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A-ladder-of-mass-m-is-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-1-and-that-betw

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Question Number 22479 by Tinkutara last updated on 19/Oct/17
A ladder of mass m is 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 (1) μ_1 = 0, μ_2 ≠ 0 and N_2 tan θ = mg/2 (2) μ_1 ≠ 0, μ_2 = 0 and N_1 tan θ = mg/2 (3) μ_1 ≠ 0, μ_2 ≠ 0 and N_2 = ((mg)/(1 + μ_1 μ_2 )) (4) μ_1 = 0, μ_2 ≠ 0 and N_1 tan θ = ((mg)/2)
$$\mathrm{A}\:\mathrm{ladder}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{leaning}\:\mathrm{against}\:\mathrm{a} \\ $$$$\mathrm{wall}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{in}\:\mathrm{static}\:\mathrm{equilibrium}\:\mathrm{making} \\ $$$$\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{floor}. \\ $$$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{wall}\:\mathrm{and}\:\mathrm{the}\:\mathrm{ladder}\:\mathrm{is}\:\mu_{\mathrm{1}} \:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{and}\:\mathrm{the}\:\mathrm{ladder}\:\mathrm{is}\:\mu_{\mathrm{2}} . \\ $$$$\mathrm{The}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ladder}\:\mathrm{is}\:\mathrm{N}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{is}\:\mathrm{N}_{\mathrm{2}} . \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{ladder}\:\mathrm{is}\:\mathrm{about}\:\mathrm{to}\:\mathrm{slip},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mu_{\mathrm{1}} \:=\:\mathrm{0},\:\mu_{\mathrm{2}} \:\neq\:\mathrm{0}\:\mathrm{and}\:{N}_{\mathrm{2}} \:\mathrm{tan}\:\theta\:=\:{mg}/\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\mu_{\mathrm{1}} \:\neq\:\mathrm{0},\:\mu_{\mathrm{2}} \:=\:\mathrm{0}\:\mathrm{and}\:{N}_{\mathrm{1}} \:\mathrm{tan}\:\theta\:=\:{mg}/\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mu_{\mathrm{1}} \:\neq\:\mathrm{0},\:\mu_{\mathrm{2}} \:\neq\:\mathrm{0}\:\mathrm{and}\:{N}_{\mathrm{2}} \:=\:\frac{{mg}}{\mathrm{1}\:+\:\mu_{\mathrm{1}} \mu_{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\mu_{\mathrm{1}} \:=\:\mathrm{0},\:\mu_{\mathrm{2}} \:\neq\:\mathrm{0}\:\mathrm{and}\:{N}_{\mathrm{1}} \:\mathrm{tan}\:\theta\:=\:\frac{{mg}}{\mathrm{2}} \\ $$
Commented by Tinkutara last updated on 19/Oct/17
Commented by ajfour last updated on 19/Oct/17
(3) and (4) are correct. See Q.22481 for solution.
$$\left(\mathrm{3}\right)\:{and}\:\left(\mathrm{4}\right)\:{are}\:{correct}. \\ $$$${See}\:{Q}.\mathrm{22481}\:{for}\:{solution}. \\ $$

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