Static and Dynamic Equilibrium EX 2

There are four forces acting on this block: $F_G$, $F_N$, $F_f$, $F$. The block slides up the incline and therefore the force of friction points down the incline (because friction opposes the slide between the two surfaces).

We select the x-axis parallel to the incline and then draw the free body diagram.

The components of the forces exerted on the mass $m=5$ kg are:

Forces	x-component	y-component
$F_G$	50 cos(-127°) = -30 N	50 sin(-127°) = -40 N
$F_N$	0	$F_N$
$F_f$	$-F_f$	0
$F$	40 cos(-37°) = 32 N	40 sin(-37°) = -24 N

Note that we used $g=10m/s^2$ in our computations.

First we have to deal with the y-component.

Writing

$-40-24+F_N=0$

we see that $F_N=64$ N.

Now, looking at the x-component, we write

$-30+32-F_f=0$

solving for $F_f$ we find $F_f=2$ N.