Unformatted text preview: is along the y axis, thus it has y component only. F 1 = 50.0 N j b) Find the second force in unitvector notation. We apply the Newton’s 2 nd law to the block and solve it for F 2 . We get m a = F 1 + F 2 ⇒ F 2 = m aF 1 = 4 × (6.92 i  4.00 j ) 50.0 j = 27.7 N i66.0 N j c) What are the magnitude and direction (relative to the + x axis) of the second force? Using Pythagorean Theorem, the magnitude is F 2 = √(F 1x 2 + F 1y 2 ) = √((27.7) 2 + (66) 2 ) = 71.6 N The angle is given by the inverse tangent function. Since F 2 is in the 3 rd quadrant, we add 180º to the answer your calculator displays. θ 2 = tan1 (F 1y /F 1x ) +180º = 247º...
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 Fall '08
 Field
 Physics, Pythagorean Theorem, Work, Inverse function, Euclidean geometry, Inverse trigonometric functions, unitvector notation

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