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Acceleration of mass on table

by kasse
Tags: acceleration, mass, solved, table
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kasse
#1
May23-08, 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?
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Texag
#2
May23-08, 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.
Dick
#3
May23-08, 10:35 PM
Sci Advisor
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P: 25,228
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.

kasse
#4
May23-08, 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 m2 with (m1+2+3). 1.9m/s^2. I don't know if my method is right, but it gave me the correct answer.
Dick
#5
May23-08, 11:02 PM
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P: 25,228
Quote Quote by kasse View Post
I solved it in the same way, replacing the static coefficient with the dynamic, and replacing m2 with (m1+2+3). 1.9m/s^2. I don't know if my method is right, but it gave me the correct answer.
That works. As Texag said, since everything is accelerating at the same rate, you can treat the system as one large mass and just add up the external forces. This does save you the step of dealing with the internal tensions.


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