Dee Ford wins AFC Defensive Player of the Week

The Kansas City Chiefs defense gets some positive press with news that pass rusher Dee Ford earned the conference’s player of the week award.

The Kansas City Chiefs defense came up big in a couple key spots in their second win over the Denver Broncos this season. One player, however, stood out from the rest as Dee Ford had a tremendous game overall. Now he has been honored as the AFC Defensive Player of the Week for Week 8 in the NFL.

Ford had three sacks and two forced fumbles against the Broncos at Arrowhead on Sunday and had eight total quarterback pressures. On the year, Ford leads the National Football League in total pressures with 45 through eight games, seven more than Brandon Graham of the Philadelphia Eagles. Ford has nine total sacks on the season for the Chiefs.

Ford’s career high in sacks is 10 and it seems pretty safe to say that he will eclipse that total at this point. In years past, Ford has been criticized for inconsistent play and injuries have often taken over his storyline as well. This year, however, in a contract season, Ford has been able to put it all together in a year that the Chiefs defense has been incredibly subpar.

What makes Ford’s success in 2018 so much more important to the Chiefs is the fact that their primary pass rusher, Justin Houston, has been missing since Week 5 and is only now starting to return to limited practice. The Chiefs are trying to utilize rookie Breeland Speaks and second-year project Tanoh Kpassagnon as pass rushers opposite Ford, but fortunately Ford has been quick enough to beat even the double teams he’s faced since Houston left the line-up.

New Orleans Saints cornerback P.J. Williams received the same award for the NFC in Week 8.