Force F acts on the frame such that 2


Force F acts on the frame such that its component acting along member AB is 650 lb, directed from B towards A, and the component acting along member BC is 500 lb, directed from B towards C. Determine the magnitude of F and its direction ϴ. Set Φ = 60°.

Force F acts on the frame such that

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

Solution:

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We will first draw the vector components and set Φ = 60° as stated by the question.

Force F acts on the frame such that

(note the colors which correspond to the angles which shows where each angle is taken from, and were found using the alternate interior angles theorem, corresponding angles theorem and consecutive angles theorem.)

We can next draw each component tail to tail as follows:

Force F acts on the frame such that its components

We can now use the law of cosines to figure out force F.

F^2\,=\,500^2+650^2-2(500)(650)\cos105^0

(Take the square root of both sides)

F\,=\,\sqrt{500^2+650^2-2(500)(650)\cos105^0}

F\,=\,916.9 lb

 

We can now use the law of sines to figure out the angle \theta. We will use the value we just found to do so:

\dfrac{\sin \theta}{500}\,=\,\dfrac{\sin105^0}{916.9}

\sin \theta\,=\,0.5267

\theta\,=\,\sin^{-1}(0.5267)

\theta\,=\,31.8^0

 

Final Answers:

F\,=\,916.9 lb

\theta\,=\,31.8^0

 

This question can be found in Engineering Mechanics: Statics, 13th edition, chapter 2, question 2-13.

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