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Homework Help: Application of Vectors: Work (calc 3)

  1. Jul 5, 2005 #1
    Hey all,

    I'm a bit stuck on a problem on my online homework for my calc 3 class, hopefully someone can help me out.

    Suppose that you push with a horizontal force on a box, to push it up a horizontal ramp, as shown in https://instruct2.math.lsa.umich.edu/webwork2_course_files/ma215u05/tmp/gif/2-prob3-pimages/sfig13-3-3.gif [Broken]

    If your force is F=22 lbs., the ramp angle a(alpha)=17 degrees above the horizontal and you push the object a distance L=6 feet how much work is done on the box?

    Thanks
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 5, 2005 #2

    dextercioby

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    What ideas do you have...? I bet the problem is not that difficult.

    Daniel.
     
  4. Jul 5, 2005 #3
    Well I've tried the plug n' chug method with the numbers give:

    W=F(dot)D=|F||D|cos a
    W=(22 lbs.)(6 ft.)(cos 17)
    W=126.232 ~ 126

    But this answer is incorrect, unless the answer registered in the system is incorrect.
     
  5. Jul 5, 2005 #4

    dextercioby

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    I don't know about your weird American system of units, but the formula is the right one.

    Daniel.
     
  6. Jul 5, 2005 #5
    I got the same answer. It looks fine to me. Did you put in the correct units?
     
  7. Jul 5, 2005 #6
    You forgot about Vectors.

    Force is a vector.

    It has X and Y components.

    Also remember about Mass x Gravity.
     
  8. Jul 5, 2005 #7
    Yes but the formula W=|F||D|cos a can be used instead of W=F(dot)D in order to avoid vectors and instead use magnitudes only.

    Also gravity is neglected in this particular problem and units are not required.
     
  9. Jul 5, 2005 #8

    Eh? That`s odd. I`ve never heard of a problem neglecting gravity before.

    also isn`t (dot) multiplication.

    NOTE: to a helper who is reading this: I really need help with my topci before 2 hours, please! :confused:
     
  10. Jul 5, 2005 #9
    Is that an attempt to mock me for my choice of words? If you don't have anything relevant to say don't post under this thread.
     
  11. Jul 5, 2005 #10
    Wow! Take it easy! I wasn`t trying to make fun of you. Geez..

    I literally have never seen a problem like that before. How can that be taken as an insult?

    anyway, there is one more thing to try.

    Are you sure your calculator is in degrees/radians mode? (for whatever the problem calls for)
     
  12. Jul 5, 2005 #11

    lurflurf

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    You are given the x-component of the force
    namely
    22 lbs.=(Force)(cos(17 degrees))
    You want work
    work=force*distance
    In other words your cos(17 degrees) should have be sec(17 degrees) or 1/cos(17 degrees).
     
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