The vertical force on the ladder at the bottom is (70 + 20)g = 90 g = 882 N.
There is a clockwise torque about the wall support point due to this force, and it equals 882N*6 sin30 = 2646 N*m
There is also a counterclockwise torque about the same point, equal to
fs*6sin30 + 20g*6 cos30 + 70g*x cos30
where x is the distance of the man from the top of the ladder.
The largest value of fs needed to obtain equilibrium is required when x is least, e.g., when x = 0 and the man is at the top of the ladder. At that time,
2646 N = 3 fs + 1018N
Solve for fs
A person with mass 70 kg is climbing a ladder (length 6m, mass 20 kg) that leans against a smooth wall (no friction). A frictional force fs between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is 30 degrees.
What is the magnitude of fs so that the person can reach the top fo the ladder?
1 answer
1 answer