Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

It's that time again: Help by tonight

  1. Oct 6, 2003 #1
    Thanks in advance for the help. I'm stuck on problem #12(please see the attached file). I did this:
    r= cos(26)*2.8(is this right?)
    r= 2.52
    a= mg/(cos(ang))

    a= 5.9(9.8)/cos(26)
    a= 64.3

    v= sqrt(2.52*64.3)
    v= 12.7

    What's wrong with this?

    Attached Files:

    • hw.pdf
      File size:
      49.4 KB
  2. jcsd
  3. Oct 6, 2003 #2


    User Avatar
    Science Advisor

    "r= cos(26)*2.8(is this right?)"

    No, it's not. r is the "opposite" side of the right triangle:
    r= sin(26)*2.8
  4. Oct 7, 2003 #3


    User Avatar
    Science Advisor

    "a= mg/(cos(ang))"

    No, that's wrong. For one thing, there is no "m" in the acceleration. For another thing, the acceleration you want, in order to calculate speed around the circle is the acceleration toward the center, that's directed along the opposite side of your triangle, not the hypotenuse. The formula you want is a/g= tan(ang) (a= opposite side, g= near side).

    You do want v2= r* a which is
    v2= (2.8 sin(26))(9.8 tan(26)).

    I get v= 2.42 m/s.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook