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. than
A
`mu_1 = 0, mu_2 != and N_2 tan theta = (mg)/(2)`
B
`mu_1 != 0, mu_2 =0 and N_1 tan theta = (mg)/(2)`
C
`mu_1 != 0, mu_2 !=0 and N_2 = (mg)/(1+mu_1 mu_2)`
D
`mu_1 = 0, mu_2 != 0 and N_1 tan theta = (mg)/(2)`
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The correct Answer is:
C, D
(c,d) when `mu_1 != 0 and mu_2!= 0 `N_1 =mu_2 N_2 [:. Horizontal equilibrium]` `mg = N_2 + mu_1 N_1 [:. Vertical equilibrium]` Solving the above equation we get `N_2 = (mg)/(1+mu_1 mu_2)` :. (c ) is the correct option. When `mu_1 = 0` Taking torque about P we get `mg xx(I)/(2) cos theta = N_1xx I sin theta` `N_1 tan theta = (mg)/(2)` :. (d) is correct
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