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?
What is force?
When she's on the verge of slipping, the max friction available will add up to her weight. The normal force, N, that you use in the friction equation will be the same at her back and her feet. (Newton's 3rd)
So write the equations for the 2 frictions, add them and set that sum equal to her weight. Solve for N.
Knowing N, you can figure the friction at her feet and calculate the fraction of her weight.
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