A boy of mass 50 kg is climbing a vertical pole at a constant speed. If coefficient of fricition between his palms and the pole is 0.75, then the normal reaction between him and the pole is `("take g "=10m//s^(2))`

Options

A. 700 N
B. 625.67 N
C. 550 N
D. 666.67 N

Answer

To solve the problem, we need to find the normal reaction force (N) between the boy and the vertical pole as he climbs at a constant speed.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Boy:**
- The weight of the boy (W) acting downwards:
\[ W = mg \]
where \( m = 50 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
- The frictional force (F_f) acting upwards, which is responsible for counteracting the weight of the boy.
- The normal reaction force (N) exerted by the pole on the boy's palms.

2. **Calculate the Weight of the Boy:**
\[ W = mg = 50 \, \text{kg} \times 10 \, \text{m/s}^2 = 500 \, \text{N} \]

3. **Understand the Motion:**
- Since the boy is climbing at a constant speed, the net force acting on him is zero. Therefore, the upward frictional force must equal the downward weight:
\[ F_f = W \]
\[ F_f = 500 \, \text{N} \]

4. **Relate Frictional Force to Normal Reaction Force:**
- The frictional force can be expressed in terms of the normal reaction force (N) and the coefficient of friction (μ):
\[ F_f = \mu N \]
where \( \mu = 0.75 \).

5. **Set Up the Equation:**
- From the previous equations, we have:
\[ 500 \, \text{N} = 0.75 N \]

6. **Solve for Normal Reaction Force (N):**
\[ N = \frac{500 \, \text{N}}{0.75} = \frac{500}{0.75} = 666.67 \, \text{N} \]

### Final Answer:
The normal reaction force (N) between the boy and the pole is \( 666.67 \, \text{N} \).

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