Force F has magnitude 21.0 kN. It acts at point A and is directed towards point
ID: 1869927 • Letter: F
Question
Force F has magnitude 21.0 kN. It acts at point A and is directed towards point C, shown in the figure A (2, 8, 7) m C (4, -4, 4) m B(6, 6, 0) m Express force F as a Cartesian vector in kilonewtons kN Find the angle in degrees between force F and a line passing through points A and B. (Enter your answer to at least one decimal place.) Find the angle in degrees between force F and a line passing through points B and C. (Enter your answer to at least one decimal place.) Finally, find the magnitudes in kilonewtons of the projections of force F along a line passing through points A and B and a line passing through points B and C. kN AB = kN BC=Explanation / Answer
vector Ac = OC - OA = (4 - 2)i + (-4 - 8)j + (4 - 7) k
vector Ac = 2i - 12j - 3k
unit vector along AC = (2i - 12 j - 3k)/sqrt(2^2+12^2+3^2) = (-2i - 12 j - 3k)/12.5
F = 21* (-2i - 12 j - 3k)/12.5 kN
F = 1.68*(2i - 12 j - 3k) kN
F = 3.36 i - 20.16 j - 5.04 k kN <<<<----ANSWER
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vector AB = oB - OA = (6-2)i + (6-8)j + (0-7)k
AB = 4i - 2j - 7k
angle between F and AB
costheta1 = AB.AC/(lABl*lACl)
magnitude of AC = 12.5
magnitude of AB = sqrt(4^2+2^2+7^2) = 8.31
costheta1 = (4i - 2j - 7k).(-2i - 12 j - 3k)/(8.31*12.5)
costheta1 = (-8 + 24 + 21)/(8.31*12.5)
theta1 = 69.1 <<<-------------ANSWER
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vector BC = oC - OB = (4-6)i + (-4-6)j + (4-0)k
BC = -2i - 10j + 4k
angle between F and AB
costheta2 = BC.AC/(lABl*lACl)
magnitude of AC = 12.5
magnitude of AB = sqrt(2^2+10^2+4^2) = 11
costheta2 = (-2i - 10j + 4k).(-2i - 12 j - 3k)/(11*12.5)
costheta2 = (4 + 120 - 12)/(11*12.5)
theta2 = 35.4 <<<<----ANSWER
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FAB = 21*costheta1 = 21*cos69.1 = 7.49 kN <<<<----ANSWER
FBC = 21*costhet2 = 21*cos35.45 = 17.1 kN <<<<----ANSWER
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