As shown a system of cables supports a bucket The dimensions

As shown, a system of cables supports a bucket. The dimensions in the figure are as follows: x1 = 5.45 ft , x2 = 1.20 ft , y1 = 1.90 ft , y2 = 4.50 ft , z1 = 2.85 ft , and z2 = 3.15 ft . If the bucket and its contents have a combined weight of W = 11.5 lb , determine TA, TB, and TC, the tensions in cable segments DA, DB, and DC, respectively.

Solution

wieght of the bucket W = -11.5 z^

we can write the position vectors of all the points A, B, C, D as

A^ = 5.45x^ +2.85z^

B^ = 1.2 x^

C^ = 4.5y^ +3.65 z^

D^ = 1.2 x^ + 1.9 y^

DA^ = A^-D^ = 4.33x^ -1.9y^ +2.85z^

DB^ = B^-D^ = -1.9y^

DC^ = C^-D^ = -1.2x^ +2.6y^ +3.15 z^

Tenssions in cables DA, DB and DC will be along the same directions as the above vectors, hence we can wrtie the tensions as

Ta^ = a (DA^) = 4.33ax^ -1.9ay^ +2.85az^

Tb^ = b(DB^) = -1.9b y^

Tc^ = c(DC^) = -1.2cx^ +2.6c y^ +3.15c z^

Ta^ + Tb^ + Tc^ + W = 0, as the system is inequilibrium

equating the x, y, z components we will have

2.85a +3.15 c = 11.5

4.33a -1.2c =0

-1.9a -1.9b -1.2c =0

solving the above we get

a = 0.81, b= 3.19, c= 2.92

Ta^ = 0.81(4.33x^ -1.9y^+2.85z^) = 3.5x^-1.54y^+2.31z^

Tb^ = 3.19(-1.9y^) = -6.06 y^

Tc^ = 2.92(-1.2x^ +2.6y^ +3.15z^) = -3.5x^ +7.59y^ +9.2 z^