A boy pulls a wagon with a force of 20N on a handle of .50 meters long. The end he holds is .25 meters higher than the end attached to the wagon. What is the magnitude of the horizontal component that acts to pull the wagon forward?

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  1. Will someone answer this please. Thank you.

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  2. No haha

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  3. Divide .25m/.50m for a total of .50
    This is sort of like cosθ=adj/hyp=.25/.50=.600

    Cos^-1 for .50 and your angle is 60°.

    After that do 20N·sin 60°
    This is hyp· sin θ

    Your answer being 17N

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