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Three blocks and two pulleys

by Squeezebox
Tags: blocks, pulleys
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Squeezebox
#1
Apr10-09, 08:39 AM
P: 58
1. The problem statement, all variables and given/known data
The coefficient of kinetic friction between the two blocks is 0.30. The surface of the table and pulleys are frictionless.

m1=2 kg
m2=3 kg
m3=10 kg

2. Relevant equations
[tex]\Sigma[/tex]F=ma
T2-m3g=m3a
T1-T2=m2a
friction-T1=m1a


3. The attempt at a solution
T2=m3g+m3a
T1=friction-m1a

friction-m1a-m3g-m3a=m2a

(friction-m3g)/(m1+m2+m3)=a
a=-6.15m/s2

The answer in my book say it is -5.7m/s2, though.
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method_man
#2
Apr10-09, 02:31 PM
P: 69
Quote Quote by Squeezebox View Post
1. The problem statement, all variables and given/known data
The coefficient of kinetic friction between the two blocks is 0.30. The surface of the table and pulleys are frictionless.

m1=2 kg
m2=3 kg
m3=10 kg
The answer in my book say it is -5.7m/s2, though.
T2=m3g-m3a
T1=m1a+m1
T2=T1+m1gμ+m2a
a=(m3g-m1gμ-m1gμ)/(m1+m2+m3)=5.7552


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