Question

AT&T 3:46 PM Done Canonical Problem 2 Fall 2018.pdf (A) Canonical Problems for Intro Mechanics Subjcct: Intro Mechanics// Topics: Torque, Statics, Forces, Friction Problem Title: Romeo is a Jerk Romeo (weight W) is climbing a ladder to see Juliet. His ladder (length L) and has a weight of W. It is leaned against the wall at an angle of above the horizontal. The wall is greased such that there is no friction but it still exerts a force normal to the surface (N). The ground exerts a normal force on the ladder (N) and has a coefficient of static fniction ps A.) (2pts) force from the ground on the ladder (N. Your answer may include only the variables W, W and ps as needed. Write an expression for the normal B.) (2pts) Write an expression for the normal force from the wall on the ladder (N) Your answer may include only the variables W, W, and ps as needed C) (2pts) Write an equation for the sum of the torques acting on the ladder about the point that is in contact with the ground. Your answer should include only the variables WL, WR.L and 0 and s as needed D.) (4pts) How far from the bottom of the ladder (variable D) can Romeo climb before the ladder slips? Answer in terms of L, W W, and s as needed.

Answer 1

(A) Sum of all vertical forces must be zero at equilibrium i.e. sigma F_v = 0 N_2 - w_L - w_g = 0 implies N_2 = W_L + w_g B) At equilibrium, sum of all horizontal force must be zero. i.e. sigma F_H = 0 F_s - N_1 = 0 [when F_s in frictional force at the bottom of the ladder) N_1 = F_s Now F_s, max = mu_s N_2 implies N_1, max = mu_s N_2 = mu_s [W_L + w_g] \r\n C) At equilibrium, sum of torques about any point must be zero. therefore About P^f = 0 (bottom of the ladder) we have N_1 sin theta - w_L L/2 cos theta - w_g cos theta = 0 F_s sin theta - W_L L/2 cos theta - W_g D cos theta = 0 equation (1) D) From eq. (1) atom W_g D cos theta = F_s sin theta - W_L 1/2 cos theta D = F_s/W_g tan theta - W_L L/2W_g As D increases so does