- #1
mybrohshi5
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Homework Statement
A uniform ladder 5.1 m long rests against a frictionless, vertical wall with its lower end 3.1 m from the wall. The ladder weighs 165 N. The coefficient of static friction between the foot of the ladder and the ground is 0.39 . A painter weighing 739 N climbs slowly up the ladder.
1) What is the actual size of the frictional force when the painter has climbed 0.9393 m along the ladder?
2) How far along the ladder can the painter climb before the ladder starts to slip?
Homework Equations
[itex]\tau=r*Fsin(\theta[/itex])
The Attempt at a Solution
[itex] \theta = cos^{-1}\frac{3.1}{5.1} = 52.57\degree [/itex]
This is the angle where the ladder meets with the floor
For the first question i found that
[itex] \tau_{man} + \tau_{ladder} - \tau_{friction} = 0 [/itex]
I used my pivot point to be at the bottom of the ladder where it meets with the floor
[itex]739N(0.9393m)(cos(52.57)) + 165N(\frac{5.1m}{2})(cos(52.57)) - f_s(5.1m)(cos(37.43)) = 0[/itex]
[itex] f_s = 167.3 N [/itex]
I know this is the right answer but i am a little confused on when to use sine or cosine and what angle to use with it. Could anyone explain this to me? I have a test in a few days and i really need to know this :)
I Could not figure out part 2 so after i figure out how to use sine and cosine for part 1 i will give part 2 a shot.
Thank you for any explanations on this :)