
#1
May2308, 09:49 PM

P: 464

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

Sci Advisor
HW Helper
Thanks
P: 25,170

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: 464

[SOLVED] 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

Sci Advisor
HW Helper
Thanks
P: 25,170




Register to reply 
Related Discussions  
[SOLVED] Two blocks  one on a table and one hanging  acceleration with friction  Introductory Physics Homework  28  
A spool of mass M sits upright on a table...  Introductory Physics Homework  13  
Acceleration of hollow sphere rolling down table.  Introductory Physics Homework  3  
Pt. Mass and Spring rotating on table  Introductory Physics Homework  1  
probability mass function of a hash table?  Introductory Physics Homework  1 