JEE Advanced 2014 · Paper 1 · Physics Q8

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PhysicsMultiple correct+3Question 8 of 60

In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $\mu_1$ and that between the floor and the ladder is $\mu_2$ . The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$ . If the ladder is about to slip, then

A ladder leaning against a vertical hatched wall (friction μ₁) from a hatched floor (friction μ₂), making angle θ with the floor.

\mu_1=0,\ \mu_2\ne0

and

N_2\tan\theta=\dfrac{mg}2

\mu_1\ne0,\ \mu_2=0

and

N_1\tan\theta=\dfrac{mg}2

\mu_1\ne0,\ \mu_2\ne0

and

N_2=\dfrac{mg}{1+\mu_1\mu_2}

\mu_1=0,\ \mu_2\ne0

and

N_1\tan\theta=\dfrac{mg}2

Answer and worked solution

Official answer: C, D

Forces at limiting friction. Floor: $N_2$ up, friction $\mu_2N_2$ towards the wall. Wall: $N_1$ away from the wall, friction $\mu_1N_1$ upwards.

$N_1=\mu_2N_2,\qquad N_2+\mu_1N_1=mg\implies N_2=\frac{mg}{1+\mu_1\mu_2}.$

Torque about the foot (ladder length $l$ ):
$N_1l\sin\theta+\mu_1N_1l\cos\theta=mg\frac l2\cos\theta.$

(C) Both coefficients non-zero: $N_2=\dfrac{mg}{1+\mu_1\mu_2}$ . True.
(D) With $\mu_1=0$ the torque equation gives $N_1\tan\theta=\dfrac{mg}2$ . True.
(A) With $\mu_1=0$ , $N_2=mg$ , so $N_2\tan\theta=mg/2$ would need $\tan\theta=\frac12$ in general. False.
(B) With $\mu_2=0$ , $N_1=\mu_2N_2=0$ ; then nothing balances the torque of the weight. The ladder cannot be in equilibrium. False.

(C), (D)

Topics:Equilibrium of rigid bodies Friction

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JEE Advanced 2014 · Paper 1, Physics question 8