Question

If the force which acting parallel to an inclined plane of angle α just sufficient to draw the weight up in n times the force which will just let it be on the point of sliding down, the coefficient of friction will be

(A) \(\mu = \frac{(n-1)}{n+1}\)tan\(\alpha\)

(B) \(\mu = \frac{(n+1)}{n-1}\) tan\(\alpha\)

(C) \(\mu\) = n  tan\(\alpha\)

(D) \(\mu\) = (n+1) tan\(\alpha\)

Answer 1

Correct option is (A) \(\mu = \frac{(n - 1)}{n + 1} \tan \alpha\)

As force F tends to push the mass upwards, friction will tend to oppose it. So, it will act downwards.

∴ F = f + mg sin α

f = m N = µ mg cos α

⇒ F₁ = µ mg cos α + mg sin α … (i)

Now when pushing downwards, friction will be acting upwards,

∴ F₂ + f = mg sin α

F₂ = mg sin α – f

f = µ mg cos α

⇒ F₂ = mg sin α – µ mg cos α   … (ii)

Given that F₁ = nF₂

∴ µmg cos α + mg sin α = n(mg sin α – µmg cos α)

⇒ \(\mu = \frac{(n - 1)}{n + 1} \tan \alpha\)

Answer 2

Correct answer is  (A) \(\mu = \frac{(n-1)}{n+1}\) tan\(\alpha\)