The 49 kg climber is supported in the "chimney" by the friction forces exerted on his shoes and back. The climber has reduced push to verge of slip. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 1.2 and 0.8, respectively. What is the magnitude of the push he must exert? What fraction of the climbers weight is supported by the frictional force on his shoes?

Answer 1

Well, if he is at the point of slip, then force friction shoes=X*mg
where x is some fraction of weight.
This means force normal=K, and mu*K= Xmg

Now if K is the normal force, then K must be acting on the back, so friction on the back is .8*mg*(1-x)

we know the two frictions have to equal weight,

.8(mg)(1-x)+1.2*X*mg=mg
.8-.8x+ 1.2x=1
.4x=.2
x= .5
so half his weight is supported by his shoes.

Answer 2

force of push = (mg)/(u1+u2), so
(49*9.8)/(1.2+.8)=240.1N

fraction of weight= (u1)/(u1+u2),so
(1.2)/(1.2+.8)=.6