Unformatted text preview: , l . 2 tan n and ) ( 2 wx l w θ l x l w l w fl + =+ = In terms of the coefficient of friction , s μ . tan 2 2 3 tan ) ( 2 2 s θ x l x l θ x x l n f l l += ++ = Solving for x , cm. 2 30 tan tan 3 2 s s . θ μ μ θ l x = +c) In the above expression, setting gives for solving and cm 10 s = x . 625 . 20 1 tan ) 20 3 ( s = +l θ l μ...
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 Force, Friction Force, Tan, nl tan

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