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

From part (a), we have established a relation for T. Now using the horizontal component of the force:
The horizontal force at A due to the wall is given as 2Mg:

$2M g = T \cdot \cos a$

Substituting our previously found expression for T:

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

Using $\cos a = \sqrt{1 - \sin^2 a} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$, we can now find the value of x:

We found the components of the forces acting on the plank. The direction of this horizontal component relates to the angle \beta. To find \tan \beta, we can use the opposite and adjacent side lengths. We know that the vertical component due to the tension acts upwards, so:

$\tan \beta = \frac{Y}{X} = \frac{5Mg}{2Mg} = \frac{5}{2}$

This indicates how steeply the force at A will act. Given that the maximum force allowable contributes directly to the positioning of P, any increase in P's position would increase the tension and might violate the tension limit of 5Mg:

Thus, there exists a limit where:

$5Mg(3x + a) \leq 5Mg$

From this inequality, we can derive the maximum limit for x that maintains equilibrium.

For the rope not to break, the tension must not exceed 5Mg. Therefore, we set up our inequality:

$5Mg(3x + a) \leq 5Mg$

Dividing both sides by 5Mg gives:

$3x + a \leq 1$

Rearranging gives:

$x \leq \frac{1 - a}{3}$

This means that the maximum position of P on the plank is dependent on a. If x exceeds the derived limit, tensions will spike due to gravitational forces, leading to potential failure of the rope.