Question

A uniform steel ring 60 in. in diameter and weigh-
ing 600 lb is lifted by the three cables, each 50 in.
long, attached at points A, B, and C as shown. Com-
pute the tension in each cable.
​

Answer 1

Explanation:

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Answer 2

Answer:

The tension in each cable is

$|T _{AD}| = 274.5lb \\ |T _{BD}| = 158.5lb \\ |T _{CD}| = 317lb$

Explanation:

The free body diagram of the steel ring is given in the attachment.

Given that,

Length of cable=50inches

Diameter of steel ring=60inches

=>Radius of steel ring=30inches

From Pythagorous Theorem,we have

$OD = \sqrt{50 {}^{2} - 30 {}^{2} } = 40in$

The coordinates of A,B,C&D are given as follows

$A = (30,0,0) \\ B = (0,30,0) \\ C = ( - 30cos30, - 30sin30,0) \\ D = (0,0,40)$

Position vectors can be found as follows

$AD=OD-OA \\ = (0i + 0j + 40k) - (30i + 0j + 0k) \\ = - 30i + 40k \\ unit \: vector \: AD {}^{ 1} = \frac{ - 30i + 40k}{ \sqrt{30 {}^{2} + 40 {}^{2} } } \\ = - 0.6i + 0.8k$

We can find tension vector as follows

$T _{AD} {}^{ } = |T _{AD}| ( - 0.6i + 0.8k) \\ - > (1)$

Similarly unit vector BD,CD and their tensions can be found as follows

$BD = OD - OB\\ = > BD {}^{1} = - 0.6j + 0.8k \\ T _{BD} = |T _{BD}| (- 0.6j + 0.8k ) \\ \\ CD = OD - OC \\ = > CD {}^{1} = 0.5196i + 0.3j + 0.8k \\ T _{CD} = |T _{CD}| (0.5196i + 0.3j + 0.8k)$

Weight vector is given as

$W = - 600k$

For equilibrium,sum of tension in all cables and weight of the steel ring is equal to zero

$= > |T _{AD}| ( - 0.6i + 0.8k) +|T _{BD}| (- 0.6j + 0.8k ) +|T _{CD}| (0.5196i + 0.3j + 0.8k)-600k = 0$

Equating i,j,k coefficients we get

$- 0.6|T _{AD}| + 0.5196|T _{CD}| = 0 - > (2) \\ - 0.6|T _{BD}| + 0.3|T _{CD}| = 0 - > (3) \\0.8 |T _{AD}| + 0.8|T _{BD}| + 0.8|T _{CD}| = 600 - > (4)$

Solving the above equations we get

$|T _{AD}| = 274.5lb \\ |T _{BD}| = 158.5lb \\ |T _{CD}| = 317lb$

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