A roller gate as shown in Fig. P. 3.17 has a span of 5 m. Determine the magnitude and direction of the resultant force on the cylinder when water just begins to overflow. Neglect weight of the gate.
The horizontal force equals the force on the projected area,
γ A \bar{h} = 9810 × 4 × 5 × 2 = 392400 N
This acts at a distance of 1.3333 m from bottom twoards the right.
The vertical upward force is equal to the weight of the water displaced
= 5 × 9810 × π × 2^{2}/2 = 308190 N
It acts at a distance of (4r/3π) = 0.84883 m left of centre and upwards
Resultant = (392400^{2} + 308190^{2})^{0.5} = 498958 N
Taking moments about the centre,
392400 × (2 – 1.3333) – 0.84883 × 308189 = 0.
As the net moment about the centre is zero the resultant passes through the centre. The direction with horizontal is given by \tan θ = 308190 / 392400, θ = 37.58°.