An object with a mass of 19 kg is on a ramp at an incline of pi/12 . If the object is being pushed up the ramp with a force of  6 N, what is the minimum coefficient of static friction needed for the object to remain put?

Aug 6, 2017 Given that

• the angle of inclination of the ramp $\theta = \frac{\pi}{12}$.

• the mass of the object kept on the ramp $m = 19 k g$.

• the object being pushed up by a force $F = 6 N$

Now

• the component of gravitational force acting on the object against applied force along the ramp $= m g \sin \theta$

• normal reaction acting on the object $m g \cos \theta$

• upward static frictional force acting on the object $f = \mu m g \cos \theta$, where $\mu$ is the coefficient of static friction.

Cinsidering the equilibrium of forces we get following relation

$m g \sin \theta - f = F$

$\implies m g \sin \theta - \mu m g \cos \theta = F$

$\implies \mu = \frac{m g \sin \theta - F}{m g \cos \theta}$

$\implies \mu = \frac{19 \times 9.8 \times \sin \left(\frac{\pi}{12}\right) - 6}{19 \times 9.8 \times \cos \left(\frac{\pi}{12}\right)}$

$\implies \mu = \tan \left(\frac{\pi}{12}\right) - \frac{6}{19 \times 9.8 \times \cos \left(\frac{\pi}{12}\right)}$

$\implies \mu \approx 0.23$