Determine the force in cables DA DB and DC if the bucket wei

Determine the force in cables DA, DB, and DC if the bucket weighs 20 pounds.

Solution

DB = B - D = -1.5j

DBcap = - j

DA = A - D = (4.5, 0, 3) - (1.5, 1.5, 0) = 3i - 1.5j + 3k

DAcap = 3i - 1.5j + 3k / sqrt(3^2 + 1.5^2 + 3^2) = 0.667i - 0.333j + 0.667k

DC = C - D = (0, 2.5, 3) - (1.5, 1.5, 0) = -1.5i + 1j + 3k

DCcap = (-1.5i + j + 3k) / sqrt(1.5^2 + 1^2 + 3^2) = -(1.5/3.5)i + (1/3.5)j + (3/3.5)k

suppose magnitude of tension is DA is T1 and in DB is T2 and in DC is T3 then

Fnet = 0

T1 + T2 + T3 + W = 0

- T1j + 0.667T2i - 0.333T2j + 0.667T2k -(1.5T3/3.5)i + (T3/3.5)j + (3T3/3.5)k - 20k = 0

0.667T2 - 1.5T3/3.5 = 0

- T1 - 0.33T2 + T3/3.5 = 0

0.667T2 + 3T3 / 3.5 = 20

solving these

T1 = 10/9 = 1.11 lb

T2 = 10 lb

T3 = 140/9 = 15.56 lb