force for each of the individual parts, but an internal force to the whole system. Writing Eh: = mix for each part of the system we nd: Mass Hanger: WI; T = mi: ax Cart: 1" Wax = mg a: We can eliminate the tension by adding our two equations. We'll add the left sides together and the right sides together. For the two-mass system we have Wk Wu = (ms + mg) (I; (1) Equation 1 looks just right. It says that the net external force equals the total mass of the system times its acceleration. In our particular quest to find the angle where the system will sit still or move at a constant speed, we can go one step further since for our equilibrium situation, (I: = 0. Thus, WhWa=(mh+mc)ax=0 so Wh=ch From our diagrams we know that for the hanger W]; = m g And for the cart W61 = mg g sin(l9) So, equating Win and War, mi: 3 = mg g Siam) 1. Let's try it. With a SO-g hanger and a 450-g cat, the angle for equilibrium, 6 = Show calculations here