Recent content by GrantFuhrer

  1. G

    Newton's second law (friction and pulley)

    Hmm.. T - 1.18 = 1N T = 2.18N / 9.81m/s^2 = 0.222kg Same answer I had before. It's not right because when I verify: (0.222*9.81) - 1.18 = 1 / (0.222 + 1) = 0.81 Acceleration has to be 1.00m/s^2 and not 0.81.
  2. G

    Newton's second law (friction and pulley)

    Fr is frictional force. Yes I know my mistake was that it is the same direction as the net force. Indeed the pulley is massless and frictionless. This is what I do so that the force of friction is different from the net force: -Fnet = (x) - 1.18 -1N = (x) - 1.18 x = 0.18N 0.18N/9.81 = 0.018kg...
  3. G

    Newton's second law (friction and pulley)

    Homework Statement Homework Equations F=ma, Fr = μN The Attempt at a Solution 1kg(1m/s^2) = 1N Fnet = (x) - 1.18 1N = x - 1.18 x = 2.18N 2.18N/9.81m/s^2 = 0.223kg I know this is not right because I was told that in my calculations Fnet and Fr go in the same direction. Also I didn't use...
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