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Magnitude of acceleration of a stationary mass

  1. Jan 30, 2009 #1
    1. The problem statement, all variables and given/known data
    An initially stationary block of mass m lies on the floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a) μs = 0.620 and μk = 0.520 and (b) μs = 0.410 and μk = 0.310?


    2. Relevant equations
    Fx=.5mgcos20
    Fy=.5mgsin20

    3. The attempt at a solution
    I first found my force in the x and y directions by using: Fx=Fcos(theta) = 0.5mgcos(theta)
    Fy=Fsin(theta) = 0.5mgsin(theta). Since I had to find the normal force (in order to find out the static friction) i used: Fnet,y=may => Fn-Fg-Fy=may and since its not accelerating in the y direction ay=0 making: Fn=Fg+Fy => mg+0.5mgsin(theta), then i found the max static friction by using: Mu*normal force when i compared my max static friction to my Force in the x direction (since thats the only direction its moving), my static friction was less than my Fx, making a=0. And then i did that again for the second set of numbers for part b, got the same thing, a=0, but got it wrong. So i dont know what to do from there.
     
    Last edited: Jan 30, 2009
  2. jcsd
  3. Jan 30, 2009 #2

    rl.bhat

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    mg acts in the downward direction. Take its component perpendicular and parallel to the inclined plane. Perpendicular component will be the normal reaction. From this find the frictional force which is along the inclined plane. Now porceed.
     
  4. Jan 30, 2009 #3

    rl.bhat

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    Sorry. Ignore the post #2.
    Since force is acted in the upward direction with angle 20 degree with horizontal, mg and Fy must be in the opposite direction. So Fn must be mg - Fy
     
    Last edited: Jan 30, 2009
  5. Jan 30, 2009 #4

    nvn

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    peaceandlove: Although your approach is currently incorrect, you got the answer for part (a) correct, right? Only your answer for part (b) was wrong, true?

    Fn is not equal to m*g + 0.5*m*g*sin(theta). Try again. You also said, "My static friction was less than my Fx, making a = 0." Are you sure this statement is correct?
     
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