An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the surface and the insect is 1/3 if the line joining the hemispherical surface to the insect makes the angle \alpha with the vertical the maximum possible value of ] \alpha is????????
harshitarawat80:
Do any one know
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19
See diagram.
It is not clear from the question, what the angle alpha is. i assume alpha as shown in the figure. change ur answer accordingly.
Let the ant be at P and be walking slowly upwards. the speed is negligible. Suppose the ant climbs to a height at P, the ant starts sliding down. It means that the friction force will act tangentially but towards up and gravitational force will act downwards and vertical. The normal reaction force N from the hemispherical surface acts as shown. Just before the ant slips down, the components of forces along the radius and tangent to surface are balanced.
N = m g cos α
Friction F = mg sin α
So F = Mu N = Mu mg cos α = mg sin α
So tan α = Mu = 0.3
α = inverse tan (0.3) = 16.7 degrees
this the maximum angle alpha is 16.7 deg.
beta = 90 - alpha/2 = 81.65 deg
It is not clear from the question, what the angle alpha is. i assume alpha as shown in the figure. change ur answer accordingly.
Let the ant be at P and be walking slowly upwards. the speed is negligible. Suppose the ant climbs to a height at P, the ant starts sliding down. It means that the friction force will act tangentially but towards up and gravitational force will act downwards and vertical. The normal reaction force N from the hemispherical surface acts as shown. Just before the ant slips down, the components of forces along the radius and tangent to surface are balanced.
N = m g cos α
Friction F = mg sin α
So F = Mu N = Mu mg cos α = mg sin α
So tan α = Mu = 0.3
α = inverse tan (0.3) = 16.7 degrees
this the maximum angle alpha is 16.7 deg.
beta = 90 - alpha/2 = 81.65 deg
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Answered by
21
Suppose the insect can crawl upto a point P
N=wcosα-----(1)
f=fictional force=wsinα-----(2)
from (1) and (2)
tanα=f/N
In limiting case... tanα=μ
Therefore tanα=1/3
cotα=3
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