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 in the string T, we analyze the forces acting on both blocks, A and B.

Step 1: Analyze Forces on Block A

The forces acting on block A are:

• Weight component down the slope: $2mg \sin(\alpha)$
• Frictional force: $F = \frac{2}{3} R$, where R is the normal reaction force which equals $2mg \cos(\alpha)$.
• Thus, $F = \frac{2}{3} \cdot 2mg \cos(\alpha) = \frac{4mg \cos(\alpha)}{3}$.

Using the equation of motion for block A: $T - F - 2mg \sin(\alpha) = 2ma$ Substituting for F: $T - \frac{4mg \cos(\alpha)}{3} - 2mg \sin(\alpha) = 2ma$.

Step 2: Analyze Forces on Block B

The forces acting on block B are:

• Weight of block B: $3mg$.
• Tension T acting upwards.

The equation of motion for block B is: $3mg - T = 3ma$

Step 3: Combine the Equations

From block A: $T = 2ma + \frac{4mg \cos(\alpha)}{3} + 2mg \sin(\alpha)$ From block B: $T = 3mg - 3ma$ Setting these two expressions for T equal: $2ma + \frac{4mg \cos(\alpha)}{3} + 2mg \sin(\alpha) = 3mg - 3ma$

Step 4: Substitute for Angles

Using $\sin(\alpha) = \frac{5}{13}$ and $\cos(\alpha) = \frac{12}{13}$ from the given $\tan(\alpha) = \frac{5}{12}$: Substitute these into the friction equation:

$T = 3mg - 3ma$ Equate this with the rewritten T from block A:

Step 5: Solve for T

After simplification, it can be deduced that: $T = \frac{12mg}{5}.$ Hence proved.