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Example #13:

Begin with a drawing. Include all forces with labels, showing any forces that need to be broken into components. (Also go ahead and do any easy calculations, like F g .). Example #13:. F app = ?. F N. Example #13:. F gy. F fric. θ. F gx. F g = 294N.

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Example #13:

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  1. Begin with a drawing. Include all forces with labels, showing any forces that need to be broken into components. (Also go ahead and do any easy calculations, like Fg.) Example #13:

  2. Fapp = ? FN Example #13: Fgy Ffric θ Fgx Fg = 294N

  3. The first step in my typical strategy is to think about how the problem will eventually be solved, and set up an equation that will eventually be used to finish the problem. • In this case, since it’s in equilibrium, I would think about x-direction balanced forces. Example #13:

  4. Fapp = Ffric + Fgx • So this will be great to solve the problem, as soon as we figure out Ffric and Fgx as two little ‘side problems’. Example #13:

  5. Fgx is relatively simple, since it’s just a component of Fg. • Fgx = Fgsinθ = 294sin49o = 221.9N Example #13:

  6. Ffric is calculated with the equation • Ffric = μFN • But FN is calculated by realizing that the object is in equilibrium in the y-direction. Example #13:

  7. Ffric = μFN = μFgy = (0.38)(294cos49o) = 73.29N Example #13:

  8. So now we can go back to the original equation: Fapp = Ffric + Fgx • Fapp = 73.29 + 221.9 = 295.2N • Woo Hoo! All done! Example #13:

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