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:
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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}.
(Solve for F_{BA})
F_{BA}\,=\,3924 N
Now we will write an equation of equilibrium for x-axis forces.
(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