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M1 friction help

  1. Jan 1, 2008 #1
    Hi.

    Could someone please help me with the following M1 question? I would really appreciate any help at all as I am completely stuck at the moment.

    I have drawn a diagram.

    A heavy package is held in equilibrium on a slope by a rope. The package is attached to one end of the rope, the other end being held by a man standing at the top of the slope. The package is modelled as a particle of mass 20 kg. The slope is modelled as a rough plane inclined at 60 degrees to the horizontal and the and the rope as a light inextensible string. The string is assumed to be parallel to a line of greatest slope of the plane. At contact between the package and the slope, the coefficient of friction is 0.4.

    (a) Find the minimum tension in the rope for the package to stay in equilibrium on the slope.


    Thank you.

    Cathy
     
  2. jcsd
  3. Jan 1, 2008 #2

    Doc Al

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    Staff: Mentor

    As always, draw yourself a free body diagram showing the forces acting on the package. Since it's in equilibrium, the sum of the forces (in any direction) must equal zero.
     
  4. Jan 1, 2008 #3
    Thank you for your help.

    If I resolve parallel to and up the plane, I that T - 20g x sin60 but this does not give T as the correct answer of 131 N. Am I missing the point of something?

    Cathy
     
  5. Jan 1, 2008 #4

    Doc Al

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    There are three forces acting parallel to the plane. What are they?
     
  6. Jan 1, 2008 #5
    Is the third force friction?

    Cathy
     
  7. Jan 1, 2008 #6

    Doc Al

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    Of course! :smile:

    What does it equal? Which way does it act?
     
  8. Jan 1, 2008 #7
    I wrote that T - 20g x sin60 - F = 0

    R - 20g x cos60 = 0 so R = 98 N.

    F = 0.4R so T = 209 N (3 s. f.).

    Can you see where I have gone wrong as the answer is 131 N and 209 N is the answer to the next part of the question - the tension in the rope?

    Thanks for helping.

    Cathy
     
  9. Jan 1, 2008 #8

    Doc Al

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    Realize that friction helps prevent the object from sliding down the incline. Which way does it act? What's its sign?
     
  10. Jan 1, 2008 #9
    Friction acts up the plane.

    Thank you - I have the correct answer now.

    Cathy
     
  11. Jan 1, 2008 #10
    Hi.

    I have no idea how to even begin the following question so I would really appreciate it if someone could please help me in any way.

    A particle P moves in a horizontal plane. The acceleration of P is (-i + 2j) m/s^2. At time t = 0, the velocity of P is (2i - 3j) m/s.

    (a) Find, to the nearest degree, the angle between the vector j and the direction of motion of P when t = 0.


    The answer is 146 degrees.

    Thank you.

    Cathy
     
  12. Jan 1, 2008 #11

    Doc Al

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    You are given the velocity vector (in component form). Draw yourself a diagram. The vector j points in the +y axis.
     
  13. Jan 1, 2008 #12
    Thanks for your help.

    I plotted the coordinates of the velocity and drew a right-angled triangle to join this point to j (+1 on the y-axis). I then used tan x = 2/4 but this did not give me the correct answer (which is 146 degrees) for the angle. Could someone please suggest what I am doing wrong?

    Thank you.

    Cathy
     
  14. Jan 1, 2008 #13

    Doc Al

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    Where did this come from? The velocity vector is 2i - 3j. Draw this. (It will be a vector pointing from the origin into the fourth quadrant.) What angle does this make with the x-axis? The +y axis?
     
  15. Jan 1, 2008 #14
    Why does the vector point from the origin? I just redrew the line starting from the point x = 2.

    Thank you for all your help.

    Cathy
     
  16. Jan 1, 2008 #15

    Doc Al

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    What does the point x = 2 have to do with anything?

    You are just comparing the directions of two vectors, so you must draw them from a common origin. A vector is just an arrow. Draw two arrows, one point up to represent the vector j (the +y axis), and the other pointing in the direction of the velocity vector. Make sure the tails of those arrows meet at a common point. (I call that point the origin for convenience.)
     
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