How Is the Acceleration of the Center of Mass Calculated with Differing Forces?

AI Thread Summary
The discussion revolves around calculating the acceleration of the center of mass for three balls with different masses and applied forces. The user initially applies the formula F/m = a incorrectly by not considering the individual forces acting on each ball. The correct approach involves calculating the net force on the system and dividing it by the total mass of the system. The acceleration of the center of mass is determined to be (1j - 2j)/6 m/s², confirming the correct magnitude. The clarification helps resolve the user's confusion regarding the calculation method.
Ridonkulus
Messages
5
Reaction score
0
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.
 
Physics news on Phys.org
You've come up with the right magnitude. The accelaration is (1j - 2j)/6 ms-2
 
Oh, ok thanks a lot!
 
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'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top