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A 14.5-m uniform ladder weighing 495 N rests against a frictionless wall. The ladder makes a 53.0° angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 835-N firefighter has climbed 4.24 m along the ladder from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 8.95 m from the bottom, what is the coefficient of static friction between ladder and ground? Part 1 of 4 - Conceptualize Refer to the force diagram in the figure. Since the wall is frictionless, only the ground exerts an upward force on the ladder to oppose the combined weight of the ladder and firefighter, so ng=m₁g + m₂g. Based on the angle of the ladder, we expect that the force of static friction is less than the sum of the weights of the ladder and the firefighter. We estimate the coefficient of friction to be somewhere between 0 and 1. m₂g nw 15 ng A mg O Tam Ꮎ Ⓡ

Part 3 of 4 - Analyze (a) The wall 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. Fx = fs- nw = 0 ΣFy = no - 1835 ✓ N - 495 N = 0 Taking torques about an axis at the foot of the ladder, we have the following. Σr=0 835 N) (4.24 4.24 m) sin 37⁰ + (495 N) (8.95 X Your response differs from the correct answer by more than 10%. Double check your calculations. m) sin 37⁰ -nul 14.5 ✔m cos 37° Solving this equation for n, we have [(4.24 m) 835 ✔ N) + (14.5 X w = Your response differs from the correct answer by more than 10%. Double check your calculations. m)(495 N)] ta tan 37⁰ 14.5 ✔ m) N. Next substitute the value for into the F equation to find fs=nw=| N. The friction force is in the positive direction toward the wall.