Question #221469
A ladder, whose length is 10 m and whose mass is 40 kg rests against a frictionless, vertical wall. Its upper end is a distance of 7.7 m above the ground. The center of mass of the ladder is 1/3 of the way up the ladder. The coefficient of static friction between the ladder and the ground is 0.53. If a carpenter climbs 85% of the way up the ladder before it starts to slip, find the mass of the carpenter.
1
Expert's answer
2021-07-30T04:25:01-0400

Gives

L=10 m

d1=7.7mL=L3d_1=7.7m\\L'=\frac{L}{3}

Mass(m)=40kg

μk=0.53\mu_k=0.53

θ=cos1(7.710)=15°\theta=cos^{-1}(\frac{7.7}{10})=15°


Fy=0\sum{F_y}=0

0=-Mg-mg+N

N=(m+M)g=(40+M)g

FR=μkNF_R=\mu_kN

F=0.53×(40+M)F=0.53\times(40+M)

According to daigram

0+Mgd1+Mgd2Fd30+Mgd_1+Mgd_2-Fd_3

d1=L3,d2=Lsinθ2,d3=Lcosθd_1=\frac{L}{3},d_2=\frac{Lsin\theta}{2},d_3=Lcos\theta

Put value


F(cosθ)=mgsinθ3+mgL2sinθF(Lcosθ)F(cos\theta)=\frac{mgsin\theta}{3}+mg\frac{L}{2}sin\theta-F(Lcos\theta)

F=mgsin153+mgL2sin15F(Lcos15)cos15=109.4NF=\frac{\frac{mgsin15}{3}+mg\frac{L}{2}sin15-F(Lcos15)}{cos15}=109.4N

109.4=0.53×(40+M)M=65.5kg109.4=0.53\times{(40+M)}\\M=65.5kg


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