Friction Forces on rock climber

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The discussion revolves around calculating the friction forces acting on a rock climber and the dynamics of a block on a slab. For the climber, the push against the rock is determined by the static friction coefficients and normal forces, with a total weight of approximately 480.32 N. The climber's shoes and back provide frictional support, but the exact fraction of weight supported by the shoes remains unclear. In the second scenario, participants emphasize the importance of drawing force diagrams to analyze the movements of both the block and the slab under the applied force. Understanding the normal force and its relationship to the weight of the objects is crucial for solving these physics problems.
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A 52 kg rock climber is climbing a “chimney” between two rock slabs. The coefficient of static friction between her shoes and the rock is 1.33; between her back and the rock it is 0.89. She has reduced her push against the rock until her back and her shoes are on the verge of slipping. (a) What is the magnitude of her push against the rock? (b) What fraction of her weight is supported by the frictional force on her shoes?

so i figure ... fshoes = Ushoes (N1)
... fback = Uback (N2)
fback and fshoes = W

So... N = mg
= (49)(9.8)
N = 480.2
Since there are two equal normal forces... i divided by 2. So each one is equal to 240.1. Not sure if that is a correct assumption/method.

So with that assumption... fshoes = (1.2)(240.1) = 288.24
fback = (0.8)(240.1) = 192.08

From the formula - fshoes + fback = W So the magnitude is 480.32.?

and I am not sure how to find the weight supported by her shoes.


ALSO,


A slab of mass m1 = 35 kg rests on a frictionless floor, and a block of mass m2 = 10 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.68, and the coefficient of kinetic friction is 0.40. The block is pulled by a horizontal force of magnitude 134 N. What are the resulting accelerations (magnitude) of (a) the block and (b) the slab?

how do u figure which one moves? can't both?
i feel as if regardless, both with move left. and there will be 2 accelerations.
im just confused on how to figure out which one would move?


Thanks
 
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N is the Force by the surface that is perpendicular to the surface.
It is NOT equal to "mg"! It is a reaction Force, as strong as it needs to be,
in order to keep the other object (her shoes, for example) from sinking in
any farther. Here, the Normal component of the cliff's Force is the unknown.

F_onback + F_onshoes = - W.
Draw the Force vectors on a diagram (they're parallel to the cliff face)
and you'll see that these have to be the friction Forces.
 
For problem #2,
DRAW the diagram of Forces that act on the Block. compute their strengths.
Draw the diagram of Forces that act on the slab. compute their strengths.
Sum the Forces to obtain ma for each object.
You should've been doing this since day 1 ...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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