Laws of Motion 3 Question 6

6. 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 μ1 and that between the floor and the ladder is μ2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then

(2014 Adv.)

(a) μ1=0,μ20 and N2tanθ=mg2

(b) μ10,μ2=0 and N1tanθ=mg2

(c) μ10,μ20 and N2=mg1+μ1μ2

(d) μ1=0,μ20 and N1tanθ=mg2

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Answer:

Correct Answer: 6. (b)

Solution:

  1. NOTE

It is not given in the question, best assuming that both blocks are in equilibrium. The free body diagram of two blocks is as shown below,

Reaction force, R =applied force F

For vertical equilibrium of A;

f1= friction between two blocks =WA=20 N

For vertical equilibrium of B;

f2= friction between block B and wall =WB+f1=100+20=120 N



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