A plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall - Edexcel - A-Level Maths Mechanics - Question 9 - 2018 - Paper 1

Using the result from part (a), we have:

$$T = \frac{2Mg}{\cos(α)}.$$

Substituting this into the equation we have from part (a), we set:

$$\frac{5Mg(3x + α)}{6a} = \frac{2Mg}{\cos(α)}.$$

From this, we can find the value of x. Rearranging gives:

$$3x + α = \frac{12a}{5} \cos(α).$$

Solving for x:

$$x = \frac{12a}{15} \cos(α) - \frac{α}{3} = \frac{2a}{3} \cos(α) - \frac{α}{3}.$$

To find tan β, we can use the relationship between the vertical and horizontal components of the forces. Let us resolve vertically:

$$Y = 3Mg + T \sin(α).$$

Equation (1) gives us:

$$Y = 3Mg + \frac{5Mg(3x + α)}{6a} \sin(α).$$

Then, resolving horizontally:

$$X = R imes \cos(β) = 2Mg.$$

Using triangles formed by the forces, we can apply:

$$tan(β) = \frac{Y}{X}.$$

We obtain:

$$tan(β) = \frac{5Mg}{2Mg} = \frac{5}{2}.$$

For the system to remain in equilibrium, the tension T in the rope must not exceed 5Mg. Therefore, we can derive the condition under which the block remains stable:

$$\frac{5Mg(3x + α)}{6a} \leq 5Mg.$$

Cancelling Mg from both sides, we get:

$$\frac{3x + α}{6a} \leq 1.$$

Rearranging gives:

$$3x + α \leq 6a.$$

This means that the maximum possible value of x is related to a, restricting the placement of P to a specific range. Thus, we can express:

$$x \leq \frac{6a - α}{3}.$$

This relationship ensures that if x attempts to exceed ( \frac{6a - α}{3} ), the tension will surpass its limit, leading to the potential breaking of the rope.