A 15.0 m uniform ladder weighing 480 N rests against a frictionless wall. The la

Question

A 15.0 m uniform ladder weighing 480 N rests against a frictionless wall. The ladder makes a 57.0° angle with the horizontal (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 810 N firefighter is 3.80 m from the bottom. Magnitude of the horizontal force o towards the wall o away from the wall Magnitude of the vertical force Direction up down (b) If the ladder is just on the verge of slipping when the firefighter is 9.20 m up, what is the coefficient of static friction between ladder and ground?

Explanation / Answer

sum of moment equals zero

let R be the magnitude of the force of the wall

M(t) = 810(3.8cos57) + 480(7.5cos57) - R(15sin57)

0 = 1676.4 + 1960.7 – 12.58*R

R = 289.12 N directed to the left
Fx = 289.12 N directed to the left
Fy = 480 + 810 = 1290 N directed upwards

(b)

M(t) = 810(9.2cos57) + 480(7.5cos57) - R(15sin57)

M(t) = 4058.65 + 1960.7 – 12.58*R

R = 478.4857 N

Fx = 478.4857 N directed away from the wall
Fy = 480 + 810 = 1290 N

for the coefficient of static friction,

f = uN
478.4857 = u(1290)
u = 0.37

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