How Does a Rock Climber Use Friction to Climb?

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A 55 kg rock climber utilizes friction to ascend a chimney by pressing her shoes against one rock slab and her back against another. The static coefficients of friction are 1.1 for her shoes and 0.80 for her back, and she is on the verge of slipping. The climber's push against the rock can be calculated using the equation that combines the friction from both surfaces with her weight. The next step involves analyzing the horizontal forces to determine the relationship between the normal forces on her shoes and back. Understanding these dynamics is crucial for calculating the magnitude of her push and the fraction of her weight supported by her shoes.
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A 55 kg rock climber is climbing a "chimney" between two rock slabs with her shoes on one side and her back against the other. The static coefficient of friction between her shoes and the rock is 1.1; between her back and the rock it is 0.80. She has reduced her push against the rock until her back and her shoes are on the verge of slipping. What is the magnitude of her push against the rock, and what fraction of her weight is supported by her shoes?

Okay, so I think I should set it up like this (though I'm not sure if this is right):
friction(shoes) + friction(back) = weight

I plugged in the numbers and now I have:
(1.1)*(N1) + (0.8)*(N2) = 539

Where do i go from here (if this is right so far)?
 
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So far, so good. See if you can figure out the relationship between N1 and N2 by analyzing horizontal forces on the climber.
 
does she have to exert a force proportional to the coefficients of friction?
 
Analyze the horizontal forces.
 
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