It includes cosθ because work can only be from force parallel to the distance. So, if you push a box up a hill by only using a horizontal force it's not only inefficient, but you haven't done a ton of work (force applied * horizontal distance from base of hill to crest of hill).
I think that's only the case if the object is being push in a sloping environment. What if I want to push it just horizontally? Do I use the same equation or use the alternative which is ∫ f • ds? (Integral of f times ds)
A little side note, but what does the d mean in ds? I also study some classical mechanics in school and we use f • s(Physics, Chem and stuff all in Science, but sadly now I'm in plant biology. And I hate plant biology. I want human biology). Does using a slightly different equation matter?
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(This post was last modified: 12-26-2014, 05:26 PM by PutraenusAlivius.)
(12-26-2014, 04:23 PM)Julius Caesar Wrote: I think that's only the case if the object is being push in a sloping environment. What if I want to push it just horizontally? Do I use the same equation or use the alternative which is ∫ f • ds? (Integral of f times ds)
A little side note, but what does the d mean in ds? I also study some classical mechanics in school and we use f • s(Physics, Chem and stuff all in Science, but sadly now I'm in plant biology. And I hate plant biology. I want human biology). Does using a slightly different equation matter?
cosθ is still there, but θ = 0 so you're just multiplying by 1
As for ∫fds, that just means the integral of the function f with respect to s
(I'm assuming f is the function for force)
For example, ∫(1/s)ds = ln(s)
∫sds = 0.5(s^2)
∫xsin(s)ds = xcos(s), note that x does not change because that's not the variable we care about when using ∫ds
Your version (work = force*distance*cosθ) is a bit of a simplified version where the force applied is a constant rather than a function.
(This post was last modified: 12-27-2014, 02:49 AM by CarnivorousJelly.)