physics
posted by joe .
A ladder can fall for two reasons. If it is set too steep and the climber gets their mass to the left of the ladder's base, the ladder likely will fall over backwards. If the ladder is set at too shallow of an angle the required force of friction between the ladder and the ground might be too great and the base of the ladder will slip. Assume that there is no friction between the ladder and the wall and that the ladder is effectively weightless. The coefficient of friction between the base of the ladder and the ground is 0.49. The person using the ladder will be 76% of the way up the ladder. If the person climbing the ladder has a weight of 970 newtons and the ladder is 5.7 meters long, how far from the wall can the base of the ladder be placed and not slip?
Distance from wall =

sum moments about the top of the ladder.
Is this a weightless ladder? the ladder weight is not given.
clockwise positive, and my sketch has clockwise toward the wall.
d is the distance from base to wall
Theta is the base angle, so d/5.7=sinTheta
sum moments about the top wall junction point with the ladder.
W*(.24*5.7*970g sinTheta)970g*.76*5.7*cosTeta*mu=0
(the sum of the moments about any point is equal to zero)
divde through by cosTheta, then solve for tanTheta, then d=5.7cosTheta
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