2.95 a gate having the cross section shown in fig. 2.95 closes an opening 1.5 m wide and 1.2 m high in a water reservoir. the gate weighs 2224 n, and its center of gravity is 0.3

General
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2.95 a gate having the cross section shown in fig. 2.95 closes an opening 1.5 m wide and 1.2 m high in a water reservoir. the gate weighs 2224 n, and its center of gravity is 0.3 m to the left of ac and 0.6 m above bc. determine the horizontal reaction that is developed on the gate at c.

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To determine the horizontal reaction developed on the gate at point C, we need to analyze the forces acting on the gate.

Since the gate is closing an opening in a water reservoir, it is subjected to hydrostatic pressure. The hydrostatic pressure acts perpendicular to the gate’s surface.

Let’s break down the forces acting on the gate:

  1. Weight of the gate: The gate weighs 2224 N and is acting vertically downward. We can consider this force as acting at the center of gravity of the gate.

  2. Hydrostatic pressure: The hydrostatic pressure acts on the vertical surface of the gate. Since the center of gravity is 0.3 m to the left of line AC and 0.6 m above line BC, the hydrostatic pressure will generate a horizontal force component at point C.

To calculate the horizontal reaction at point C, we’ll consider the moments about point C.

Taking moments about point C:

(Moment of weight) + (Moment of hydrostatic pressure) = 0

(Moment of weight) = (Moment of hydrostatic pressure)

The moment of the weight of the gate is given by:

Moment of weight = Weight * Distance from C to the center of gravity

Moment of weight = 2224 N * 0.3 m = 667.2 Nm

The moment of the hydrostatic pressure is given by:

Moment of hydrostatic pressure = Hydrostatic pressure * Area * Distance from C to the center of pressure

The hydrostatic pressure at a depth h is given by the formula:

Pressure = density of water * acceleration due to gravity * height of fluid column

The height of the fluid column is 1.2 m in this case.

Substituting the values:

Pressure = 1000 kg/m^3 * 9.8 m/s^2 * 1.2 m = 11,760 N/m^2

The area of the gate is 1.5 m * 1.2 m = 1.8 m^2

Distance from C to the center of pressure is half the height of the gate = 0.6 m

Moment of hydrostatic pressure = 11,760 N/m^2 * 1.8 m^2 * 0.6 m = 12,676.8 Nm

Now, equating the moments:

667.2 Nm + 12,676.8 Nm = 0

12,344 Nm = 0

Therefore, the horizontal reaction at point C is 0 N.

This means that there is no horizontal force acting on the gate at point C, and the gate is in equilibrium in the horizontal direction.