Page History

Thus,

the

net

force

along

the

surface

is

zero

without

the

influence

of

static

friction,

and

so

the

static

friction

force

will

also

be

0.

In

order

to

prevent

the

box

from

moving,

then,

static

friction

would

have

to

satisfy:

\begin{large}\[ F_{s} = - \mbox{25 N }\hat{x} - \mbox{25 N }\hat{y} = \mbox{35.4 N at 45}^{\circ}\mbox{ S of W}.\]\end{large}

We must now check that this needed friction force is compatible with the static friction limit. Newton's 2nd Law for the z direction tells us:

\begin{large}\[ \sum F_{z} = N - mg = ma_{z}\]\end{large}

We know that the box will remain on the surface, so a_z = 0. Thus,

\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}

With this information, we can evaluate the limit:

\begin{large}\[ F_{s} \le \mu_{s} N = \mbox{49 N} \]\end{large}

Since 35.4 N < 49 N, we conclude that the friction force is indeed 35.4 N at 45° S of W.

\[ \sum_{F \ne F_{f}} F_{x} = 0 \]
\[\sum_{F \ne F_{f}} F_{y} = F_{1}+F_{2} = \mbox{50 N} \]\end{large}

In order to prevent the box from moving, then, static friction would have to satisfy:

\begin{large}\[ F_{s} = \mbox{0 N }\hat{x} - \mbox{50 N }\hat{y} = \mbox{50 N due south}.\]\end{large}

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:

\begin{large}\[ \sum F_{z} = N - mg = ma_{z}\]\end{large}

and know that the box will remain on the surface, so a_z = 0. Thus,

\begin{large}\[ N = mg = \mbox{98 N}\]\end{large}

With this information, we can evaluate the limit:

\begin{large}\[ F_{s} \le \mu_{s} N = \mbox{49 N} \]\end{large}

Since 50 N > 49 N, we conclude that the static friction limit is violated. The box will move and kinetic friction will apply instead!