituting these into equation (1): $T - \frac{4mg \cdot \frac{12}{13}}{3} - 2mg \cdot \frac{5}{13} = 2ma$

Now substituting for ( a ) from equation (2), we can eliminate a and simplify both equations to solve for T. Eventually, we will arrive at: $T = \frac{12mg}{5}$.