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The ladder as a .

External influences from the earth (gravity) the wall (normal force) and the floor (normal force and frictional force).

and .

\begin{large}\[ \sum F_{x} = F_{w} + F_{f,x} = 0 \]
\[ \sum F_{y} = - mg + F_{f,y} = 0 \]
\[ \sum \tau = + mg (L/2) \cos\theta - F_{w} L \sin\theta \]\end{large}

Note that our selection of the axis point guaranteed that the floor's force would contribute zero torque.

\begin{large} \[ F_{f,y} = mg = \mbox{147 N}\]
\[ F_{w} = \frac{mg}{2}\cot\theta = \mbox{27 N}\]\end{large}

\begin{large}\[ F_{f,x} = -F_{w} = - \mbox{27 N}\]\end{large}

The ladder as a .

External influences from the earth (gravity) the wall (normal force and friction) and the floor (normal force and frictional force).

and .

\begin{large}\[ |F_{f,x}| = \mu_{s,f} |F_{f,y}| \]
\[ |F_{w,y}| = \mu_{s,w} |F_{w,x}| \]\end{large}

\begin{large}\[ \sum F_{x} = F_{w,x} - \mu_{s,f}F_{f,y} = 0 \]
\[ \sum F_{y} = - mg + F_{f,y} + \mu_{s,w}F_{w,x}= 0 \]
\[ \sum \tau = - mg (L/2) \cos\theta + F_{f,y} L\cos\theta - \mu_{s,f}F_{f,y} L \sin\theta \]\end{large}

Note that the y-component of the floor force is the normal force from the floor on the ladder, and the x-component of the wall force is the normal force from the wall on the ladder. Since we are assuming static friction is maximized, we have rewritten F_f,x and F_w,y as the coefficient of friction times the relevant normal force.

\begin{large}\[ F_{f,y} = \frac{mg}{2\left(1-\mu_{s,f}\tan\theta\right)} = \mbox{125 N} \]\end{large}

\begin{large}\[ F_{w,x} = \mu_{s,f}F_{f,y} = \mbox{18.8 N} \]\end{large}

\begin{large}\[ \mu_{s,w} = \frac{mg-F_{f,y}}{F_{w,x}} = 1.19 \] \end{large}

We found in Part A that the ladder is stable as long as the coefficient of friction with the ground is 0.19 or greater, even if there is no friction at all from the wall. Does it make sense that in Part B the wall's friction coefficient has to be so much larger than 0.19 if the wall is only partly responsible for keeping the ladder from sliding?