If cable CB is subjected to a tension


If cable CB is subjected to a tension that is twice that of cable CA, determine the angle ϴ for equilibrium of the 10-kg cylinder. Also, what are the tensions in wires CA and CB?

If cable CB is subjected to a tension

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

Solution:

Let us first draw a free body diagram showing the forces.

If cable CB is subjected to a tension

We can now write our equations of equilibrium. We will choose forces going \rightarrow^+ to be positive and \uparrow+ to be positive.

\sum \text{F}_\text{x}=0

F_{CB}\text{cos}\,(\theta)\,-\,F_{CA}\text{cos}\,(30^0)=0 —————–(eq.1)

\sum \text{F}_\text{y}=0

F_{CB}\text{sin}\,(\theta)\,+\,F_{CA}\text{sin}\,(30^0)\,-\,98.1=0 —————–(eq.2)

 

Note that the question states “cable CB is subjected to a tension that is twice that of cable CA” which we can write mathematically as:

F_{CB}=2F_{CA}

 

Thus, we can replace all F_{CB} values in eq.1 and eq.2 with 2F_{CA}.

 

2F_{CA}\text{cos}\,(\theta)\,-\,F_{CA}\text{cos}\,(30^0)=0 —————–(eq.3)

2F_{CA}\text{sin}\,(\theta)\,+\,F_{CA}\text{sin}\,(30^0)\,-\,98.1=0 —————–(eq.4)

 

We can now solve for θ and F_{CA}.

 

Factor out F_{CA} from eq.3.

F_{CA}(2\text{cos}\,(\theta)\,-\,\text{cos}\,(30^0))=0

(divide both sides of the equation by F_{CA})

2\text{cos}\,(\theta)\,-\,\text{cos}\,(30^0)=0

(isolate for \text{cos}\,(\theta))

\text{cos}\,(\theta)=\frac{\text{cos}\,(30^0)}{2}

(solve for \theta)

\theta\,=\,\text{cos}^{-1}\frac{\text{cos}\,(30^0)}{2}
 
\theta\,=\,64.3^0

 

Substitute the value of \theta we found into eq.4.

2F_{CA}\text{sin}\,(64.3^0)\,+\,F_{CA}\text{sin}\,(30^0)\,-\,98.1=0

(solve for F_{CA})

F_{CA}=42.6 N

 

Since,

F_{CB}=2F_{CA}

F_{CB}=2(42.6)=85.2 N

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-11.

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