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lighter material, reduce friction/contact of other materials

And for similar reasons, explain why using higher masses in calculations might increase your accuracy.

You lose sensitivity, as the ratio of the actual weights to the mass of your lever decreases. (When it's balanced, it's not as accurate this way). Think about it, when you have a heavier bar, it requires more force to move it then a lighter bar.

lchs wrote:

Mathdino wrote:

Here's a IMA question (Click to Show Hidden Text)

Suppose that the pulley in the below image is connected to something weighing 10 kg, and I need to pull on it with 9 N of force to hold it up. What's the efficiency? [img]http://02.edu-cdn.com/files/89501_89600/89569/file_89569.jpg[/img]

I'm not so sure about this but I'll give it a shot...

Answer (Click to Show Hidden Text)

IMA = (Force out of the machine)/(Force into the machine). 
So for this scenario, the Force into the machine would be 9N, I think. And the Force out of the machine would be 10kg * 9.8m/s/s = 98N. 
Therefore, the IMA = 98N/9N = 10.89? 
I feel like I need to use the visual somehow, but this is what I ended up doing... :?

I don't believe your answer is right.
IMA is Ideal Mechanical Advantage.
IMA is calculated by distance. Not force.
It your were calculating a MA with force that would be AMA (Actual Mechanical Advantage).

Is it 9?

Isn't the IMA 9? So the ideal force required would be 98/9 or 10.9 N? If so, then how can the force required be less? Unless I have the IMA wrong......

Agreed, I see the efficiency being over 100 percent, which is obviously not right

Explain why there is a difference in the calculation of IMA in the inclined plane and the wedge, with the IMA of the inclined plane being dependent on the hypotenuse length and the wedge being dependent on the overall length.

I thought wedge was dependent on the length of the slope?

Oops, I thought it would be the overall length of wedge because that would be considered the input distance.