Question

Hi, solve this question ,
refer to the attachment ​

Answer 1

Let $\sf{m}$ be the mass of the insect.

Free Body Diagram of the insect is given below.

$\setlength{\unitlength}{1.5mm}\begin{picture}(5,5)\thicklines\put(0,0){\vector(-1,-1){7.07}}\put(0,0){\vector(1,-1){7.07}}\put(8,-9){ \sf{mg\sin\alpha} }\put(-17,-9.2){ \sf{mg\cos\alpha} }\put(0,0){\circle*{1}}\put(0,0){\vector(0,-1){14.14}}\put(0,0){\vector(1,1){7.07}}\put(0,0){\vector(-1,1){7.07}}\put(-2.5,-18){\sf{mg}}\put(8,7){\sf{R}}\put(-8.5,7.5){\sf{f}}\qbezier(-1.5,-1.5)(-1,-3)(0,-2)\put(-2.5,-5){ \alpha }\end{picture}$

From the fig. we get,

$\sf{\longrightarrow R=mg\cos\alpha\quad\quad\dots(1)}$

As the insects crawls up very slowly, it would be due to that the friction acting on it is greater than or equal to its weight component, i.e.,

$\sf{\longrightarrow f\geq mg\sin\alpha\quad\quad\dots(2)}$

But we know that,

$\sf{\longrightarrow f=\mu\,R}$

From (1)

$\sf{\longrightarrow f=\dfrac{1}{3}\,mg\cos\alpha}$

Then (2) becomes,

$\sf{\longrightarrow \dfrac{1}{3}\,mg\cos\alpha\geq mg\sin\alpha}$

Since $\alpha$ is an acute angle,

$\sf{\longrightarrow\cot\alpha\geq3}$

So the maximum possible value of $\alpha$ will be given by,

$\sf{\longrightarrow\underline{\underline{\cot\alpha_{max}=3}}}$

because value of $\cot\alpha$ decreases with increase in value of $\alpha.$

Hence (a) is the answer.

Answer 2

Explanation:

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