# A truck pulls boxes up an incline plane. The truck can exert a maximum force of 5,600 N. If the plane's incline is (2 pi )/3  and the coefficient of friction is 7/6 , what is the maximum mass that can be pulled up at one time?

May 1, 2016

979 kg

#### Explanation:

Note, by definition, an inclined plane cannot have an inclination more than $\frac{\pi}{2}$. I take the angle is measured from positive x-axis, so it is just $\theta = \frac{\pi}{3}$ the other way.

here $f$ is the applied force, NOT the frictional force.

So, as we can easily observe in the picture, the forces that oppose will be (m is expressed in $k g$):

1. gravitational pull: $m g \sin \theta = 9.8 \times \frac{\sqrt{3}}{2} m = 8.49 m N$

2. frictional force, opposite to the direction of tendency of movement: $\mu m g \cos \theta = \frac{7}{6} \times 9.8 \times \frac{1}{2} m N = 5.72 m N$

Hence total is: $\left(8.49 + 5.72\right) m N = 14.21 m N$

So, for the truck the to be able to pull it up, the maximum force it can exert must be more than this:

$5600 N > 5.72 m N \implies m < 979 k g$