Question

1) A pumping unit is used to recover oil. When the walking beam ABC is horizontal, the force acting in the wireline at the well head is 250 lb. The horse-head C weighs 60 lb and has a center of gravity at Gc. The walking beam ABC has a weight of 130 lb and a center of gravity at GB, and the counterweight has a weight of 200 lb and a center of gravity at GW . The pitman, AD, is pin connected at its end and has negligible weight (a two-force member). Determine the torque M which must be exerted by the motor in order to overcome the force at the well head through the following steps.

Answer 1

Answer:

The torque M is 313 lb*ft

Explanation:

according to the attached diagram and taking the moment at point B:

∑MB = 0

$F_{AD} sin70*(5)-(60*6)-250(6+1)=0\\F_{AD}(4.7)=360+1750\\F_{AD}=449lb$

the moment at point E

∑ME = 0

$-M+(449*3)-200*cos20(3+2.5)=0\\M=1347-1034=313lb*ft$