Posted by **Lucy X** on Monday, March 9, 2009 at 9:18pm.

(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.

---

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.

- Physics -
**bobpursley**, Monday, March 9, 2009 at 9:28pm
Yes, the acceleration is for all of them.

The tension on each string is the mass following that string times the acceleration.

- Physics -
**Anonymous**, Monday, March 9, 2009 at 9:35pm
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

??

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