h2. Part C
Suppose a 10 kg cubic box is at rest on a horizontal surface. There is friction between the box and the surface characterized by a coefficient of static friction equal to 0.50. If one person pushes on the box with a force _F_~1~ of 25 N directed due north and a second person pushes on the box with a force _F_~2~ = 25 N also directed due north, what is the magnitude and direction of the force of static friction acting on the box?
h4. Solution
{toggle-cloak:id=sysc} *System:* {cloak:id=sysc}Box as [point particle].{cloak}
{toggle-cloak:id=intc} *Interactions:* {cloak:id=intc} External influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).{cloak}
{toggle-cloak:id=modc} *Model:* {cloak:id=modc}[Point Particle Dynamics].{cloak}
{toggle-cloak:id=appc} *Approach:*
{cloak:id=appc}
{toggle-cloak:id=diagc} {color:red} *Diagrammatic Representation* {color}
{cloak:id=diagc}
To determine the force of static friction, we first find the net force _in the absence of friction_. We first draw the situation. A top view (physical representation) and a side view (free body diagram) of the box _ignoring any contribution from friction_ are shown here.
!basicstatic3.jpg!
{cloak:diagc}
{toggle-cloak:id=mathc} {color:red} *Mathematical Represenatation* {color}
{cloak:id=mathc}
The net force parallel to the surface in the absence of friction is then:
{latex}\begin{large}\[ \sum_{F \ne F_{f}} F_{x} = 0 \]
\[\sum_{F \ne F_{f}} F_{y} = F_{1}+F_{2} = \mbox{50 N} \]\end{large}{latex}
In order to prevent the box from moving, then, static friction would have to satisfy:
{latex}\begin{large}\[ F_{s} = \mbox{0 N }\hat{x} - \mbox{50 N }\hat{y} = \mbox{50 N due south}.\]\end{large}{latex}
Again, we must check that this needed friction force is compatible with the static friction limit. Again, Newton's 2nd Law for the _z_ direction tells us:
{latex}\begin{large}\[ \sum F_{z} = N - mg = ma_{z}\]\end{large}{latex}
and know that the box will remain on the surface, so _a_~z~ = 0. Thus,
{latex}\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}{latex}
With this information, we can evaluate the limit:
{latex}\begin{large}\[ F_{s} \le \mu_{s} N = \mbox{49 N} \]\end{large}{latex}
Since 50 N > 49 N, we conclude that the static friction limit is violated. The box will move and kinetic friction will apply instead!
{cloak:mathc}
{cloak:appc |