Question

An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the surface and the insect is $ \dfrac{{1}}{3}$. If the line joining the center of the hemispherical surface to the insect makes an angle $\alpha $ with vertical, the maximum possible value of $\alpha $ is given by-

(a). $\cot \alpha = 3$
(b). $\tan \alpha = 3$
(c). $\sec \alpha = 3$
(d). $\cos ec\alpha = 3$

Answer 1

- Hint: First we will determine the horizontal component and the vertical component of the force acting on the insect due to its mass. Then, we will equate the equation of the friction that will be equal to the horizontal component to the formula of friction. Refer to the solution below-

Formula used: $f = \mu N$

Complete step-by-step solution:
Force acting on the insect due to its mass is mg.
Its horizontal and vertical components due to the angle $\alpha $ that can be seen in the figure will be-
Horizontal component- $mg\sin \alpha $
Vertical component- $mg\cos \alpha $
The normal reaction N will be opposite in direction and equal in magnitude of the vertical component. So-
$ \Rightarrow N = mg\cos \alpha $ (equation 1)
Friction acting on the insect will be equal to the horizontal component. Thus-
$ \Rightarrow f = mg\sin \alpha $ (equation 2)
As we know the formula of friction is-
$ \Rightarrow f = \mu N$ (equation 3)
Now, equating equation 2 and 3, we will have-
$ \Rightarrow mg\sin \alpha = \mu N$
Putting the value of N from the equation 1 into the above equation, we get-
$
   \Rightarrow mg\sin \alpha = \mu N \\
    \\
   \Rightarrow mg\sin \alpha = \mu \times mg\cos \alpha \\
    \\
   \Rightarrow \sin \alpha = \mu \cos \alpha \\
    \\
   \Rightarrow \mu = \dfrac{{\sin \alpha }}{{\cos \alpha }} \\
$
Putting the value of $\mu $ from the question into the above equation, we have-
$
   \Rightarrow \mu = \dfrac{{\sin \alpha }}{{\cos \alpha }} \\
    \\
   \Rightarrow \dfrac{1}{3} = \dfrac{{\sin \alpha }}{{\cos \alpha }} \\
    \\
   \Rightarrow \tan \alpha = \dfrac{1}{3} \\
$
As we know that $\tan \theta = \dfrac{1}{{\cot \theta }}$,
$
   \Rightarrow \dfrac{1}{{\cot \alpha }} = \dfrac{1}{3} \\
    \\
   \Rightarrow \cot \alpha = 3 \\
$

Hence, option A is the correct option.

Note: Knowledge of basic trigonometric equations and rules like $\tan \theta = \dfrac{1}{{\cot \theta }}$ used in the solution above is necessary. Do not leave the answer without converting the trigonometric function if possible, in the solution.