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Friction Forces 
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#1
Feb1806, 05:35 PM

P: 44

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 im 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? cant 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 


#2
Feb1806, 08:46 PM

HW Helper
P: 1,117

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. 


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