When Matt Forte went down with a knee injury in Week 8 against the Minnesota Vikings, rookie running back Jeremy Langford was thought to be the man to step into the starting position. Not only has he filled in nicely to this point, but one can argue that he is actually putting up better numbers than Forte was. Those of you who were able to grab Langford off the waiver wire were probably not expecting this much production.
Langford has started two games in replacement of Forte. In those two games Langford hasn't been overly effective running the ball with only 3.8 yards per carry. Fellow running back Ka'deem Carey has quietly averaged better numbers with 21 carries for 84 yards, good for 4.0 yards per carry. However, Langford's mediocre YPC numbers are masked by the fact that he has produced 145 yards and two rushing touchdowns in both contests. In addition, it has really been his involvement in the passing game that has put him into the elite RB1 category. In those two games, he has caught 10 passes for 189 yards and one touchdown. This includes the long 83-yard touchdown pass he caught last game against a solid St. Louis Rams defense.
Veteran running back Matt Forte continues to practice in a limited fashion due to a sprained MCL in his knee. Since Forte has been practicing since last week, it wouldn't be a stretch if he was able to suit up Week 11 against the Denver Broncos. There are two scenarios to consider for this situation:
1) If Matt Forte does not play this weekend, consider Langford a RB1 as the Bears will look for him to carry the bulk of the load. Don't let the Broncos defense scare you because in the past two weeks they have given up 226 yards on the ground to opposing teams.
2) If Matt Forte can play it will create a nightmare-ish situation for Langford and Forte owners. Forte will more than likely be moderately worked back into the mix coming off an injury and especially because of Langford's effective play. Look to start Forte and Langford in your Flex RB spot because the splitting of carries will decrease both of their values.