An object with a mass of  15 kg is on a ramp at an incline of pi/8 . If the object is being pushed up the ramp with a force of  9 N, what is the minimum coefficient of static friction needed for the object to remain put?

Apr 1, 2017

Answer:

The coefficient of static friction is obtained from the ratio of the Normal force to the force parallel to the incline, in this case 9N in a direction up the incline.

Explanation:

The forces involved are:

• force of gravity on the 15 kg block, which is $15 k g \cdot 9.8 m {s}^{-} 2 = 147 N$
• Normal force on the block perpendicular to the incline, which is $147 \cdot \cos \left(\frac{\pi}{6}\right) = 147 N$
• force on the block parallel to the incline which is $147 \cdot \sin \left(\frac{\pi}{6}\right) = 1.34 N$
If there is a force of 9N pushing up the incline, and a parallel force of 1.34 N in the opposite direction, then the net force up the incline is $7.66 N$.
The coefficient of static friction therefore, is $\frac{7.66}{147} = 0.52$