A boy of mass 40 kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g = 10 m/s2, the horizontal force that he is applying on the pole is

Question

A boy of mass 40 kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g = 10 m/s2, the horizontal force that he is applying on the pole is (A) 300 N (B) 400 N (C) 500 N (D) 600 N

Tardigrade Admin · Accepted Answer

The coefficient of friction =0.8, g=10 m / s 2, mass =40 kg The frictional force =μ N Hence, μ N =m g ⇒ N=(m g/μ) = (40 × 10/0.8)=500 N

Q. A boy of mass $40 k g$ is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is $0.8$ and $g = 10 m / s^{2}$ , the horizontal force that he is applying on the pole is

300 N

400 N

500 N

600 N

The coefficient of friction $= 0.8,$
$g = 10 m / s^{2}$ ,
mass $= 40 k g$
The frictional force $= μ N$
Hence, $μ N = m g$
$\Rightarrow N = \frac{m g}{μ}$
$= \frac{40 \times 10}{0.8} = 500 N$

Q. A boy of mass 40kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g=10m/s2, the horizontal force that he is applying on the pole is