# 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?

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.

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