Example of Free Body Diagram and Application of II Law.

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2f kTNTsin20Tcos2020 040gsin3030 030 0 40gcos3040gUsing the connection between f k <strong>and</strong> the normal force we have:fkk( 339.5 −.342T) = 67.9 −.0684T( .4)= µN = 0.2EqSubstituting Eq. (4) into Eq. (2) we get:( 67.9 − .0684T) −19630. 94T−=Solving we obtain: T= 291 Newtons <strong>and</strong> N = 240 Newtons

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Example of Free Body Diagram and Application of II Law.

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