Friction Forces on rock climber

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SUMMARY

The discussion focuses on the physics of friction forces acting on a rock climber and a block resting on a slab. The climber, weighing 52 kg, experiences static friction with coefficients of 1.33 for her shoes and 0.89 for her back. Calculations reveal that her push against the rock is approximately 480.32 N, with the frictional force on her shoes supporting a fraction of her weight. Additionally, a 35 kg slab and a 10 kg block scenario illustrates the need to analyze forces to determine the resulting accelerations of both objects under a horizontal force of 134 N.

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  • Ability to calculate normal forces and weight
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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 ...
 

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