A ladder (ℓL = 7.80 m) of weight WL = 350 N leans against a smooth vertical wall. The term "smooth" means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is 885 N, stands 6.40 m up from the bottom of the ladder (this distance goes along the ladder, it is not the vertical height). Assume that the ladder's weight acts at the ladder's center, and neglect the hose's weight. What is the minimum value for the coefficient of static friction between the ladder and the ground, so that the ladder (with the fireman on it) does not slip? (Assume θ = 54.0°.)

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wagonbelleville wagonbelleville · Answer 1 · 2019-07-31T05:43:31+02:00

Answer:

[tex]N_{wall} = 654.72 N[/tex]

μ = 0.53

Explanation:

ladder length = 7.8 m

weight of ladder= 350 N

weight of fire fighter = 885 N

distance of the fire fighter from bottom of ladder = 6.40 m

θ = 54⁰

taking torque about the bottom of ladder

[tex]N_{wall}\times l_{Ladder}\ sin\theta- 885\times 6.4\ cos\theta-350\times (7.8cos\theta)/2=0[/tex]

[tex]N_{wall}\times 7.8\ sin54^{\circ}- 885\times 6.4\ cos54^{\circ}-350\times (7.8cos54^{\circ})/2=0[/tex]

[tex]N_{wall} = 654.72 N[/tex]

taking vertical forces :

[tex]N_y-350-885=0\\N_y=1235N[/tex]

now all the horizontal forces:

[tex]f\ -\ N_{wall} = 0\\f= 654.72N[/tex]

we know,

f = μ N

[tex]\mu = \dfrac{f}{N_{y}}=\frac{654.72}{1235}[/tex]

μ = 0.53