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Acceleration of mass on table 
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#1
May2308, 09:49 PM

P: 463

1. The problem statement, all variables and given/known data
http://www.badongo.com/pic/3633497 3. The attempt at a solution The resultant force of the two hanging mases is 19.6 N. This means that just after the masses are released, the friction is 0.3*mg = 4.41 N, so that the total resultant force is 15.2 N. The acceleration is then 15.2/1.5 = 10.1 m/s^2 Why am I wrong? 


#2
May2308, 10:25 PM

P: 19

First of all, The coefficient of friction you have is for static friction.
Secondly, you must account for the entire system's mass since they are all accelerating. 


#3
May2308, 10:35 PM

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P: 25,250

You got an answer larger than 9.8m/s^2. That should be sending up alarms. No matter how heavy m3 is it's maximum acceleration down is 9.8m/s^2. Now try and do the problem right. There is a tension T1 and T2 in each of the two strings. You have to do a force balance for each mass and then set all of the accelerations equal.



#4
May2308, 10:50 PM

P: 463

Acceleration of mass on table
I solved it in the same way, replacing the static coefficient with the dynamic, and replacing m_{2} with (m_{1}+_{2}+_{3}). 1.9m/s^2. I don't know if my method is right, but it gave me the correct answer.



#5
May2308, 11:02 PM

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P: 25,250




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