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What is the minimum coeffecient of static friction μmin required between the ladder and the ground so that the ladder does not slip? Express μmin in terms of m1

Answer :

Answer:

[tex]\mu = \frac{1}{2tan\theta}[/tex]

Explanation:

let the ladder is of mass "m" and standing at an angle with the ground

So here by horizontal force balance we will have

[tex]\mu N_1 = N_2[/tex]

by vertical force balance we have

[tex]N_1 = mg[/tex]

now by torque balance about contact point on ground we will have

[tex]mg(\frac{L}{2}cos\theta) = N_2(L sin\theta)[/tex]

so we will have

[tex]N_2 = \frac{mg}{2tan\theta}[/tex]

now from first equation we have

[tex]\mu (mg) = \frac{mg}{2tan\theta}[/tex]

[tex]\mu = \frac{1}{2tan\theta}[/tex]

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