All agricultural machines are a combination of these five simple machines. The analysis of the forces on a complex machine exceeds the objectives of this text, but Figure 4.17 is an example of how two or more of these machines can be used together.
This example is also useful for discussing the problem solving process. The first inclination is to try and solve for the amount of weight first. If this approach is
used it will soon become evident that an important piece of data is missing. What is the value for the applied force? To solve this problem it is important to realize that the first step is solving the amount of force produced by the jack.
Fr = —-a = -r-r- = -= 5775.36 lb r Ar 1ft 0.052 ...
[Note: in this example the answer was not rounded to the appropriate number of significant figures because it is just the first step in a multiple step problem.]
Ignoring friction and the strength of the materials, the jack will be able to produce 5,775.36 lb of force. The force produced by the jack is the applied force for the lever. The amount of weight that can be lifted by the lever is determined by:
Fa x Aa 5775.36 lb x 2.76 ft 15939.9936
The machines in the illustration are capable of lifting 22,000 lb with an application of 150.4 lb on the jack handle.
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