# Three chains act on the bracket such that

Three chains act on the bracket such that they create a resultant force having a magnitude of 500 lb. If two of the chains are subjected to known forces, as shown, determine the angle of the third chain measured clockwise from the positive x axis, so that the magnitude of force F in this chain is a minimum. All forces lie in the x–y plane. What is the magnitude of F? Hint: First find the resultant of the two known forces. Force F acts in this direction.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first figure out the resultant force with respect to the 200 lb force and the 300 lb force. To do so, we will draw a vector component diagram.

We can now draw these vectors tail to tail like so:

Now, using the law of cosines, we can figure out $F_{R1}$.

$(F_{R1})^2\,=\,300^2+200^2-2(300)(200)\cos60^0$

(Take the square root of both sides)

$F_{R1}\,=\,\sqrt{300^2+200^2-2(300)(200)\cos60^0}$

$F_{R1}\,=\,264.6$ lb

Now, we can use law of sines and the value of $F_{R1}$ we found to figure out the angle $\theta$.

$\dfrac{\sin(30^0+\theta)}{200}\,=\,\dfrac{\sin60^0}{264.6}$

$\sin(30^0+\theta)\,=\,0.655$

$30^0+\theta\,=\,\sin^{-1}(0.655)$

$30^0+\theta\,=\,40.9^0$

$\theta\,=\,10.9^0$

The question states that force F is directed along force $F_{R1}$. That means we only need to find the other component of the force F. In a diagram, we can represent the vectors as follows:

Mathematically we can express this as:

$F_R\,=\,F_{R1}+F_{min}$

$F_{min}\,=\,F_R\,-\,F_{R1}$

$F_{min}\,=\,500\,-\,264.6$

(Remember $F_R$ is 500 lb as stated in the question)

$F_{min}\,=\,235.4$ lb

$\theta\,=\,10.9^0$
$F_{min}\,=\,235.4$ lb