# 1.5 m long cord AB can withstand a maximum 2

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:

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.

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

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

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

(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.

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

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

(Solve for $F_{BC}$)

$F_{BC}\,=\,2898.2$ N

Show me the free body diagram

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

$F_{BC}\,=\,2898.2$ N