Static and Dynamic Equilibrium EX 3

There are four forces acting on this block: $F_{G}$, $F_{N}$, $F_{f}$, $F$. The block is at rest on the incline and therefore we are not certain which way the force of friction points. It depends on the magnitudes of all the other forces acting on the block. For this reason we will not include the force of friction in the initial drawing of forces and in our calculation we will add the x-component of the force of friction to the x-components of other forces.

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 $m=8$ kg are:

Forces	x-component	y-component
$F_{G}$	80 cos(-120°) = -40 N	80 sin(-120°) = -69 N
$F_{N}$	0	$F_{N}$
$F_{f}$	${F_{f}}_{x}$	0
$F$	30 cos(-30°) = 26 N	30 sin(-30°) = -15 N

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

First we have to deal with the y-component.

Writing

$-69-15+F_{N}=0$

we see that $F_{N}=84$ N.

Now, looking at the x-component, we write

$-40+26+{F_{f}}_{x}=0$

From this equation we find that the x-component of the horizontal force does not balance the x-component of the gravitational force and therefore, the force of friction has to point in the direction of the positive x-axis because the ${F_{f}}_{x}=14$ N is positive.