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/s2 )

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/s2 ) (A) 700 textN (B) 625.67 textN (C) 550 textN (D) 666.67 textN

Tardigrade Admin · Accepted Answer

Let the normal reaction is N. Since, the boy is climbing with constant speed. ∴ Fn e t=0 (on the boy) Free body diagram <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-wevzhol1smt8.png" /> mg=fr (friction force) mg=μ N (∵ fr = μ N) ⇒ N=(m g/μ )=(50 × 10/0.75) =666.67 text N

Q. A boy of mass $50 k g$ 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 $g = 10 m / s^{2}$ )

$700 N$

$625.67 N$

$550 N$

$666.67 N$

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

$m g = f_{r}$ (friction force)
$m g = μ N$ $(∵ f_{r} = μ N)$
$\Rightarrow N = \frac{m g}{μ} = \frac{50 \times 10}{0.75}$
$= 666.67 N$

Q. A boy of mass 50kg 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=10m/s2 )

700N

625.67N

550N

666.67N