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Acceleration of the lower block 
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#1
May1509, 02:22 PM

P: 138

1. The problem statement, all variables and given/known data
In the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 58.6 N. The coefficient of kinetic friction between the lower block and the surface is 0.204. The coefficient of kinetic friction between the lower block and the upper block is also 0.204. What is the acceleration of the lower block, if the mass of the lower block is 4.68 kg and the mass of the upper block is 2.11 kg? 2. Relevant equations F=ma / friction = [tex]\mu[/tex]mg 3. The attempt at a solution so i wrote the forces that affect each box For the upper box ::[tex]\sum[/tex] F= T[tex]\mu[/tex]m1g for the lower box :: [tex]\sum[/tex] F= FT friction1 friction2 i solved the first equation for tension and then substituted it into the 2nd equation and i solved for acceleration...can u pls. tell me what i'm doin wrong thanx :)) 


#2
May1509, 04:52 PM

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#3
May1509, 05:14 PM

P: 138

for friction 1 i used (coeffecient of frictionxmass1xgravity) and for friction 2 i use (coeffecient of frictionxmass2xgravity)
and the equation i got before isolating for acceleration is :: a(m2+m1)= Fapplied(coeffecient of frictionxmass1xgravity)(coeffecient of frictionxmass1xgravity)(coeffecient of frictionxmass2xgravity) so then a = [Fapplied(coeffecient of frictionxmass1xgravity)(coeffecient of frictionxmass1xgravity)(coeffecient of frictionxmass2xgravity)] /(m1+m2) 


#4
May1509, 05:18 PM

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Acceleration of the lower block



#5
May1509, 05:21 PM

P: 138

is it f 2 = coeffecient of frictionxmass2+ mass1xgravity...???



#6
May1509, 05:31 PM

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#7
May1509, 05:40 PM

P: 138

ok so i changed that and my answer is still coming out wrong is there anything else wrong i cld have done in the initial equation??



#9
May1509, 05:50 PM

P: 138

my final equation wld look like ::
m2a = [Fappliedm1a(coeffecient of frictionxmass1xgravity)(coeffecient of frictionxmass1=mass2xgravity)(coeffecient of frictionxmass2xgravity)] a = [Fapplied(coeffecient of frictionxmass1xgravity)(coeffecient of frictionxmass1=mass2xgravity)(coeffecient of frictionxmass2xgravity)] /(m1+m2) Fapplied is the force given in the question 


#10
May1509, 07:04 PM

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#11
May1509, 07:39 PM

P: 138

m2a= [F m1a (μm1g) (μg(m1+m2))  (μm2g)]
a = [F  (μm1g)  (μg(m1+m2))  (μm2g)]/(m1+m2) iss that clearer ?? 


#12
May1509, 07:45 PM

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#13
May1509, 09:29 PM

P: 138

this is what i have as my equations for m1 and m2
m1:: m1a= T μm1g m2 :: m2a = Fapp  T μ(m1+m2)g μm2g i dont see the extra m2 in my equations can u be a bit more specific 


#14
May1609, 04:46 AM

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P: 41,459




#15
May1609, 12:48 PM

P: 138

so this m2 :: m2a = Fapp  T μ(m1+m2)g μm2g wld be this m2 :: m2a = Fapp  T μ(m1+m2)g μm1g....
then the final equation wld look like a = [Fapp  (μm1g)  (μg(m1+m2))  (μm1g)]/(m1+m2) is that correct?? 


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