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Hammer - Forces!

  1. Apr 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A hammer slides 21 m down the side of the roof inclined at 60 deg. above the horizontal, and falls down to the lawn 11m below the edge of the roof.

    How far away from the wall will this hammer impact with the lawn?

    m = 1 kg
    mu = 0.23


    2. Relevant equations



    3. The attempt at a solution

    Fgx = mg sin 60
    = (1) (9.8)sin60
    = 8.487N

    Ff = UFn = 1.127 N

    Fnet = Fgx - Ff = 8.487 - 1.127 = 7.36 N

    Fnet = ma
    a = Fnet/m = 7.36 m/s^2

    to find t sub y = 11

    y = yo + voyt - 1/2gt^2
    0 = 4.9t^2 + 15.23t - 11
    t = -15.23 +- (15.23^2 - 4(4.9) (-11)))^sq root

    t = -15.23 +- (15.23^2 - 4(4.9) (-11)))^sq root / 2 (4.9)

    t = -15.23+- (447.5529)^1/2/ 9.8

    then t = 0.06004

    sub t = 0.604 into x = Vxt
    x = 17.582cos60 (0.604)
    x = 5.3 m

    5.3 m away from the wall

    ?? i rly doubt this is rite...??
     
  2. jcsd
  3. Apr 26, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    It appears to be correct.
     
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