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**1. Homework Statement**

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.

**2. 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]

**3. 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.

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