In the figure, a ladder of length L = 14 m and mass m = 43 kg leans against a sl

Question

In the figure, a ladder of length L = 14 m and mass m = 43 kg leans against a slick (frictionless) wall. The ladder's upper end is at height h = 11 m above the pavement on which the lower end rests (the pavement is not frictionless). The ladder's center of mass is L/3 from the lower end. A firefighter of mass M = 72 kg climbs the ladder. Let the coefficient of static friction mu_5 between the ladder and the pavement be 0.51. How far (in percent) up the ladder must the firefighter go to put the ladder on the verge of sliding?

Explanation / Answer

Fire fighter from base S

Then Sx = S cos

Fpy = normal force and Fpx = static friction

FW = Fpx

Fpy = ( M + m ) g

Fpy = ( 43 + 72 ) 9.8 = 1127 N

FW = ( m / 3 + M / L )g cos

And = sin-1 ( h / L )

= sin-1 ( 11 / 14 )

= 57.78

and cot

= cot ( 55.78 )

= 0.68

FW = ( m / 3 + M / L ) g cot

FW = ( 43 / 3 + 72 / 14 ) X 9.8 X 0.63

FW = ( 14.33 + 5.14 ) X 6.17

FW = 120.12 N

To calculate S value

By doing and substituting all force values

S = ( ( µ ( M + m ) g L tan ) / M g ) – ( m / M ) ( L / 3 )

S = ( (0.51 X ( 72 +43 ) X 14 X tan ( 57.78 ) ) / 72 ) – ( 43 / 72 ) ( 14 / 3 )

S = 18.09 – 2.75

S = 15.34 m

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