tProblemID 6187005 Force Acting along a Line Part A - Determining components of
ID: 1636901 • Letter: T
Question
tProblemID 6187005 Force Acting along a Line Part A - Determining components of The tension in cable DA has a magnitude of rba directed from D to A 7.40 lb. Find the Cartesian components of tension TDA, which is Learning Goal: To magnitude and determine position and resultant Express your answers, separated by commas, to three significant figures Hints As shown, a 2.9-lb ball is suspended at point D w = 9.30 ft. d-6.40 ft, and h-4.40 ft. The ball is held in place by three cables anchored at points A, B, and C on the surface of the box TDA5.7,-4.6,3.3 Give Up Incorrect; Try Again Part B-Determining an unknown force the tensions in cables DA and DB areTpA = 7.40 lb and TDB-521 lb, respectively. what is the tension in cable DC? If point D is the origin of the Cartesian coordinate system, point A is located at (-4.40,-3.50, 2.70) ft, point B is located at (2.00,-3.50, 1.80) ft, and point C is located at (2.00,5.80, -1.70) frt 32017Explanation / Answer
given
coordinates of A = (-4.4,-3.5, 2.7)
coordinates of B = (2,-3.5,1.8)
coordinates of C = (2,5.8,-1.7)
coordinates of D = (0,0,0)
as tension is always pulling away from D,
tension along DA = T1
along DB = T2
ALong DC = T3
now, unit vector in DA direction = (4.4i + 3.5j - 2.7k)/sqroot(4.4^2 + 3.5^2 + 2.7^2) = 0.705i + 0.5611j - 0.432k [ where i , j and k are unit vectors along x y and z axis)]
unit vector in DB direction = (-2i + 3.5j - 1.8k)/sqroot(4 + 3.5^2 + 1.8^2) = -0.45i + 0.792j - 0.4077k
unit vector in DC direction = (-2i - 5.8j + 1.7k)/sqroot(4 + 5.8^2 + 1.7^2) = -0.31i - 0.911j + 0.267k
a. given Tda = 7.4 lb
castesian components of TDA = 7.4*(0.705i + 0.5611j - 0.432k) = (5.217, 4.15214, -3.1968)
b. Tda = 7.4 lb
Tdb = 5.21 lb
Tdc = ?
so, from force balance
7.4(0.705i + 0.5611j - 0.432k) + 5.21(-0.45i + 0.792j - 0.4077k) + Tdc(-0.31i - 0.911j + 0.267k) - 2.9k = 0
Tdc = 9.266 lb
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