# A block weighing 6 kg is on a plane with an incline of (3pi)/8 and friction coefficient of 2/5. How much force, if any, is necessary to keep the block from sliding down?

Apr 10, 2017

The force is $= 63.3 N$

#### Explanation: Let the force be $F$

The frictional force is

${\mu}_{s} = {F}_{r} / N$

$N = m g \cos \theta$

${F}_{r} = {\mu}_{s} \cdot m g \cos \theta$

Resolving in the direction parallel to the plane ↗^+

The force is

$F = {F}_{r} + m g \sin \theta$

$F = {\mu}_{s} m g \cos \theta + m g \sin \theta$

$= \frac{2}{5} \cdot 6 \cdot 9.8 \cos \left(\frac{3}{8} \pi\right) + 6 \cdot 9.8 \cdot \sin \left(\frac{3}{8} \pi\right)$

$= \left(9 + 54.3\right) N$

$= 63.3 N$

The force is $= 63.3 N$

Apr 10, 2017

$\text{it is necessary a force 45.38 Newtons}$

#### Explanation:

$\text{given data :}$
$m : 6 \text{ "kg" mass of block}$
$g = 9.81 \text{ } \frac{N}{k g}$
$\beta = \frac{3 \pi}{8}$
${\mu}_{k} = \frac{2}{5}$

$\text{The block appears as if it were given on an inclined plane.}$
$\text{Please notice that the weight of of block is perpendicular}$
$\text{to the ground.}$ $\text{You should split the weight of block into two components }$ $\text{The green vector "G_x " is parallel to the inclined plane.}$

$\text{The orange vector "G_y " is perpendicular to the inclined plane.}$ ${G}_{x} = m g \cdot \sin \beta$

${G}_{x} = 6 \cdot 9.81 \cdot \sin \beta$

$\textcolor{g r e e n}{{G}_{x} = 54.38} \text{ } N$

${G}_{y} = m g \cdot \cos \beta$

${G}_{y} = 6 \cdot 9.81 \cdot \cos \beta$

$\textcolor{\mathmr{and} a n \ge}{{G}_{y} = 22.52} \text{ } N$

$\text{since it is perpendicular to the surface ,The " G_y " component causes friction.}$ $\textcolor{red}{{F}_{f} = {\mu}_{k} \cdot {G}_{y}}$

${F}_{f} = \frac{2}{5} \cdot 22.52$

${F}_{f} = 9 \text{ } N$ $\text{Since "G_x " greater than "F_f ," block is sliding down.}$
${F}_{\text{net}} = {G}_{x} - {F}_{f} = 54.38 - 9$

F_("net")=45.38 " "N

$\text{it is necessary a 45 Newtons of force at opposite direction}$
$\text{so that it does not slide.}$