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Variable Force

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data
    The figure shows a cord attached to a 2 kg cart that starts from rest and can slide along a frictionless horizontal rail. The right end of the cord is pulled over a frictionless pulley at height h = 2.0 m above the point of attachment of the cord to the mass, so the cart slides from x1 = 4.0 m to x2 = 1.0 m where x is measured positively to the left from a line that passes vertically through the center of the pulley. During the move, the tension T, in the cord is a constant 49 N. What is the speed of the cart (in m/s to two decimal places) when it reaches x2 ?

    Image: http://drbensonphysics.org/file.php/3/AP_images/variable_force_constant_tension_.jpg [Broken]

    2. Relevant equations

    F=ma
    v^2=2aΔx

    3. The attempt at a solution

    I understand that the force on the object varies due to the changing angle of the applied force, and that the relevant component of the force is equal to the cosine of the angle. I found this to be Fcos(arctan(2/x2-x1-D)) where D is the distance the mass has traveled at any given point in time. I don't have any idea where to go from here because it seems that I must keep substituting the same equation into itself if I use a = F/m. Thanks for any help.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 24, 2015 #2

    gneill

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    Staff: Mentor

    Hi jvalencia, Welcome to Physics Forums.

    Unfortunately the image at the url you've given is behind a username/password login. Can you upload the image here instead? (use the UPLOAD button)
     
  4. Nov 24, 2015 #3
    Thank you for telling me that. I shall figure out how to do so shortly
     
  5. Nov 24, 2015 #4
    Here it is

    variable_force_constant_tension_.jpg

    (Mentor's note: edited post to expand thumbnail image)
     
    Last edited by a moderator: Nov 24, 2015
  6. Nov 24, 2015 #5

    gneill

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    Staff: Mentor

    You can either integrate the horizontal component of the force over the distance that the cart moves, or perhaps do something clever...

    Notice that the horizontal part of the rope after the pulley also has a force pulling it to the right. That force is constant and always horizontal. Does that suggest anything to you?
     
  7. Nov 24, 2015 #6
    My understanding of calculus is pretty shaky; I'm only in my first semester. Would this be the way to set up that integral?
    $$ \int F \cos\theta dx $$
    $$ \theta = \arctan\frac{2}{x_1 - x_2 - x} $$

    Also, I don't really see what you are hinting at with that last bit. I know that the tension force is always constant but the angle varies. Consequently, the component of the force doing work on the object changes, assuming gravity exists and holds the cart to the rail.
     
  8. Nov 24, 2015 #7

    gneill

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    Staff: Mentor

    Your integral outline is okay. The angle varies of course, and you should be able to use the basic trig function definitions (ratios side lengths of the triangle) to write an expression for ##cos( \theta )## in terms of x.

    As to what I was hinting at, note that by conservation of energy, the energy that's put into the cart must be the same as the work that's done pulling on the rope. The horizontal part of the rope moves a certain distance, say from x'1 to x'2, while the cart moves from x1 to x2. But no angles are involved here: it's a straight line motion, and the force is constant... can you determine how far the rope moves?

    Fig1.png
     
  9. Nov 24, 2015 #8
    That is vastly more intuitive!!! Thank you.

    Any point in the rope moves a distance equal to the difference in the hypotenuses of triangles x1 and x2. Therefore
    ## \sqrt\frac {2F(A-B)} {m} = v ##
    Where A and B are the hypotenuses of the aforementioned triangles.
     
  10. Nov 24, 2015 #9

    gneill

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    Staff: Mentor

    Looks good :smile:
     
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