Learning Goal To determine Cartesian force components from m

Learning Goal: To determine Cartesian force components from magnitude and determine position and resultant vectors by summing Cartesian components. As shown, a 3.1-lb ball is suspended at point D inside a box with dimensions w = 9.30 ft, d = 6.70 ft, and h = 4.10 ft. The ball is held in place by three cables anchored at points A, B, and C on the surface of the box. : s If point D is the origin of the Cartesian coordinate system, point A is located at (-4.30,-3.50, 2.40) ft, point B is located at (2.40,-3.50, 1.80) ft, and point C is located at (2.40, 5.80,-1.70) ft.

Solution

DA = A - D = (-4.3, -3.5, 2.40) - (0,0,0) = -4.3i - 3.5j + 2.40k

this is the vector in same direction as Tda is.

so DAcap = DA/ |DA| = (-4.3i - 3.5j + 2.40k ) / sqrt(4.3^2 + 3.5^2 + 2.40^2) = (-4.3i - 3.5j + 2.40k ) / 6.04

suppose magnitdue of Tda is 9.62 lb then

vector Tda =|Tda| DAcap = - 6.85i - 5.57j + 3.82k

similarly,

DB = B - D = 2.40i - 2.50j + 1.80k

DBcap = (2.40i - 2.50j + 1.80k ) / sqrt(2.40^2 + 2.50^2 + 1.80^2)

= (2.40i - 2.50j + 1.80k ) / 3.91

Tdb = |Tdb| DBcap = 3.34i - 3.48j + 2.51k lb

DC = C - D = 2.40i + 5.80j - 1.70k

DCcap = (2.40i + 5.80j - 1.70k ) / 6.50 = 0.369i + 0.892j - 0.261k

suppose magnitude of Tdc is T then

Tdc = T ( 0.369i + 0.892j - 0.261k)

and weight force, W = - 3.1k lb

D is in equilibrium so net force on it will be zero.

Fnet = Tda + Tdb + Tdc + W = 0

- 6.85i - 5.57j + 3.82k + 3.34i - 3.48j + 2.51k lb + Tdc - 3.1k =0

Tdc = 3.51i + 9.05j - 3.23k

magnitude = sqrt(3.51^2 + 9.05^2 + 3.23^2) = 10.23 lb ..........Ans