1. The problem statement, all variables and given/known data A decorative light fixture in an elevator consists of a 2.0 kg light suspended by a cable from the ceiling of the elevator. From this light, a separate presses the emergency stop button. During the stop, the upper cable snaps. The elevator engineer says that the cable could withstand a force of 40 N without breaking. 2. Relevant equations v2 = v1 + a12(t2-t1) r2 = r1 + v1(t2-t1)+.5(a12)(t2-t1)^2 ƩF=ma 3. The attempt at a solution My biggest problem is getting a value for t2, r2 or a12. I can't seam to solve for any of them. If it had said that the force that did break the cable WAS 40 N I would be fine. But the wording makes it sound like the actual force could be any number higher then 40. But here is what I have tried. I knew I was going to have problems) First I tried solving for a12 using v2 = v1 + a12(t2-t1) 0 = 4 + a12(t2-0) -4 = a12(t2) a12 = -4/t2 Then went for r2 using r2 = r1 + v1(t2-t1) +.5(a12)(t2-t1)^2 r2 = 0+4(t2-0) + .5(-4/t2)(t2-0)^2 r2 = 4t2 - 2t2(t2^2) r2 = 4t2 -2t2 r2 = 2t2 My next try was trying to see if pluging into ƩF = ma for both lights would help The bottom one -F B.Cable + F Gravity = ma12 -F B.Cable +(0.8)(9.8) = 0.8a12 -F B.Cable + 7.84 = 0.8a12 a12 = (7.84- F B.Cable)/0.8 The top one -F T.Cable + F B.Cable + F Gravity = ma12 -F T.Cable + F B.Cable + 2(9.8) = 2 ((7.84 - F B.Cable)/0.8) -F T.Cable + F B.Cable + 19.6 = 2.5(7.84 -F B.Cable) -F T.Cable + F B.Cable + 19.6 = 19.6 - 2.5 F B.Cable F T.Cable = 3.5 F B.Cable So what else should I do? Should I assume that F T.Cable = 40 N and solve from there?