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Newton's laws

  1. Mar 11, 2004 #1
    A horizontal force of 100N pushes a 12kg block up a frictionless
    incline that makes an angle of 25 degree with the horizontal. a)
    what is the normal force that the incline exerts on the block? b)
    what is the acceleration of the block?


    I don't know when to chose the force of the x axes or the y axes.
    After drawing a free body diagram I get really lost when the cos and sin are switched. for example, when Fy = cos F instead Fy = sin F.

    I know this is easy, but I am trying to pinpoint the pit falls so I can master the concept. I thank you all in advance.

    I draw a freebody diagram and taking x=axes to be forming 25 degree with the theta.

    Sigma Fy = Fny + Wy + Fy = May
    Fny - W + F sin(25) = 0
    Fn = 12(9,81) - 100(sin 25)
    Fn = 75.5N

    Please help me
    Last edited: Mar 12, 2004
  2. jcsd
  3. Mar 12, 2004 #2


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    Science Advisor
    Gold Member

    Let's start out by being crystal clear about the orientation of each axis (well, as clear as we can be without pictures). Let's make the x-axis point along the ramp (parallel to the ramp/tangent to the ramp; however you like to say it). The y-axis is then rotated from that direction 90 degrees CCW (so that it is perpendicular to the ramp/in the same direction as the normal force).

    Now, I'll rewrite your first equation:

    [tex] \Sigma F_y = N - W_y + F_{a,y} = ma_y [/tex]

    (Fa,y is just the component of the applied force in the horizontal direction.)

    First rhetorical question for you: what is ay? It's 0: the block stays on the ramp the whole time.

    OK, now we have to do a little geometry/trig to find the components of the forces in our equation. I'm calling the angle that the ramp is to the horizontal θ.
    [tex] W_y = mg\cos\theta [/tex]
    You should see this by making similar triangle-type arguments until you get to the angle you're looking for. Reason it out a few times on these types of problems and before long, you'll recognize which angle is which without thinking about it (which is admittedly a little dangerous).
    Now for the applied force (which is horizontal), you've got to work out the geometry again to find that:
    [tex] F_{a,y} = F_a\tan\theta [/tex]

    (Fa is the applied force)

    So, we're done with that axis; onto the x-axis:

    [tex] \Sigma F_x = ma_x = F_{a,x} - W_x [/tex]

    And from geometry:

    [tex] W_x = mg\sin\theta [/tex]
    [tex] F_{a,x} = \frac{F_a}{\cos\theta} [/tex]

    I hope that helps. Work it out for yourself to make sure I didn't make any mistakes and to make sure you understand each step. Keep going back and forth between the picture and the equations until it makes sense. Have fun.
  4. Mar 12, 2004 #3
    so now Fn,y = mg(cos 25) - F (tan 25) = 0
    Fn,y= 12kg(9,81m/s^2)(cos 25) - 100N(tan 25)
    Fn,y = 106.7 - 46.6 = 60N?

    f/(cos 25) - mg(sin 25) = 12k ax
    = 110,33 - 49.75 = 12 ax
    ax = 60.58/12 = 5m/s^2?

    why fa,x = fa/(cos 25) instead of fa(cos 25)?
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