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?

#### Solution:

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

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

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

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

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

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}.

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.

(solve for F_{CA})

F_{CA}=42.6 N

Since,

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