How do I find the acceleration of a mass on a table?

AI Thread Summary
To find the acceleration of a mass on a table, consider the resultant forces acting on the system, including friction and tension. The initial calculation of 10.1 m/s² was incorrect due to using static friction instead of dynamic friction and not accounting for the total mass of the system. The maximum possible acceleration cannot exceed 9.8 m/s², indicating an error in the initial approach. By treating the entire system as a single mass and summing external forces, a corrected acceleration of 1.9 m/s² can be achieved. This method simplifies calculations by eliminating the need to analyze internal tensions.
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[SOLVED] acceleration of mass on table

Homework Statement



http://www.badongo.com/pic/3633497


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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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.
 
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.
 
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.
 
kasse said:
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
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