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Ramps and Newton's Second Law

  1. Nov 6, 2008 #1
    1. The problem statement, all variables and given/known data
    The 60 kg block starting at rest is pushed 7.8 m up a ramp at an angle θ = 26.5° to the horizontal in 15 s. If the coeficient of kinetic friction is 0.17:
    What is the acceleration of block?
    What is the force used to push the block up the ramp?

    2. Relevant equations
    Fnet = ma
    Delta X = Vit + (1/2)at^2
    Fkinetic = μ * N

    3. The attempt at a solution
    I think I understand the first part,
    7.8 = 0*15 + (1/2)(a)(15^2)
    7.8 = 112.5a

    But the second part is whats giving me trouble
    Do I need to make an Fnet = ma equation?
    If I do I know that it would look like this
    Force of Push - Force of friction = ma
    Force of Push - Force of friction = (60)(.069)
    Can I substitute in this equation? Fkinetic = μ * N for Fk?
    Force of Push - (μ * N) = 60* .069
    Fpush - (.17 * N) = 4.14

    But what is my N force? I think that it would just be mass times acceleration of gravity and in that case:
    Fpush - (.17 * (60 * 9.81)) = 4.14
    Fpush - 100.06200 = 4.14
    Fpush = 104.20200N

    I'm not sure if this is right, I appreciate your help in advance.
  2. jcsd
  3. Nov 6, 2008 #2
    You did the first part right.

    For the second part, you did everything right except your calculation of the normal force. Try drawing a force diagram of the block. On an inclined ramp, gravity doesn't act perpendicular to the plane of motion.
  4. Nov 6, 2008 #3
    Do I just need to find the horizontal component of gravity?
  5. Nov 6, 2008 #4
    Not the horizontal component, though you do need to break gravity into components. You need to find the component of gravity that is perpendicular to the ramp.
  6. Nov 7, 2008 #5
    So then do I just do
    Cosine (26.5) = A/H
    and in this case adjacent is the force im looking for and H is gravity
    Is the gravity Fg? so I can substitute Fg=ma
    60 * 9.81 = Fg

    Then I have
    Cosine (26.5) * (60 * 9.81) = A

    Is this correct?
  7. Nov 7, 2008 #6
    Alright well thats wrong and I'm not sure why
    Cos(26.5) * H = A
    Cos(26.5) * (60 x 9.81) = A
    which ends up being 118.71N
    When I plug that in I get
    Fpush - (.17 x 118.97) = 4.14
    Fpush = 24.36
    And I this doesn't work, so I'm not sure whats going wrong
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