Question

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

• A. $700 \, \text{N}$
• B. $625.67 \, \text{N}$
• C. $550 \, \text{N}$
• D. $666.67 \, \text{N}$

Answer 1

Answer: (D)

Let the normal reaction is $N.$
Since, the boy is climbing with constant speed.
$\therefore F_{n e t}=0$ (on the boy)
Free body diagram

$mg=f_{r}$ (friction force)
$mg=\mu N$ $\left(\because \, f_{r} = \mu N\right)$
$\Rightarrow N=\frac{m g}{\mu }=\frac{50 \times 10}{0.75}$
$=666.67 \, \text{ N}$