# 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°. 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 draw the vector components and set Φ = 60° as stated by the question. (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: 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:

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

$\sin \theta\,=\,0.5267$

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

$\theta\,=\,31.8^0$

$F\,=\,916.9$ lb

$\theta\,=\,31.8^0$

## 2 thoughts on “Force F acts on the frame such that”

• Gilbert

I guess the given angle there should be phi = 60 degrees and not theta

• questionsolutions Post author

You are absolutely right. We fixed the typo. Thank you for letting us know, we really appreciate it! 🙂