2. A house painter stands 3.0 m above the ground on a 5.0-m-long ladder that lea
ID: 2193418 • Letter: 2
Question
2. A house painter stands 3.0 m above the ground on a 5.0-m-long ladder that leans against the wall at a point4.7 m above the ground. The painter weighs 680 N and the ladder weighs 120 N. Assuming no friction between the house and the upper end of the ladder, find the force of friction that the driveway exerts on the bottom of the ladder.
Explanation / Answer
1. Weight of ladder acts at middle of ladder (2.5 m along ladder's length). So the lever arm is: (2.5m)(cos70). Torque due to ladder's weight = (2.5m)(cos70)(120N) 2. Weight of man. He is 3.0m above the ground (NOT 3.0m along the ladder). So the lever arm is: (3.0m)/(tan70). Torque due to man's weight = (3.0m)/(tan70)(680N) Counter-clockwise torque: 1. Force of wall pushing toward left: Call this "F_wall". Lever arm is: 4.7m. Torque due to wall is: (4.7m)(F_wall) Torques balance, so: (4.7m)(F_wall) = (2.5m)(cos70)(120N) + (3.0m)/(tan70)(680N) Now balance the forces. Notice that the only horizontal forces are: F_wall (pushing toward left); F_ground (force of friction due to ground, pushing toward right) Forces balance, so the magnitude of F_ground equals the magnitude of F_wall. |F_ground| = |F_wall| = ((2.5m)(cos70)(120N) + (3.0m)/(tan70)(680N)) / (4.7m) = 180N
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