# Question f3746

Jul 11, 2017

See below.

#### Explanation:

Here

$L =$ ladder length
$P =$ ladder weight
${F}_{a} =$ applied horizontal force
$\mu =$ static friction coefficient
$H =$ normal force in the vertical wall
$V =$ normal force in the floor
${y}_{0} =$ height ladder contact point
${x}_{0} =$width ladder contact point
$d =$ applied force distance over the ladder
$\sin \left(\theta\right) = {y}_{0} / \sqrt{{y}_{0}^{2} + {x}_{0}^{2}}$

Now we will develop the modeling for two cases:

1) ${F}_{a}$ pushing to the left

$\left\{\begin{matrix}{\sum}_{X} \to H - {F}_{a} + \mu V = 0 \\ {\sum}_{Y} \to V - P - \mu H = 0 \\ {M}_{O} \to {x}_{0} V + {F}_{a} d \sin \left(\theta\right) - P \left({x}_{0} / 2\right) - {y}_{0} H = 0\end{matrix}\right.$

2) ${F}_{a}$ pushing to the right

$\left\{\begin{matrix}{\sum}_{X} \to H + {F}_{a} - \mu V = 0 \\ {\sum}_{Y} \to V - P + \mu H = 0 \\ {M}_{O} \to {x}_{0} V - {F}_{a} d \sin \left(\theta\right) - P \left({x}_{0} / 2\right) - {y}_{0} H = 0\end{matrix}\right.$

Solving both cases we have for each case:

1) F_a = 1/2(P ((mu^2-1) x_0 - 2 mu y_0) sqrt[x_0^2 + y_0^2])/( (mu x_0 - y_0) sqrt[x_0^2 + y_0^2]+d (1 + mu^2) y_0)

putting figures we obtain the condition ${F}_{a} > 416.667$ to the left

2) F_a = 1/2(P ((mu^2-1) x_0 + 2 mu y_0) sqrt[ x_0^2 + y_0^2])/( (mu x_0 + y_0) sqrt[x_0^2 + y_0^2]- d (1 + mu^2) y_0)#

putting figures we obtain the condition ${F}_{a} > 38.889$ to the right.