Determine the tension developed in each cord 8


Determine the tension developed in each cord required for equilibrium of the 20-kg lamp.

Determine the tension developed in each cord

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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We will start off by drawing a free body diagram focusing on ring D.

Determine the tension developed in each cord solution

Let us write an equation of equilibrium for the y-axis forces.

+\uparrow \sum \text{F}_\text{y}\,=\,0

F_{DE}\text{sin}\,(30^0)\,-\,196.2\,=\,0

(Solve for F_{DE})

F_{DE}\,=\,392.4 N

 

Now, we will write an equation of equilibrium for x-axis forces.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

F_{DE}\text{cos}\,(30^0)\,-\,F_{DC}\,=\,0

(Substitute the value of F_{DE} we found)

392.4\text{cos}\,(30^0)\,-\,F_{DC}\,=\,0

(solve for F_{DC})

F_{DC}\,=\,339.8 N

 

We can now draw a free body diagram focusing on ring C.

Determine the tension developed in each cord

 

Again, we will write our equations of equilibrium.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

339.8\,-\,F_{CA}\dfrac{3}{5}\,-\,F_{CB}\text{cos}\,(45^0)\,=\,0 (eq.1)

(Remember that we already found F_{DC}\,=\,339.8 N)

 

+\uparrow \sum \text{F}_\text{y}\,=\,0

F_{CA}\dfrac{4}{5}\,-\,F_{CB}\text{sin}\,(45^0)\,=\,0 (eq.2)

 

Let us solve for F_{CA} and F_{CB}

Isolate for F_{CA} in eq.2.

F_{CA}\,=\,\dfrac{F_{CB}\text{sin}\,(45^0)}{\dfrac{4}{5}}

(Simplify)

F_{CA}\,=\,0.88F_{CB} (eq.3)

 

Substitute this value into eq.1.

339.8\,-\,0.88F_{CB}\dfrac{3}{5}\,-\,F_{CB}\text{cos}\,(45^0)\,=\,0

(solve for F_{CB})

F_{CB}\,=\,275.1 N

 

Substitute this value back into eq.3 to figure out F_{CA}.

F_{CA}\,=\,(0.88)(275.1)

F_{CA}\,=\,242.1 N

 

Final Answers:

F_{DE}\,=\,392.4 N

F_{DC}\,=\,339.8 N

F_{CB}\,=\,275.1 N

F_{CA}\,=\,242.1 N

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-28.

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