Interesting question regarding simple decomposing of vectors

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Homework Statement


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Find the x-component of the ##F_b## (parallell with ##\vec{s}## and the ground) and find the resultant total Force.
Then calculate the total work done by both over a distance of 15 m.

Homework Equations


[itex] F_{x} = |F| \cdot \cos v [/itex]
[itex] F_{y} = |F| \cdot \sin v [/itex]
[itex] W = \vec{F} \cdot \vec{s} [/itex]
[itex] W = |F| \cdot |s| \cdot \cos{v} [/itex]

The Attempt at a Solution


I am a science teacher myself, and a student recently asked me an interesting question I hadn't thought of before.

The math here isn't all that hard, and the x and y components of ##F_b## are
##F_{bx} = 500 \cdot \cos{30 } = 500N \cdot \frac{\sqrt{3}}{2} = 433.01N## (Thanks, voko)
##F_{by} = 500 \cdot \sin{30 } = 500N \cdot \frac{1}{2} = 250 N ##

Now, the question which confuses me is how and why these 2 forces doesn't match up to be equal to 500. From a purely mathematical point of view, I know that they are added via pythagoras and that the math is correct, so that isn't my issue.

Since this is actual forces we are talking about, how is the sum of the forces he is transferring to the object NOT equal to the force he is dragging on the rope with?

Again, let me stress that I know the math of how to use pythagoras to add them, but from an intuitive point of view, my student was unable to understand this, and I was unable to explain it intuively.
 
Last edited:

Answers and Replies

  • #2
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First of all, ## \cos 30^\circ = {\sqrt 3 \over 2 } ##.

Second, your question does not make a lot of sense to me. The sum of the forces (which I assume is ##\vec F_{bx} + \vec F_{by} ##) is most certainly equal to the force applied to the rope, which you know "from a purely mathematical point a view". So what is your real question?
 

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