- #1
Kamisama
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Homework Statement
A 5.00-m long uniform ladder leans against a smooth wall and its base rests on a rough floor. The ladder has a mass of 18.0 kg and its base is a distance of 2.30 m from the wall. A person of mass 70.0 kg climbs 2.50 m up the ladder. If the ladder is to remain in place, what frictional force must be exerted by the floor on the ladder?
Homework Equations
t = FR
t =0
N=mgcosθ
ΣF=0
Σt = 0
-mLgbg-mPgbP+Fn,w*a = 0
Fn,w= (mLgbg + mPgbP ) / a
ΣFx = 0
Fs - Fn,w = 0
Fs = Fn,w = (mLgbg + mPgbP ) / a
The Attempt at a Solution
wall (a)= sqrt(30.29)= 5.50m, bL=2.30m, bP=cos(67.3)*cP
c= 5.00m, cP=2.50m
θL= tan-1(5.50/2.30)= 67.3°
Fs = ((18.0)(9.81)(2.30)+(70.0)(9.81)(.965)) / 2.5 = 194 N
I'm not completely sure if I found all forces for this problem, did I find everything for the frictional force?