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How do you resolve a horizontal force of 40 newtons into two perpendicular directions one of them inclines by 30 degree with the horizontal and upwards?
Wiki User
∙ 10y ago
-- The component that's inclined 30 degrees above the horizontal is
[ 20 sqrt(3) ] = about 34.641 newtons. (rounded)
-- The other component is inclined 60 degrees below the horizontal,
and its magnitude is 20 newtons.
Wiki User
∙ 10y ago
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