If the 1.5 m long cord AB can withstand a maximum force of 3500 N, determine the force in cord BC and the distance y so that the 200-kg crate can be supported.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
Show me the final answers↓
Let us draw a free body diagram focusing on ring B.
First, we will write an equation of equilibrium for y-axis forces as this will allow us to figure out a value for θ.
3500\text{sin}\,(\theta)\,-\,1962\,=\,0
\text{sin}\,(\theta)\,=\,\dfrac{1962}{3500}
(solve for \theta)
\theta\,=\,\text{sin}^{-1}\left(\dfrac{1962}{3500}\right)
\theta\,=\,34.1^0
Let us now write an equation of equilibrium for x-axis forces.
3500\text{cos}\,(34.1^0)\,-\,F_{BC}\,=\,0
(Solve for F_{BC})
F_{BC}\,=\,2898.2 N
We can now figure out the value of y by using trigonometry.
y\,=\,0.84 m
Final Answers:
Length of y = 0.84 m
This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-25.
FBC should be 3500 cos(34.1 )−FBC=0. equals to 2898.37N
Thank you for catching the mistake. This has been fixed 🙂