Answer Happy

A 14.5-m uniform ladder weighing 485 N rests against a frictionless wall. The ladder makes a 63.0° angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 845-N firefighter has climbed 3.74 m along the ladder from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 8.75 m from the bottom, what is the coefficient of static friction between ladder and ground?
(a) The wall is frictionless, but it does exert a horizontal normal force nw. For the x and y components of the force, we have the following from Newton's second law. Ex = fs - nw = 0 F = ne ng 845 N - 485 N = 0 Taking torques about an axis at the foot of the ladder, we have the following. Στ = 0 845 m sin 270 N)(3.74 m) + (485N)( 13.74 x Your response differs from the correct answer by more than 10%. Double check your calculations. m sin 27° -n 14.5 cos 270 Solving this equation for w. we have [(3.74 m) 8.75 x n = Your response differs from the correct answer by more than 10%. Double check your calculations. m m) (485 ) 845 tan 270 14.5 N. Next substitute the value for ninto the F, equation to find N. The friction force is in the positive x direction toward the wall.