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Static friction coefficient on different slopes 
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#1
Nov110, 12:04 PM

P: 20

1. The problem statement, all variables and given/known data
Two blocks with the same mass m = 1 kg are stationary on two slopes with slope angles of 30 and 45 degrees. Both blocks are connected with the larger one in the middle via two pulleys. Made a sketch to make it clearer: [tex]\alpha[/tex]= 30° [tex]\beta[/tex]=45° m = 1 kg M = mass of the large block in the middle Question: What is the static friction coefficient of the two blocks, if both of them start sliding at the same time, when M is big enough? Coefficient is the same for both blocks. 2. Relevant equations Equation for static friction coefficient: [tex]\mu[/tex]_{s} = F_{s}/F_{N} (F_{S} is static friction force, F_{N} is normal force) F_{N} = mg[tex]\times[/tex]cos[tex]\phi[/tex] (Phi being either 30 or 45 degrees) 3. The attempt at a solution Both coefficients are the same, so the ratio of both static friction forces and normal forces must be the same. Now if the system is in equilibrium, both static friction forces must be euqal to the force with which the body in the center with mass M is pulling them upwards the slope (that would me Mxg or Mxg/2 for each block) MINUS the dynamic component of force of mass of each block (which is assisting the static friction force with pulling downwards). But here it gets messy, as I'm just not sure enough about all the forces involved and how to properly evaluate them and I have a feeling I'm really over complicating things. I apologize for any nonsense that I've might written and I'd be happy to provide some additional info if it's needed. Thanks in advance. 


#2
Nov110, 05:17 PM

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Hi Rosengrip! Welcome to PF!
The question doesn't say so, but I'd assume that there's a third frictionless pulley above mass M, and that the rope is continuous round that pulley (so the tension is the same on both sides). 


#3
Nov210, 02:13 AM

P: 20

Thanks :)
Yes you're right, this is the case, I forgot to mention that. 


#4
Nov210, 02:51 AM

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Static friction coefficient on different slopes
So, are you ok now?
If not, show us all the equations you do have. 


#5
Nov210, 03:25 AM

P: 20

Well I thought about it a bit and I came up with this:
F_{D} + F_{S} = Mg/2 (F_{D} being a dynamic component of force of mass of a block on slope) This can be written for both blocks as the force which is pulling them upwards is equal (because of that pulley above mass M), the only different thing are the angles. If I break the equation down we have: mgsin([tex]\alpha[/tex]) + [tex]\mu[/tex]_{S}*mgcos([tex]\alpha[/tex]) = Mg/2 for [tex]\alpha[/tex]=30°, same idea for the other block with [tex]\beta[/tex]=45° I could then express M from the second equation, insert it into the 1st and get the coefficient out of it, since I have all the other info. Am I going the right way here? :D 


#6
Nov210, 05:19 AM

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Hi Rosengrip!
(just got up …) Yes, that's fine, but slightly quicker is to simply put the two LHSs equal (the result won't depend on M or m). 


#7
Nov210, 08:37 AM

P: 20

Yep that's definitely faster. Anyways, problem solved, thanks for your time and help! :)



#8
Nov1810, 10:50 AM

P: 5

I have a conceptual problem with this question. How can the blocks both be released up the slopes at the same time if they have the same coefficient of friction and the same pulling force up the slope (the tension). Shoudln't it be harder (i.e. require more force) for the steeper ramp since the component of gravity against that motion is larger? I think that the only way they could release at the same time is if they have different coefficients of static friction. Is this making sense? 


#9
Nov1810, 01:52 PM

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hi jtebb!
so α and β can be either side of that (i suggest you try finding the trig formula for µ in terms of α and β to see how it all fits in … rewrite µ = cotθ, and use one of the standard trigonometric identities ) 


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