An overweight acrobat, weighing in at 115 kg, wants to perform a single hand stand. He tries to cheat by resting one foot against a smooth frictionless vertical wall. The horizontal force there is 130 N. What is the magnitude of the force exerted by the floor on his hand? Answer in N.

Options

A. 1134
B. 1257
C. 997
D. 1119

Explanation

To solve the problem, we need to analyze the forces acting on the acrobat when he is in a handstand position with one foot against a frictionless vertical wall.

### Step-by-Step Solution:

1. **Identify the Weight of the Acrobat:**
The weight (W) of the acrobat can be calculated using the formula:
\[ W = m \cdot g \]
where \( m = 115 \, \text{kg} \) (mass of the acrobat) and \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
\[ W = 115 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 1127 \, \text{N} \]

2. **Identify the Horizontal Force:**
The horizontal force (F_H) exerted by the wall is given as:
\[ F_H = 130 \, \text{N} \]

3. **Set Up the Equilibrium Condition:**
For the acrobat to be in equilibrium, the net force acting on him must be zero. The forces acting on him are:
- The weight (W) acting downwards.
- The normal reaction force (R_H) from the floor acting upwards.
- The horizontal force (F_H) from the wall acting horizontally.

The vertical forces must balance:
\[ R_H = W \]

4. **Calculate the Resultant Force:**
The resultant force (R) acting on the acrobat can be calculated using the Pythagorean theorem since the forces are perpendicular:
\[ R_H = \sqrt{W^2 + F_H^2} \]
Substituting the values:
\[ R_H = \sqrt{(1127 \, \text{N})^2 + (130 \, \text{N})^2} \]
\[ R_H = \sqrt{1270729 + 16900} \]
\[ R_H = \sqrt{1287629} \]
\[ R_H \approx 1134 \, \text{N} \]

5. **Conclusion:**
The magnitude of the force exerted by the floor on his hand is approximately:
\[ R_H \approx 1134 \, \text{N} \]

### Final Answer:
The magnitude of the force exerted by the floor on his hand is approximately **1134 N**.