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.

#### Solution:

Show me the final answerâ†“

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

#### Final Answers:

F_{BA}\,=\,3924 N

F_{BC}\,=\,3398 N