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# An object with a mass of 5 kg is on a ramp at an incline of pi/8 . If the object is being pushed up the ramp with a force of  2 N, what is the minimum coefficient of static friction needed for the object to remain put?

Apr 30, 2018

$\textcolor{b l u e}{0.37} \textcolor{w h i t e}{8888}$ 2 d.p.

#### Explanation:

Since the $\boldsymbol{2 N}$ force is acting upwards the frictional force $\boldsymbol{F r}$ is acting in the opposite direction.

For the body to remain static, we have:

$F r = \mu R$

Where $\mu$ is the coefficient of friction.

Resolving forces perpendicular to the plane:

$R = 5 g \cos \left(\frac{\pi}{8}\right)$

Resolving forces horizontal to the plane:

$F r = 5 g \sin \left(\frac{\pi}{8}\right) - 2 N$

Substituting these two equations into$\setminus \setminus \setminus \setminus \setminus F r = \mu R$:

$5 g \sin \left(\frac{\pi}{8}\right) - 2 = \mu \cdot 5 g \cos \left(\frac{\pi}{8}\right)$

$\therefore$

$\mu = \frac{5 g \sin \left(\frac{\pi}{8}\right) - 2}{5 g \cos \left(\frac{\pi}{8}\right)}$

Taking $g = 9.81$

$\mu = \frac{5 \cdot \left(9.81\right) \sin \left(\frac{\pi}{8}\right) - 2}{5 \cdot \left(9.81\right) \cos \left(\frac{\pi}{8}\right)} = 0.3700793240 N$

$\mu = 0.37$ 2 d.p.