# Determine the force in each cord for equilibrium

Determine the force in each cord for equilibrium of the 200-kg crate. Cord BC remains horizontal due to the roller at C, and AB has a length of 1.5 m. Set y = 0.75 m.

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

#### Solution:

Let us draw a free body diagram around ring B.

The angle was found using the inverse of sine (arcsin)

angle = $\text{sin}^{-1}\left(\dfrac{0.75}{1.5}\right)\,=\,30^0$

Let us write our equations of equilibrium. We will start off by writing an equation of equilibrium for the y-axis forces, as this will give us a direct solution to $F_{BA}$.

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

$F_{BA}\text{sin}\,(30^0)\,-\,1962\,=\,0$

(Solve for $F_{BA}$)

$F_{BA}\,=\,3924$ N

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

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

$F_{BA}\text{cos}\,(30^0)\,-\,F_{BC}\,=\,0$

(Substitute the value of $F_{BA}$ we found)

$3924\text{cos}\,(30^0)\,-\,F_{BC}\,=\,0$

(Solve for $F_{BC}$)

$F_{BC}\,=\,3398$ N

Show me the free body diagram

$F_{BA}\,=\,3924$ N
$F_{BC}\,=\,3398$ N