2a - F 1 sin 40° = 150 sin 40° = 96.4 F 2 F 2 cos 110° = 120 cos 110° =-41.0 F 2 sin 110° = 120 sin 110° = 112.8 F 3 F 3 cos 160° = 80 cos

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PHY2053 – Section 7392 Quiz #2 Friday, January 23, 2004 Name_______________________________ UFID___________________ Please read the problem carefully and be sure to show all of your work. Partial credit will be given. Three people pull on a large crate as shown in the figure below. Find the resultant force on the crate. x y F 1 = 150 N F 2 = 120 N F 3 = 80 N 70° 40° 20° SOLUTION: Resolve F 1 , F 2 , and F 3 into x and y components and add them up to find the resultant force: x y F 1 F 1 cos 40° = 150 cos 40° = +114.9
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Unformatted text preview: F 1 sin 40° = 150 sin 40° = +96.4 F 2 F 2 cos 110° = 120 cos 110° = -41.0 F 2 sin 110° = 120 sin 110° = + 112.8 F 3 F 3 cos 160° = 80 cos 160° = -75.2 F 3 sin 160° = 80 sin 160° = + 27.4 F R +114.9 – 41.0 -75.2 = -1.3 +96.4+112.8 + 27.4= 236.6 F R = - 1.3 x + 236.6 y | F R | = (236.6) 1.3) (-2 2 + = 236.6; Direction of F R: tan θ = 236.6/(-1.3) ! θ = -89.7° but clearly the resultant vector lies in the second quandrant, so must add 180° ! θ = -89.7° + 180° = 90.3°...
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This note was uploaded on 09/25/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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