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.

#### 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.

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

3500\text{sin}\,(\theta)\,-\,1962\,=\,0

\text{sin}\,(\theta)\,=\,\dfrac{1962}{3500}

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.

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

3500\text{cos}\,(34.1^0)\,-\,F_{BC}\,=\,0

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.

\text{sin}\,(34.1^0)\,=\,\dfrac{y}{1.5}
y\,=\,1.5\,\text{sin}\,(34.1^0)

y\,=\,0.84 m

#### Final Answers:

F_{BC}\,=\,2898.2 N

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 🙂