We were working in the lab today talking about and building simple machines. In our discussion of Ideal Mechanical Advantage (IMA) and Actual Mechanical Advantage (AMA) a problem arose that I am not quite sure how to handle. We built a simple class 1 lever using a meter stick with two weight hangers attached to each end. The stick also had a sliding pivot point that could be secured at any point along the stick. I asked one group to make this lever so that it would have an IMA of 8. They did so. Then I asked them to calculate the AMA. They began hanging weights on both sides and stopped when it balanced. after calculating the effort force and the resistance force they calculated AMA. It turned out to be 10 or 11 or something along those lines. Yesterday I had said in lecture that they should expect IMA to be less than AMA because of frictional losses etc. They were surprised to see the reverse of what I had suggested should happen. As we discussed the situation it became clear that one factor that had not been considered was the mass of the weight hanger and another was the mass of the meter stick itself. My question is, should the weight of the hangers and the meter stick itself be considered when calculating AMA? It seems to me that the effort force should only include the force applied to the machine by the operator not any force supplied by the weight of the machine itself. If this is the case then AMA very easily could be considerably greater than IMA in fact AMA would depend on the size of the resistance force. How should this be handled? Am I forgetting to consider something?
Cliff Parker
Never express yourself more clearly than you can think. Niels Bohr