In the figure, a ladder of mass is shown leaning against a wall. It is in static equilibrium making an angle with the horizontal floor. The coefficient of friction between the wall and the ladder is and that between the floor and the ladder is . The no

Question

In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $θ$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $μ_{1}$ and that between the floor and the ladder is $μ_{2}$ . 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 [2014]

Smart Achievers · Accepted Answer

(3, 4)

$When μ_{1} \neq 0 and μ_{2} \neq 0$

Horizontal equilibrium, $N_{1} = μ_{2} N_{2}$

Vertical equilibrium, $m g = N_{2} + μ_{1} N_{1}$

Solving the above equations we get

$N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$

$When μ_{1} = 0; Torque about P$

$m g \times \frac{l}{2} \cos θ = N_{1} \times l \sin θ$

$\Rightarrow N_{1} \tan θ = \frac{m g}{2}$

Solution

1 $μ_{1} = 0, μ_{2} \neq 0 and N_{2} \tan θ = \frac{m g}{2}$

2 $μ_{1} \neq 0, μ_{2} = 0 and N_{1} \tan θ = \frac{m g}{2}$

3 $μ_{1} \neq 0, μ_{2} \neq 0 and N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$

4 $μ_{1} = 0, μ_{2} \neq 0 and N_{1} \tan θ = \frac{m g}{2}$

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Solution

1 μ1=0, μ2≠0 and N2tanθ=mg2

2 μ1≠0, μ2=0 and N1tanθ=mg2

3 μ1≠0, μ2≠0 and N2=mg1+μ1μ2

4 μ1=0, μ2≠0 and N1tanθ=mg2

Ans. (3, 4) When μ1≠0 and μ2≠0 Horizontal equilibrium, N1=μ2N2 Vertical equilibrium, mg=N2+μ1N1 Solving the above equations we get N2=mg1+μ1μ2 When μ1=0; Torque about P mg×l2cosθ=N1×lsinθ ⇒ N1tanθ=mg2

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1 $μ_{1} = 0, μ_{2} \neq 0 and N_{2} \tan θ = \frac{m g}{2}$

2 $μ_{1} \neq 0, μ_{2} = 0 and N_{1} \tan θ = \frac{m g}{2}$

3 $μ_{1} \neq 0, μ_{2} \neq 0 and N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$

4 $μ_{1} = 0, μ_{2} \neq 0 and N_{1} \tan θ = \frac{m g}{2}$