Application of Vectors: Work (calc 3)

  • Thread starter ek530n
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  • #1
ek530n
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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
 
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Answers and Replies

  • #2
dextercioby
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What ideas do you have...? I bet the problem is not that difficult.

Daniel.
 
  • #3
ek530n
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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.
 
  • #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.
 
  • #5
Jameson
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I got the same answer. It looks fine to me. Did you put in the correct units?
 
  • #6
Scirel
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You forgot about Vectors.

Force is a vector.

It has X and Y components.

Also remember about Mass x Gravity.
 
  • #7
ek530n
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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.
 
  • #8
Scirel
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ek530n said:
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.


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:
 
  • #9
ek530n
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Scirel said:
Eh? That`s odd. I`ve never heard of a problem neglecting gravity before.

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.
 
  • #10
Scirel
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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)
 
  • #11
lurflurf
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ek530n said:
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.
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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