Question

A uniform ladder is 14 m long and weighs 210 N. In the figure, the ladder leans against a vertical, frictionless wall at height h = 7.0 m above the ground. A horizontal force r is applied to the ladder at distance d = 1.2 m from its base (measured along the ladder). If force magnitude F = 40 N, what are (a)-, (b)y-components of the force of the ground on the ladder? If F 130 N, what are (c)x-, (d)y-components of the force of the ground on the ladder? (e) Suppose the coefficient of static friction between the ladder and the ground is 0.44; for what minimum value of the force magnitude F will the base of the ladder just barely start to move toward the wall? (a) Number (b) Number (c) Number Units Units Units

Answer 1

Theta = sin^-1 (h/l) = sin^-1 (7/14) = 30 degrees Ladder is in equilibrium, therefore sigma F_x = 0 sigma F_y = 0 sigma F_M = 0 sigma F_x = 0 F_x + F - N = 0 equation (1) sigma F_y = 0 F_y - W = 0 equation (2) sigma M = 0 M times h - W times l/2 cos theta - F times d sin theta = 0 equation (3) F_x + 40 - N = 0 F_y - 210 = 0 N times 7 - 210 times 14/2 times cos (30) - 40 times 1.2 times sin (30) = 0 N = 205.86 N (a) F_x = N - 40 = 205.86 - 40 = 165.86 N F_x = 165.86N (b) F_y = 210 N (c) N times 7 - 210 times 14/2 cos (30) - 130 times 1.2 times sin (30) = 0 N = 259.86 N F_x = N - 130 = 259.86 - 130 = 129.86 N F_x = 129.86 N (d)