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There are two forces on the 2.61 kg box in the overhead view of the figure but only one is shown. For F1 = 13.2 N, a = 14.4 m/s2, and θ = 28.1°, find the second force (a) in unitvector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)
 one year ago
 one year ago
There are two forces on the 2.61 kg box in the overhead view of the figure but only one is shown. For F1 = 13.2 N, a = 14.4 m/s2, and θ = 28.1°, find the second force (a) in unitvector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)
 one year ago
 one year ago

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\[a \sin \theta = ycomponent\]\[a \cos \theta = xcomponent\] \[theta\] from the +x axis just add 180 to \[theta\] so you get 28.1+180=208.1 so you have the direction, you already know the magnitude and you know the magnitude already as it is given. The first part is a little right triangle trigonometry so make sure you are up on that.
 one year ago
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