If a ladder makes an angle with the wall, has a mass m and is uniform in design, show that the minimum force of friction which must exist between the ladder and floor to keep the ladder from slipping is given by(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

F_f = \frac{1}{2}mg\tan\theta

[/tex]

Now, [tex]\mu = \frac{F_g_x}{F_g_y}[/tex], and in a previous calculation I got [tex]F_g_x = \frac{\frac{L\cos\theta}{2}mg}{L\sin\theta}[/tex] and [tex]F_g_y = mg[/tex], with [tex]L[/tex] being the length of the ladder.

When I plug all this together, I get [tex]\frac{\cot}{2}[/tex], not even close to what I'm intended to get. Can somebody tell me what I'm doing wrong?

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# Ladder equilibrium proof

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