If the the centre of gravity of the bar is at distance a and b from the ends A & b respectively, show that at impending motion, the inclination of thee bar with the horizontal will be:

$θ= \tan^{-1} (\dfrac 1{μ_F} \dfrac {a-bμ_F μ_W}{a+b})$

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : MAY 2016

$$∑f_x=0$$

$N_2-μ_F N_1=0 \\ N_2=μ_F N_1 ----(1) \\ ∑f_y=0 \\ μ_w N_2+N_1-W=0 $

From (1)

$μ_w.μ_F N_1+N_1=W \\ N_1 = \dfrac W{1+μ_w μ_F } \text { and } N_2 = \dfrac {μ_F}{1+μ_w μ_F} W \\ M_A=0 \\ w×a \cos θ-μ_w N_2 (a+b)\cos θ-N_2 (a+b)\sin θ=0 \\ \cos θ.[W.a-(a+b) μ_w. \dfrac {μ_F}{1+μ_w μ_F} W] = \dfrac {μ_f}{1+μ_w μ_F} W(a+b)\sin θ \\ \dfrac {\sin θ}{\cos θ} = [\dfrac {a-\dfrac {μ_w μ_F (a+b)}{1+μ_w μ_f }}{\dfrac {μ_F}{1+μ_w μ_F} (a+b) }] \\ θ = \tan^{-1} [\dfrac {a(1+μ_F μ_w )-(a+b) μ_w μ_F}{μ_F (a+b)} ] \\ θ = \tan^{-1} [\dfrac {a+aμ_F μ_w-aμ_F μ_w-b μ_F μ_w}{μ_F (a+b) }] \\ θ=\tan^{-1} [\dfrac 1{μ_F} . \dfrac {a-bμ_F μ_w}{(a+b) }] $

Hence proved.