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A 16.0-m uniform ladder weighing 480 N rests against a frictionless wall. The ladder makes a 55.0° angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 850-N firefighter has climbed 3.80 m along the ladder from the bottom.
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

Respuesta :

Answer:

horizontal force component 535.00 N

vertical force component 1330 N

[tex]\mu= 0.38[/tex]

Explanation:

from equilibrium condition at A

[tex]N_2(16 sin55) = 850*(3.80cos55) +480(\frac{16}{2}cos55)[/tex]

[tex]N_2(13.12) = 535.00 N[/tex]

FROM equilibrium condition

N_1 = mg +850

       = 480 + 850

       = 1330 N

horizontal force component

f_2 = N_2

      = 535.00 N

vertical force component

N_1 = 1330 N

B) Coefficient of friction between ladder and ground when fire fighter is 9.4 m away from ground

[tex]N_2(16 sin55) = 850*(9.40cos55) +480(\frac{16}{2}cos55)[/tex]

N_2 =517.71 N

Thus coefficient of friction is

[tex]\mu = \frac{f_2}{N_1}[/tex]

      =[tex] = \frac{517.71}{1330}[/tex]                     [f_2 = N_2]

       = 0.38

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