Yes, the acceleration is for all of them.
The tension on each string is the mass following that string times the acceleration.
(Apparently, I'm not allowed to post URLs, but this question is better illustrated by a diagram, i39. tinypic. com/ 17e336. jpg)
<--P--[BOX A]<--Q--[BOX B]<--R--[BOX C]
There are three boxes attached together with string and being pulled across a frictionless surface. Box A has a mass of 2 kg, box B has a mass of 4 kg and box C has a mass of 4 kg. The strings are labelled P, Q and R respectively.
The force of tension for string P is 250 N. I have to find the force of tension for strings Q and R.
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On a system diagram of the entire system, I already found the acceleration (a = F/m = (250/10) = 25 m/s^2). This acceleration is the same for the whole system, right? But I don't know how to find the forces of tension for strings Q and R.
Any help?
Thanks.
2 answers
Then for FTQ (force of tension of string Q):
F=ma
=(8)(25)
=200 m/s^2
FTR
F=ma
=(4)(25)
=100 m/s^2
??
F=ma
=(8)(25)
=200 m/s^2
FTR
F=ma
=(4)(25)
=100 m/s^2
??