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## Homework Statement

A a uniform plank of L = 6.10 m and weight 445 N, rests on the ground and on a frictionless roller at the top of a wall of height 3.05 m. The plank remains in equilibrium for any value of theta less than 70 degrees, but slips if theta bigger or equal to 70 degrees. What is the coefficient of static friction between the ladder and the ground?

## Homework Equations

N_2 = normal force between plank and ground. N_1 equals the normal force of the roller onto the plank.

F with no subscripts = f

_{s}, aka. the friction force (which is pointing to the right)

[tex]\mu[/tex]

_{s}=F

_{s}/N

_{2}

## The Attempt at a Solution

F_net(x) = F - N

_{1}[tex]\theta[/tex]

F

_{net(y)}= N

_{2}+

_{1}cos[tex]\theta[/tex] -W (the weight) =0

T(Torque)

_{taken about the middle}= F

_{s}h - N

_{2}htan[tex]\theta[/tex]=0

N

_{1}=F/tan[tex]\theta[/tex]

N

_{2}+ F/(sin[tex]\theta[/tex]cos[tex]\theta[/tex]) =W [comes from F

_{net(y)}with F/sin[tex]\theta[/tex] being plugged in for N

_{1}.

W = N

_{2}+ Fcot[tex]\theta[/tex]

W = F/tan[tex]\theta[/tex] + Fcot[tex]\theta[/tex] (plugging in N

_{2}= F/tan[tex]\theta[/tex])

W = F ( 1/tan[tex]\theta[/tex] + cot ([tex]\theta[/tex])

....It keeps going on until you reach that the coefficient of friction equals tan[tex]\theta[/tex], which is the wrong answer

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