As shown, a mass is being lifted by a strut that is supported by two cables AC a

Question

As shown, a mass is being lifted by a strut that is supported by two cables AC and CD. The dimensions given are a=2.04 ft, b=1.7 ft, c=6.0 ft,d=1.0 ft, and e=2.0 ft.

A weight of 120 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. What is the weight of an unknown hanging mass when the compressive force in strut BC is 700 lb? By convention compressive forces are expressed as negative values and tensile forces are expressed as positive values. Find the maximum weight that the cable and strut system can support if the magnitude of the compressive force in the strut cannot exceed 1000 lb and the magnitude of tension in the cables cannot exceed 300 lb. As shown, a mass is being lifted by a strut that is supported by two cables AC and CD. The dimensions given are a=2.04 ft, b=1.7 ft, c=6.0 ft,d=1.0 ft, and e=2.0 ft.

Explanation / Answer

A(1, 0, 0)

B(0, 2.04, 0)

C(0, 2.04+1.7, 6) = (0, 3.74, 6)

D(-2, 0, 0)

Vector CA = (1, -3.74, -6)

Vector CB = (0, -1.7, -6)

Vector CD = (-2, -3.74, -6)

Magnitude CA = sqrt (1^2 + 3.74^2 + 6^2) = 7.14 ft

Magnitude CB = sqrt(1.7^2 + 6^2) = 6.236 ft

Magnitude CD = sqrt (2^2 + 3.74^2 + 6^2) = 7.348 ft

Unit Vector along CA = (1, -3.74, -6) / 7.14

Unit Vector along CB = (0, -1.7, -6) / 6.236

Unit Vector along CD = (-2, -3.74, -6) / 7.348

Forces equilibrium:

F_CA + F_CB + F_CD + W = 0

F_CA* (i -3.74 j - 6k) / 7.14 + F_CB*(-1.7 j - 6 k) / 6.236 + F_CD*(-2 i - 3.74 j - 6 k) / 7.348 + (-120 k) = 0

Thus,

F_CA/7.14 - 2/7.348*F_CD = 0

-3.74/7.14*F_CA - 1.7/6.236*F_CB - 3.74/7.348*F_CD = 0

(-6/7.14)*F_CA + (-6/6.236)*F_CB + (-6/7.348)*F_CD = 120

Solving them,

F_CA = 79.33 lb

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Forces equilibrium:

F_CA + F_CB + F_CD + W = 0

F_CA* (i -3.74 j - 6k) / 7.14 + (-700)*(-1.7 j - 6 k) / 6.236 + F_CD*(-2 i - 3.74 j - 6 k) / 7.348 + (-W k) = 0

Thus,

(1/7.14)*F_CA - (2/7.348)*F_CD = 0

(-3.74/7.14)*F_CA + (700*1.7/6.236) + (-3.74/7.348)*F_CD = 0

(-6/7.14)*F_CA + (700*6/6.236) + (-6/7.348)*F_CD = W

Solving them, W = 367.37 lb

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Forces equilibrium:

F_CA + F_CB + F_CD + W = 0

F_CA* (i -3.74 j - 6k) / 7.14 + F_CB*(-1.7 j - 6 k) / 6.236 + F_CD*(-2 i - 3.74 j - 6 k) / 7.348 + (-W k) = 0

Thus,

F_CA/7.14 - 2/7.348*F_CD = 0

-3.74/7.14*F_CA - 1.7/6.236*F_CB - 3.74/7.348*F_CD = 0

(-6/7.14)*F_CA + (-6/6.236)*F_CB + (-6/7.348)*F_CD = W

Solving them,

F_CD = 0.5145*F_CA

F_CB = -2.882*F_CA

F_CA = 0.6611*W

Therefore,

F_CA = 0.6611*W

F_CB = -1.905*W

F_CD = 0.34*W

Thus, for given W, highest force is in CB.

If F_CB = -1000 lb, we get W = 524.9 lb, F_CA = 347 lb, F_CD = 178.5 lb

We notice that F_CA > 300 lb limit.

Thus we get that F_CA = 300 lb.

Therefore, W = 453.8 lb

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