Learning Goal: Part A Finding the tension in cable AC To apply the condition of
ID: 1873494 • Letter: L
Question
Learning Goal: Part A Finding the tension in cable AC To apply the condition of equilibrium to three-dimensional systems and solve for unknown forces A weight of 190 lb acts at C on the strut. Find the magnitude of the tension in cable AC Express your answer to three significant figures and include the appropriate units. As shown, a mass is being lifted by a strut that is supported by two cables AC an given are a = 7.10 ft, b 4.30 ft, c = 9.50 ft. d 3.30 ft, and e-5.70 ft. d CD. The dimensions View Available Hint(s) Figure 1) Tac Value Units gure 1 of 1 Submit Part B - Finding the unknown weight What is the weight of an unknown hanging mass when the compressive force in strut BC is 525 lb? Express your answer to three significant figures and include the appropriate units. View Available Hint(s)Explanation / Answer
let the magnitude of tension in AC be T1
tension in CD be T2
normal reaction in strut be T
hence from force balance at point C
T1(di - (a + b)j - ck) + T2(-ei - (a + b)j - ck) + N(bj + ck)/sqroot(b^2 + c^2) - mg*k = 0
now,
a = 7.1
b = 4.3
c = 9.5
d = 3.3
e = 5.7
hence
T1(3.3i - 11.4j - 9.5k) + T2(-5.7i - 11.4j - 9.5k) + N(4.3j + 9.5k)/10.427 - mgk = 0
from moment balance
(11.4j + 9.5k) x T1(3.3i - 11.4j - 9.5k) + (11.4j + 9.5k) x T2(-5.7i - 11.4j - 9.5k) + (11.4j + 9.5k) x (-mg*k) = 0
T1(-37.62k 31.35j) + T2(64.98k - 54.15j ) + mg(-11.4i) = 0
comparing
mg = 0
but
mg = (Mg + W)
W = -190 lb
hence
unknown weight = 190 lb
and
3.3T1 -5.7T2 = 0
-11.4T1 -11.4T2 + 4.3N = 0
-9.5T1 - 9.5T2 + 9.5N = 0
T1 + T2 = N
-11.4(T1 + T2) + 4.3(T1 + T2) = 0
hence T1 = -T2
9T1 = 0
T1 = 0 N
T2 = 0 N
extra weigth = 190 lb
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