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2074763332 - .707(734.958 N 300 = T => 819.615 N = T...

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MECH 234- Section ______ Name ______________________________ MECH 235- Section ______ Quiz 2. An applied force of 600 N is acting on point A. Determine the tension in cable AC and the magnitude of the resultant force directed along the axis of member AB. Solve using an analytical method. Define each vector: x-comp. y-comp. Vector R = - R cos 45 ° î - R sin 45 ° ĵ = - .707 R î - .707 R ĵ Vector T = - T î = - T î Vector F = 600 sin 30 ° î - 600 cos 30 ° ĵ = 300 î - 519.615 ĵ Group like terms from the basic equation: R = T + F î terms: - .707 R = - T + 300 eq. (1) ĵ terms: - .707 R = 0 - 519.615 eq. (2) You can solve for R from eq. (2) R = 519.615 = 734.958 N = R .707 Back substitute in eq. (1) to get T:
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Unformatted text preview: -.707 (734.958 N) - 300 = - T => 819.615 N = T ____________________ Alternate math approach would be to simply subtract the two equations to solve since the coefficients for the R terms are the same. î terms:- .707 R =- T + 300 eq. (1) ĵ terms: -- .707 R =- 519.615 eq. (2) T = 819.615 Back substitute in eq. (1) to solve for R:- .707 R =- (819.615) + 300 R =- 519.615 / - .707 = 734.958 = R Quiz2-sp2006.doc 45 ° 30 ° A C B 600 N T pulls at A R Assumed direction F Begin with basics: R = ∑ F = T + F Define each vector in rectangular coordinates. Group like terms. 45 °...
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