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 $g = m / s^{2}$ . The horizontal force that he is applying on the pole is:

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Solution

$f = m g$
$μ n = m g$
$N = \frac{m g}{μ} = \frac{40 \times 10}{0.8} = 500 N$

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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=m/s2. The horizontal force that he is applying on the pole is:

f=mgμn=mgN=mgμ=40×100.8=500N

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 $g = m / s^{2}$ . The horizontal force that he is applying on the pole is:

$f = m g$
$μ n = m g$
$N = \frac{m g}{μ} = \frac{40 \times 10}{0.8} = 500 N$