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.

#### Solution:

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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}.

(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.

\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_{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

#### Final Answers:

F_{min}\,=\,235.4 lb