A fireman of mass 60kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman slide down (g = 10 m/s²)

The correct answer is 5 m/s².

When a fireman slides down a pole, two vertical forces act on him:

1. His weight acting downward.
2. The frictional force acting upward (opposing the downward sliding motion).

The fireman presses against the pole with a horizontal force — this is the normal reaction between his hands/body and the pole.

Step 1: Identify the given values

Mass of fireman: m = 60 kg
Normal force (pressing force): N = 600 N
Coefficient of friction: μ = 0.5
Acceleration due to gravity: g = 10 m/s²

Step 2: Calculate the weight of the fireman

Weight acts vertically downward:

W=m\times g=60\times 10=600\text{N}

Step 3: Calculate the frictional force

The frictional force opposes the sliding motion (acts upward along the pole) and is given by:

F_{\mathrm{friction}}=\mu \times N=0.5\times 600=300\text{N}

Step 4: Apply Newton's Second Law along the vertical direction

Taking downward as the positive direction, the net force acting on the fireman is:

F_{\mathrm{net}}=W-F_{\mathrm{friction}}=600-300=300\text{N (downward)}

Step 5: Calculate the acceleration

Using Newton's Second Law, F_{\mathrm{net}}=m\times a:

a=\frac{F_{\mathrm{net}}}{m}=\frac{300}{60}=5\text{m/s²}

Conclusion: Since the net force is directed downward, the fireman accelerates downward at 5 m/s². The friction force is large enough to slow his descent (compared to free fall at 10 m/s²), but not enough to hold him stationary, so he slides down with an acceleration of 5 m/s².