# Application of Vectors: Work (calc 3)

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

Daniel.

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.

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

Daniel.

I got the same answer. It looks fine to me. Did you put in the correct units?

Force is a vector.

It has X and Y components.

Also remember about Mass x Gravity.

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.

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? Thats odd. Ive never heard of a problem neglecting gravity before.

also isnt (dot) multiplication.

NOTE: to a helper who is reading this: I really need help with my topci before 2 hours, please! Scirel said:
Eh? Thats odd. Ive 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.

Wow! Take it easy! I wasnt 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)

lurflurf
Homework Helper
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).