A uniform rod, AB, has length 7 metres and mass 4 kilograms - AQA - A-Level Maths Mechanics - Question 13 - 2020 - Paper 2

To find the weight $W$, we need to calculate the moment of force about the pivot point C. The moment is given by the formula:

$ext{Moment} = ext{Force} imes ext{Distance}$

For the weight at A, the distance from C to A is 2 m. Therefore, the moment due to the weight $W$ is:

$ext{Moment}_{W} = W imes 2$

For the rod, which has a mass of 4 kg, the weight is:

$ext{Weight of rod} = 4 imes 9.81 = 39.24 ext{ N}$

The center of mass of the rod is located at the midpoint, which is 3.5 m from A (i.e., 1.5 m from C since AC = 2 m). Thus, the moment due to the weight of the rod is:

$ext{Moment}_{ ext{rod}} = 39.24 imes 1.5$

Setting the moments equal for equilibrium:

$W imes 2 = 39.24 imes 1.5$

Now, solve for $W$:

W = rac{39.24 imes 1.5}{2} = 29.43 ext{ N}

Therefore, W = rac{3g}{2}, using $g eq 9.81$ if required.