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Two Masses, a Pulley, and an Inclined Plane 
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#1
Mar2906, 10:48 PM

P: 6

Block 1, of mass m1, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is mu.
Find the ratio of the masses m1/m2. 


#2
Mar2906, 11:00 PM

P: 3

Let the tension in the thread be T. ok. now eq of motion of the 2 masses are...
(m1)g  T = (m1)a T  (m2)gsin(theta)  (mu)(m2)gcos(theta) = (m2)a Solving them by addng the two eqns u get, m1/m2 = (g(sin(theta) + (mu)cos(theta)))/(g  a) 


#3
Mar2906, 11:07 PM

P: 6

Thanks rammstein, but after inputting m1/m2, I get "aren't you missing a term?" Thanks a bunch for the help though!



#4
Mar3006, 06:21 AM

P: 335

Two Masses, a Pulley, and an Inclined Plane
Dan 


#5
Apr206, 10:28 PM

P: 6

I tried setting the two equations equal to each other through T, but was unable to discover the missing term...



#6
Apr306, 07:15 AM

P: 335

[tex](m_1gm_1a)m_2gsin \theta\mu m_2gcos \theta=m_2a[/tex] [tex]m_1gm_1a=m_2gsin \theta+\mu m_2gcos \theta+m_2a[/tex] [tex]m_1(ga)=m_2(gsin \theta+\mu gcos \theta+a)[/tex] [tex]\frac{m1}{m2}=\frac{gsin \theta+\mu gcos \theta+a}{ga}[/tex] rammstein left out the last "a" in the numerator. Dan 


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