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/s^2, the horizontal force that he is applying on the pole is

Answer 1

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/s^2, the horizontal force that he is applying on the pole is 500N.

Explanation:

μ=0.8

Frictional force=μN₁

=mg/μ

=400/0.8

=500N

Answer 2

Answer:

Explanation:

The normal force acts perpendicular to the surface so F applied =normal

But friction = weight

Since one act along up and other down and minimum force acts when they are at rest wrt each other so uN=mg

N=mg/u

Since N=F

F=mg/u

=40x10 /0.8

500