Find force??????

A 64 kg rock climber is climbing a "chimney" between two rock slabs. The static coefficient of friction between her shoes and the rock is 1.06; between her back and the rock it is 0.84. She has reduced her push against the rock until her back and her shoes are on the verge of slipping. What is her push against the rock?





What fraction of her weight is supported by the frictional force on her shoes?

Find force??????
The normal force on each surface is F


summing forces in the vertical


F*(0.84+1.06)=64*9.8


solve for F


F=64*9.8/(0.84+1.06)


that is the push





The shoes support


F*1.06/(64*9.8)


fraction of her weight





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