Center of Mass F/m = a

  • Thread starter Ridonkulus
  • Start date
  • #1
For some reason I can't seem to figure this question out. I get the feeling it's much easier than I'm making it. But here it is:

Ball 1, ball 2, and ball 3 have masses 1 kg, 2 kg, and 3 kg respectively, and are initially arranged at x = 1 m, x = 2 m, and x = 3 m respectively. y = 1 m for all three balls. Now suppose that a net force of 1 N is applied to ball 1 in the +y direction and a net force of 2 N is applied to ball 2 in the -y direction. The net force on ball 3 is zero. What is the magnitude of the acceleration of the center of mass of the three-ball system?

I've been using the formula F/m = a, where a is the acceleration of the center of mass. For F I've been using (1N-2N), and for mass I've been using (1kg+2kg+3kg). This gives me an answer of .1667 m/s^2, which I think is incorrect (because the formula doesn't take into account which part of the mass is moving in which direction with which force). Any help is greatly appreciated, thanks for taking the time to consider my question.
 

Answers and Replies

  • #2
2,076
2
You've come up with the right magnitude. The accelaration is (1j - 2j)/6 ms-2
 
  • #3
Oh, ok thanks a lot!
 

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