mohabitar
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Whats wrong with my answer? I've tried just about everything, but its not accepting it. I've tried +N too, even though I think what I have in there should be right.
Look at the definition of the axes below the free body diagram. y is pointed perpendicular (away) from the surface (i.e. normal to the surface). x points parallel to the surface (down the slope).mohabitar said:Whats wrong with my answer? I've tried just about everything, but its not accepting it. I've tried +N too, even though I think what I have in there should be right.
Sure, if you want to define them that way, and if you are given the freedom to do so. But in this problem, the x-axis and y-axis are defined for you. One doesn't have a choice on how to define them, for this particular problem.thrill3rnit3 said:Shouldn't up the y-axis be positive and down the y-axis be negative?
Yes, okay I see what you're saying now.mohabitar said:I don't get it. It should just be the sum of all the forces in the y direction, and that's simply the Normal force and the y component of mg, which is mgcos30?
collinsmark said:(My previous line of reasoning is that these two force components add up to something very trivial, and that trivial thing is what is equal to may. But maybe that's not the form of what this online question is asking.)
Well, yes I suppose. If I had to use something of that form, "N - mgcos30" would be better.mohabitar said:I think I've tried all possible combinations of signs (+,-) to see if it works, but it wouldn't accept anything. It should be N-mgcos30 then right?
collinsmark said:Or maybe the program is barfing because you are inputting "N - mgcos30" instead of "N - mgcos(30)" or something like that.