How Do You Calculate Mass 2 in Different Friction Scenarios?

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To calculate mass 2 in different friction scenarios, the discussion focuses on three cases: when mass 2 travels uniformly upward, downward, and when both masses are at rest. For the case of rest, the relationship m2 = μs * m1 is established, while for upward motion, μk * m1 is used. The downward scenario leads to the equation m2 = m1 / μk. The conversation also highlights the importance of understanding the setup, which may involve angles and additional forces not initially mentioned. The participants emphasize the need for clarity in the problem setup to accurately solve the equations.
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Homework Statement

mass 1 is 2.0 kg, the static and kinetic coefficients of friction are .30 and .20 respectively.

a) What is mass 2 if it travels uniformly upward?
b) Downward?
c) mass 1 and 2 are at rest?



Homework Equations





The Attempt at a Solution



Part c is the easiest. Friction and Tension are going to be the same so m_{2} = \mu_{s}m_{1}.

Part a, I did the same thing about except I used \mu_{k}m_{1}.

Part b, m_{1}g = \mu_{k}m_{1}g
m_{2} = m_{1}/\mu_{k}

Is this right?

Why is it that I can ace Abstract Algebra but this gives me a headache?
 
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Hi lockedup! :smile:

erm :redface: … what is the set-up? :confused:
 
I'm guessing its a triangular slope of some sort with one mass on either side.
 
Hi inutard! :smile:
inutard said:
I'm guessing its a triangular slope of some sort with one mass on either side.

ah, but then we'd have heard about a θ :wink:
 
Good point. Maybe its in the diagram that he is neglecting to show us (As he also mentions tension but he neglected to talk about a rope or string in his question).
 
Damn, I forgot to attach the picture! I'm a she BTW... lol
 

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(just got up :zzz: …)

Hi lockedup (f.)! :wink:

mmm … inutard :smile: guessed right! :biggrin:

ok then, there's a θ = 37º, which isn't in your equations …

put it in now …

what do you get? :smile:
 
Pardon me about the gender mixup! Remember that there's not only frictional force opposing direction of motion but also a force to the bottom-left (depending on the angle) that is caused by the normal force on block m1
 
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