A uniform rod, AB, has length 3 metres and mass 24 kg - AQA - A-Level Maths Pure - Question 11 - 2017 - Paper 2

To find the value of M, we will take moments about point C. The clockwise moment due to the weight of the rod can be calculated as:

$24 imes g imes 0.7$

where $g$ is the acceleration due to gravity (approximately 9.81 m/s²).

The counterclockwise moment due to the particle of mass M is:

$M imes g imes 0.8$.

For the rod to be in equilibrium, the clockwise moments must balance the counterclockwise moments. Therefore, we can set up the equation:

$M imes g imes 0.8 = 24 imes g imes 0.7$

Dividing both sides by $g$, we simplify to:

$M imes 0.8 = 24 imes 0.7$

This gives us:

$M = \frac{24 imes 0.7}{0.8}$.