Question

Since the floor is rough, it exerts both a normal force N, and a frictional force / on the ladder. However, since the wall is frictionless, it exerts only a normal force N on the ladder. The ladder has a length of L-43m aweight of w -67.0 N, adrests against the wall a distance d.3.75 m above the floor. If a p son with amass rm" (a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.) which includes all forces acting on the ladder. See if you can apply the first to acting on n N. N

Answer 1

Given that L = 4.3 m, W_L = 67 N d = 3.75 m, L = 4.3 m m = 90 kg consider equilibrium equation sigma F_x = 0 N_1 - mg - W_L = 0 N_1 = (90 times 9.81) + 67 N_1 = 949.9 tidleequal 950 N N_1 = 950 N sigma F_x = 0 N_2 = f_1 sum of moment = 0 sigma M = 0 (N_2 times 3.75) - (1/2 times 67x) - (1/2 times 90 times 9.81x) = 0 N_2 = (67/2 x) + (45 times 9.81x)/3.75 therefore x = squareroot l^2 - d^2 rightarrow Horizontal x = squareroot 4.3^2 - 3.75^2 = 2.104 m N_2 = (33.5 times 2.104) + (45 times 9.81 times 2.104)/3.75 N_2 = 266.43 N f_1 = N_2 f_1 = 266.48 N \r\n person is three fourth of the way up the ladder N_1 = 950 N sigma M = 0 (N_2 times 3.75) - (1/2 times 67 times 2.104) - 3/4 times 90 times 9.81 times 2.104 = 0