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h3. Part A

A 3.0 m long ladder is leaned against a vertical wall at a 70° angle above the horizontal.   The wall is frictionless, but the (level) floor beneath the ladder does have friction.  Assuming the laddercenter of ismass essentiallyof athe uniform15.0 rodkg withladder ais massin ofits 15.0exact kgcenter, find the forces exerted on the ladder by the wall and by the floor.  

System:  The ladder as a [rigid body] subject to external influences from the earth (gravity) the wall (normal force) and the floor (normal force and frictional force).

Model:  [Rotation and Translation of a Rigid Body].

Approach:  This is a [statics] problem, so the object is to ensure that the ladder does _not_ rotate or translate.  Thus, we know that all accelerations and angular accelerations are zero.  We first sketch the situation and set up coordinates, which includes selecting an axis.  Selecting the position of the axis is technically arbitrary, but often it is actually specified by the unknowns in the system.  In this case, as we shall see, the forces in the x-direction are less constrained than the y-direction forces.  Thus, it is a good idea to position the axis to cancel the torque from one of the x forces.  We choose to put the axis at the point of contact with the floor.

PICTURE


We then write the equations of linear and rotational equilibrium:

{latex}\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}{latex}

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

The y-direction equation and the torque equation are immediately solvable to obtain:

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

Once we have _F_~w~, we can find the x-component of the floor's force:

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

We can now construct the total floor force:

PICTURE

Which gives a total force of 150 N at 80° above the horizontal.

h3. Part B 

Consider the same basic situation as in Part A, but now suppose that the wall has friction.  If the coefficient of static friction between the floor and the ladder is only 0.15, what is the minimum coefficient of static friction needed between the wall and the ladder to prevent the ladder from slipping?  

System:  The ladder as a [rigid body] subject to external influences from the earth (gravity) the wall (normal force and friction) and the floor (normal force and frictional force).

Model:  [Rotation and Translation of a Rigid Body].

Approach:  The minimum coefficient of friction will occur when both friction forces are maxima.  Thus, we have the important relationships:

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

Because we know the floor's static friction coefficient, we can streamline the solution by switching our choice of axis.

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The resulting equations are:

{latex}\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}{latex}

The only equation that can be solved immediately is torque balance, which gives:

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

We can then substitute this result into the equation of x-force balance to find:

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

Finally, we can find the required coefficient of static friction to be:

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