# Determine the force in each cable and the force F

Determine the force in each cable and the force F needed to hold the 4-kg lamp in the position shown. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will analyze point B and C separately. First, we will draw a free body diagram for point B and write our equations of equilibrium. Let us now write the equations for equilibrium for the x-axis forces and y-axis forces. Forces going right, $\rightarrow^+$ will be considered positive and forces going up,$\uparrow+$ will be considered positive.

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

$T_{BC}\text{cos}\,(30^0)\,-\,T_{BA}\text{cos}\,(60^0)=0$ —————-(eq.1)

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

$T_{BA}\text{sin}\,(60^0)\,-\,T_{BC}\text{sin}\,(30^0)\,-\,39.24=0$ —————-(eq.2)

We will now solve for $T_{BA}$ and $T_{BC}$.

Isolate for $T_{BC}$ in eq.1.

$T_{BC}=\frac{T_{BA}\text{cos}\,(60^0)}{\text{cos}\,(30^0)}$

(simplify)

$T_{BC}=0.577T_{BA}$ —————-(eq.3)

Substitute this value into eq.2.

$T_{BA}\text{sin}\,(60^0)\,-\,0.577T_{BA}\text{sin}\,(30^0)\,-\,39.24=0$

(solve for $T_{BA}$)

$T_{BA}=67.9$ N

We can use this value to figure out $T_{BC}$ by substituting it back into eq.3.

$T_{BC}=(0.577)(67.9)\,=\,39.2$ N

Now we will draw a free body diagram for point C. As before, we will write our equations of equilibrium for x-axis forces and y-axis forces. Considerations for positive forces will be the same as before.

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

$T_{CD}\text{cos}\,(30^0)\,-\,T_{BC}\text{cos}\,(30^0)=0$ —————-(eq.4)

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

$T_{CD}\text{sin}\,(30^0)\,+\,T_{BC}\text{sin}\,(30^0)\,-\,F=0$ —————-(eq.5)

Remember that we now know that $T_{BC}=39.2$ N.

Isolate for $T_{CD}$ and substitute the value of $T_{BC}$ in eq.4.

$T_{CD}=\frac{39.2\text{cos}\,(30^0)}{\text{cos}\,(30^0)}$

(Solve for $T_{CD}$)

$T_{CD}=39.2$ N

Substitute the value of $T_{CD}$ we found and the value of $T_{BC}$ into eq.5.

$39.2\text{sin}\,(30^0)\,+\,39.2\text{sin}\,(30^0)\,-\,F=0$

(Solve for F)

$F=39.2$ N