What Are the Limits of m1 to Prevent System Acceleration with µ=0.42?

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The discussion focuses on determining the minimum and maximum values of mass m1 to prevent system acceleration, given a coefficient of friction (µ) of 0.42. The equations of motion and friction are applied, specifically using F = m(a) and Ff = µk(N). The calculations reveal that the net force must equal zero to maintain equilibrium, leading to the conclusion that the correct approach involves considering forces along the diagonal direction, which was initially overlooked.

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Homework Statement



What are the minimum and maximum values of m1 in the figure to keep the system from accelerating? Take µs = µk = 0.42.

5-30.gif


________ kg < (or equal to) m1 < (or equal to) ________ kg


Homework Equations



F = m(a)
Ff=Mk(N)
Ff=Ms(n)

The Attempt at a Solution



Fy(net)=0=Fn-Fgy
Fgy=Fgcos30
Fg=9.8m
Fn=9.8mcos30

Ff=.42(9.8mcos30)

Does Ff equal 49 since Fg of mass 2=49 and thus Ft must equal 49 so mass 2 doesn't move and then that means the Ft on the mass 1 also equals 49 so Ff must also equal 49??

49=.42(9.8mcos30)

I get m=13.75 which is wrong. :(
 
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You didn't write the net force along the diagonal direction. If you did, you'd notice that something was missing.
 

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