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zdickz
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This is for an online homework assignment that apparently is not crediting the proper answer. I'm just curious to know if I got this right or if I'm way off. Thanks in advance.
If the coefficient of static friction is 0.362, and the same ladder makes a 61.0° angle with respect to the horizontal, how far along the length of the ladder can a 70.2-kg painter climb before the ladder begins to slip?
[tex]\mu[/tex]s = [tex]\frac{P}{N}[/tex]
P = force of wall on ladder
N = weight of ladder + painter
dx = [P(dy)sin[tex]\theta[/tex]a]/(Nsin[tex]\theta[/tex]b)
theta a = the angle between the ladder and the floor
theta b = the angle between the ladder and the wall
Force of wall on ladder
P = 0.362(743.66) = 268.84N
Horizontal distance of painter
dx = [268.84(10.6)sin 61]/[(743.66)sin 29] = 6.91 m
Homework Statement
If the coefficient of static friction is 0.362, and the same ladder makes a 61.0° angle with respect to the horizontal, how far along the length of the ladder can a 70.2-kg painter climb before the ladder begins to slip?
Homework Equations
[tex]\mu[/tex]s = [tex]\frac{P}{N}[/tex]
P = force of wall on ladder
N = weight of ladder + painter
dx = [P(dy)sin[tex]\theta[/tex]a]/(Nsin[tex]\theta[/tex]b)
theta a = the angle between the ladder and the floor
theta b = the angle between the ladder and the wall
The Attempt at a Solution
Force of wall on ladder
P = 0.362(743.66) = 268.84N
Horizontal distance of painter
dx = [268.84(10.6)sin 61]/[(743.66)sin 29] = 6.91 m
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