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Arrrgh! sliding friction problem

  1. Apr 11, 2003 #1
    Ok, I have been going over this problem and the answer I got just doesn't make sense....help!

    It's a really easy problem (apparently) but the answer that I keep getting doesn't make sense to me...please let me know what I'm doing wrong or forgetting or whatever....

    I thought it'd be easier to answer if you could see the problem I was given so I put it up with my interpretation as to how to answer it on a yahoo! website :


    Thanks in advance!!!!
  2. jcsd
  3. Apr 11, 2003 #2


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    Staff Emeritus
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    Gold Member

    Welcome to the forums!

    Your mistake lies with how you drew your yellow triangle. If you call the 40 degree angle A, the one on the 100N side B and the one on the R side C, then the 90 degree angle needs to be at point C.

    Your tipoff that you made a mistake is this: The normal force is pushing the block with a force larger than the block weighs.

    The easiest way I've found to do these is to draw a new coordinate system:

    Make the line which the block is sliding on the X axis and the direction which N points (it should be pointing toward the block, being the normal force...) the +y axis. Now, when you break any forces which are at an angle into their component parts, you know that they will form perpendicular angles with that coordinate system.

    So, you will have 100cos(40) for the Y and 100sin(40) for the X.

    Make sense?
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