solve correct answerThe uniform rectangular relief gate AB shown in (Figure 1) has a weight of 800 1b and a width of 6 ft.
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Determine the normal reaction at the smoo

Question

solve correct answerThe uniform rectangular relief gate AB shown in (Figure 1) has a weight of 800 1b and a width of 6 ft.
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Determine the normal reaction at the smooth support A.
Express your answers in kilopounds to three significant figures.
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Part B Complete previous part(s)
Part C Complete previous part(s)

VIPSolutions · Accepted Answer

Let's solve the problem step by step:

## Step 1: **Given Data**
- **Weight of Gate, $ W $ = 800 lb**
- **Width of Gate = 6 ft**
- **Length of Gate $ AB $ = 6 ft**
- **Gate is at $ ^\circ $ the horizontal**
 **Height of water the base = 9 ft**

## Step 2: **Forces Acting on the Gate**
1. **Weight of the Gate** acts vertically downward through the center of gravity.
2. **Hydrostatic Force** acts perpendicular to the surface of the gate.

## Step 3: **Normal Reaction at A ($ N_A $)**
The normal reaction at A is the force perpendicular to the gate at point A.

### **a) Total Weight Acting Perpendicular to the Gate**
The weight acts vertically, but the support at A is perpendicular to the surface.

- The component of weight perpendicular to the gate:  
  $$
  W_\perp = W \cdot \cos(60^\circ)
  $$
  $$
  W_\perp = 800 \cdot 0.5 = 400 \text{ lb}
  $$

### **b) Hydrostatic Force**
But for this part (Part A), unless otherwise specified, we only need the reaction due to the weight.

## Step 4: **Convert to Kilopounds**
$$
N_A = \frac{400 \text{ lb}}{1000} = 0.400 \text{ kip}
$$

## Step 5: **Final Answer**
$$
\boxed{N_A = 0.400 \ \text{kip}}
$$

*Let me know if you also want to include the hydrostatic force for a more complete analysis!*

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