Answer Happy

Let us assume a scenario that:
Below you find parts that can be used to realize a ladder logic implementation of a machine control. Mark the parts that are need.
Overview Description:
A worker is throwing a box onto the conveyer belt. The conveyer belt transports the box to the right. If another box is on the incline the conveyer belt stops when the box has reached the middle. Then the flap opens and the box, that is waiting on the incline drops onto the first one. The conveyer belt transports the pile of two boxes (or just the first box if there had been no box on the incline) to the right end of the conveyer belt. There a decision is made. One box alone leaves the conveyer belt at this side and drops into the right container. A pile of two boxes is transported by the conveyer belt to the left side again and drops into the left container. After all boxes have left the conveyer belt the conveyer belt stops and a new box can be thrown onto the conveyer belt by a worker.
Components of the machine and hints to the implementation:
CB: A conveyer belt that transports boxes to the right (CB_right is active) or to the left (CB_left is active). Flap: flap opens if it is active and closes if its deactive. PS: A pressure sensor that notices if there is a box on the incline. F S1: A sensor sensing if a box has left the incline. L1, L2, L3, L4, L5 Light barriers (Logic of all light barriers: true if object detected). Sk, Sl, Sm, Sw, Sx (random order): internal states, which represents: the transport of the first box to the middle, the stacking process, the transport to the right end, the transport to the left hand, the moment a box has returned to the left end and is about to drop into the container. Stop: a variable that is used to send an internal signal.
시 THIH 이 베베베베스 E) S F) 사 Listop L stop +1)+(1) L stop 사 G) {}{\} H)- 1) stop 수 stop 사 L₂ stop AHHH L₂ S, Ps S 시대가 K) 니다. M). Ps S 나 stop (FN) stop 이 R) { ÛÛ ÛÒ 5) Û 사 PS4 ㅔ Ps Sm CB_URCB UHH H Flap with FS1 12 Flap 3_right CB_left CB_right