Problem Layout with Given Information
FreeBody Diagram of Bull 

A good place to start for all statics problems is a freebody diagram (FBD). However, in this problem, because there are numerous objects, it can be confusing which object (or objects) to use in a FBD. Since the bull seems to be central to the problem, it would be a good place to start (note, there is no "wrong" place to start). The FBD of the bull is shown at the left. Summing forces in the ydirection gives,
ΣF_{y}
= F_{scale } + T_{CH} + T_{EG}  W_{bull}
= 0
W_{bull} = 980 lb + T_{CH} + T_{EG}
The rope tensions T_{CH} and T_{EG} must be equal since
it is assumed that the sling holding the bull is symmetrical. This gives,
W_{bull} = 980 lb + 2T_{CH} (1)
However,
you will notice that there are still two unknowns, T_{ch} and
W_{bull}. More information is needed.

Pulley E and D Diagrams 

The next logical object to examine is pulley C since it is connected to the bull. Its FBD is shown at the left along with the rope connected to the counterweight. Summing forces on pulley C gives,
ΣF_{y} = T_{BC} + T_{DC}  T_{CH} = 0
The rope must have the same tension since the pulleys are assumed massless,
T_{BC} = T_{DC} = F = 75 lb
The rope tension T_{CH} can be calculated as,
T_{CH} = 150 lb
Substitute T_{EG} and T_{CH} into Eq. 1 to find the
weight of the bull,
W_{bull} = 1280 lb 


In part a), the bull's weight was found. Now, the problem is reversed, the bull weight is known, 1,368 lb, and the counterweight needs to be determined to make the scale register 1000 lb.

FBD of Cow for Part b) 

This time it is easier to start with the bull and work though the pulleys to find the counterweight. The FBD for the bull is shown at the left.
Summing forces in the ydirection gives the tension in the ropes T_{EG} and T_{CH},
ΣF_{y} = F_{scale } + T_{CH} + T_{EG}  W_{bull} = 0
T_{CH} + T_{EG} = 368
lb
It can be assumed (as done in part a) that the bull's weight is equally distributed between the two ropes holding the cow. Thus T_{CH} equals T_{EG} giving,
T_{CH} = T_{EG} = 368/2 = 184 lb 
Pulley C FreeBody Diagram


The next step is to analyze the pulley C since the tension is rope CH is known. The FBD of the pulley is given at the left. Summing forces gives,
ΣF_{y} = T_{BC} + T_{DC} + T_{CH} = 0
The tensions T_{BC} and T_{DC} are equal,
T_{BC} = 184/2 = 92 lb
Since the rope has the same tension as it goes from pulley C to the counterweight,
F = T_{FA} = T_{AB} =
T_{BC}
The magnitude of the counterweight needed to just register 1000 lb on the scale is
F = 92 lb 