Suppose a horse leans against a wall as in the figure below. (a) Fall 1.2 m CG W 0.35 m (b) 1.4 m Calculate the force exerted on the wall assuming that force is horizontal and using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 800 kg. Take the data to be accurate to three digits.

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### Horse Leaning Against a Wall - Equilibrium Analysis

#### Problem Statement

Suppose a horse leans against a wall as shown in the figure below.

#### Visual Descriptions:
- **Figure (a)**: An illustration showing a horse leaning sideways against a red wall. The horse is seen from behind, and its rider holds onto the wall.
- **Figure (b)**: A schematic representation of the horse's interaction with the wall.

The schematic (Figure b) provides specific measurements and forces involved in the scenario:
- **F_wall**: The horizontal force exerted on the wall.
- **CG**: Center of gravity of the horse and rider system.
- **w**: Weight force acting vertically downward from the center of gravity at 1.4 meters high.
- The distance to the wall from the center of gravity horizontally is marked as 0.35 meters.
- Height from the ground to the point where the force \( F_{wall} \) acts is 1.2 meters.

#### Problem Requirement

Calculate the force exerted on the wall assuming that the force is horizontal and using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium.

**Given**:
- The total mass of the horse and rider is 800 kg.
- Take the data to be accurate to three digits.

### Calculation Steps

1. **Determine the Weight of the Horse and Rider**:
   Weight, \( w \) = mass × gravitational acceleration
   \( w = 800 \, \text{kg} \times 9.81 \, \text{m/s}^2 \)
   \( w = 7848 \, \text{N} \)

2. **Equilibrium Conditions**:
   To keep the system in equilibrium, the sum of moments about any point should be zero.
   Take moments about the point where the horse’s hooves contact the ground.

   Assume Clockwise moments are positive:

   Moment due to weight (\( w \)):
   \( M_w = w \times 0.35 \)

   Moment due to horizontal force (\( F_{wall} \)):
   \( M_{F_{wall}} = F_{wall} \times 1.2 \)

   Since the system is in equilibrium:
   \( M_w = M_{F_{
Transcribed Image Text:### Horse Leaning Against a Wall - Equilibrium Analysis #### Problem Statement Suppose a horse leans against a wall as shown in the figure below. #### Visual Descriptions: - **Figure (a)**: An illustration showing a horse leaning sideways against a red wall. The horse is seen from behind, and its rider holds onto the wall. - **Figure (b)**: A schematic representation of the horse's interaction with the wall. The schematic (Figure b) provides specific measurements and forces involved in the scenario: - **F_wall**: The horizontal force exerted on the wall. - **CG**: Center of gravity of the horse and rider system. - **w**: Weight force acting vertically downward from the center of gravity at 1.4 meters high. - The distance to the wall from the center of gravity horizontally is marked as 0.35 meters. - Height from the ground to the point where the force \( F_{wall} \) acts is 1.2 meters. #### Problem Requirement Calculate the force exerted on the wall assuming that the force is horizontal and using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. **Given**: - The total mass of the horse and rider is 800 kg. - Take the data to be accurate to three digits. ### Calculation Steps 1. **Determine the Weight of the Horse and Rider**: Weight, \( w \) = mass × gravitational acceleration \( w = 800 \, \text{kg} \times 9.81 \, \text{m/s}^2 \) \( w = 7848 \, \text{N} \) 2. **Equilibrium Conditions**: To keep the system in equilibrium, the sum of moments about any point should be zero. Take moments about the point where the horse’s hooves contact the ground. Assume Clockwise moments are positive: Moment due to weight (\( w \)): \( M_w = w \times 0.35 \) Moment due to horizontal force (\( F_{wall} \)): \( M_{F_{wall}} = F_{wall} \times 1.2 \) Since the system is in equilibrium: \( M_w = M_{F_{
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