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AAMC cbt 9 #51
- Thread starter kabtq9s
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I got this problem right but I kinda made a guess on it.
I keep trying to relate it to the inclined plane formulas in the image below the problem , does it have any relation ?
I wouldn't think about inclined planes here, that's not the way to do it.
Constant speed so forces left = forces right.
F = the horizontal component of T
Horizontal component of T is T cos theta..
done.
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but isn't the answer T cos theta ?
to be honest I just looked at the theta in figure and said "that looks like a cos, since cos = adjacent/ hypotenuse" but I wasn't exactly sure what I was doing 😕
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Yea, typo there sorry about that.
T cos theta.
cos is adjacent over hypotenuse... and T is the hypotenuse.. which leaves you with just the adjacent (which is the horizontal component).
I got this problem right but I kinda made a guess on it.
I keep trying to relate it to the inclined plane formulas in the image below the problem , does it have any relation ?
I also wouldn't think about inclined planes though I guess the whole trigonometry thing is still there. In inclined planes, because the object is actually moving at an angle to horizontal, it's the force is related to it's weight while in the example you showed, we're looking at movement in the horizontal direction (weight is perpendicular to that). Though I guess tension is related to weight so I'm probably not using the best rational when I did it, 😛.
