Question

The 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. Determine the torque M which must be exerted by the motor in order to overcome this load. 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 ends and has negligible weight.

Answer 1

Answer:

449.08 lb

313.53 lb.ft

Explanation:

A free body diagram of the walking beam is shown in the first attached file below.

Applying moment equilibrium about point B.

$\sum M_B = 0$

$F_{AD} \ sin \ 70^0 \ (5) \ - \ 60 (6)\ - \ 250\ (6+1) = \ 0$

$F_{AD} \ 0.9397(5) \ - 360 \ - 1750 = \ 0$

$F_{AD} = \frac{360\ + \ 1750}{4.6985}$

$F_{AD} = \frac{2110}{4.6985}$

$F_{AD} = 449.08 \ lb$

A free body diagram of the counterweight is shown in the second part of the diagram

Now, Applying moment equilibrium about point E..

$\sum \ _ {MB} = 0$   ;

$- M + F_{AD}(3) \ - \ 200 \ cos 20 \ (3 + 2.5)= 0$

$- M + 449.08(3) \ - \ 200 \ cos 20 \ (3 + 2.5)= 0$

$- M + 1347.24- 200 (0.9397)(5.5) = 0$

$- M + 1347.24 - 200(5.168309414) = 0$

$- M + 1347.24 - 1033.661883 = 0$

$- M = - 1347.24 + 1033 . 661883$

$M = 1347.24 - 1033.661883$

$M = 313.578117$

$M = 313.58 \ lb \ .ft$