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Forces and Angle

  1. Feb 27, 2004 #1
    When you draw a free body diagram of a box, car, collar, etc... and you put the normal force, frictional force, force of gravity and any other applicable forces, how do you know where the angle goes or maybe a better question would be how do you find an appropriate angle? I hope I'm making myself clear. Thanks in advance.
     
  2. jcsd
  3. Feb 27, 2004 #2

    Doc Al

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    You'll probably need to give a specific example that causes you confusion.

    In general, your knowledge of the physics tells you what direction the forces act. For example: gravity points toward the earth (usually down), normal forces are normal (perpendicular) to the surfaces exerting them, frictional forces are parallel to the surfaces exerting them, etc.
     
  4. Feb 27, 2004 #3
  5. Feb 27, 2004 #4
    Often a vector is situated at the center of gravity when an object's gravitation or momentum is concerned, or located at the point of application of a force (as to describe a torque), or multiple vectors can describe a field (such as an electromagnetic field) over a space.

    An [horizontal] external force F on a block is balanced by a static frictional force S. As F is increased, S also increases, until S reaches a certain maximum value. The object then "breaks away," accelerating suddenly in the direction of F. If the block is now to move with constant velocity, F must be reduced [to 0] from the maximum value it had just before the block broke away.

    Halliday/Resnick/Walker, Fundamentals of Physics, p. 99, Wiley 6th ed., 2001

    [The vertical gravitational force and its reaction remain equal and opposite.]
     
  6. Feb 27, 2004 #5
    Excuse me.
     
  7. Feb 27, 2004 #6
    Your link at first glance seems fairly complete. What are you missing?
     
  8. Feb 27, 2004 #7
    The link works
     
  9. Feb 27, 2004 #8

    Doc Al

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    I'm not sure I understand your question. The normal force is always perpendicular to the surface. At the point B, the surface is horizontal (the slope of the tangent is zero) so the normal force acts vertically.

    Restate your question.

    (Oops: I read the wrong problem. Give me a minute.)
     
    Last edited: Feb 27, 2004
  10. Feb 27, 2004 #9
    Are we looking at the same question? Problem 13-80 is on page 5. I think you're looking at page 1 problem 13-81?
     
  11. Feb 27, 2004 #10
    Two equations, two unknowns.
     
  12. Feb 27, 2004 #11

    Doc Al

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    Yep. I was looking at the wrong problem.

    But now that I'm looking at the right problem, what's the problem?

    The normal force is always perpendicular to the surface (and thus the curve). So the angle of the normal force can be figured out (it's not a variable). So can the angles θ and α. The only tricky part is what to use for "r": find the radius of curvature at that point.
     
  13. Feb 27, 2004 #12
    I appreciate the help.

    I think I'm understanding it more:) But how do you find á?

    è = arctan(dy/dx)

    The radius of curvature is easy to figure out too. [1+ (dy/dx)^2]^3/2/d2y/dx2
     
  14. Feb 27, 2004 #13

    Doc Al

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    I don't understand your notation. What's "á" ?
    That's the angle of the tangent to the curve; the perpendicular is π/2 away. :smile:
    Yep. (OK, not so tricky.)
     
  15. Feb 27, 2004 #14
    oops, sorry.

    I copied and pasted it, but ¨¢ is the angle between the force of the spring and the n-axis. How do you find it?

    "theta = arctan(dy/dx)"

    "That's the angle of the tangent to the curve; the perpendicular is ¦Ð/2 away."

    So theta, from the equation above, would give me the angle between the t-axis and W. Right?

    Sorry to keep wasting your time.
     
  16. Feb 27, 2004 #15

    Doc Al

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    Here's how I'd find it. First find x & y coordinates of the point A. Then the angle the spring makes with the horizontal is arctan (y/x).
    Theta (as defined above) is the angle of the tangent to the curve with respect to the horizontal. So... yes, you are right.
    You're not wasting my time. I enjoy it. :smile:
     
  17. Feb 27, 2004 #16
    Thank-you sooooooooooooooooooooooo much:)
    I truly appreciate your assistance. You have been an extremely big help! I think I have enough information now to apply these concepts to other problems. Thanks for your help!!!!! I'll probably be back with more questions in the near future though...lol. Thank-you.
     
  18. Feb 27, 2004 #17

    Doc Al

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    Excellent. Glad I could help.
     
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