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?
3 answers
Will someone answer this please. Thank you.
No haha
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
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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