# A truck pulls boxes up an incline plane. The truck can exert a maximum force of 3,500 N. If the plane's incline is (5 pi )/12  and the coefficient of friction is 7/12 , what is the maximum mass that can be pulled up at one time?

Jan 17, 2017

Maximum mass is $438.24 \setminus k g$ (2dp)

#### Explanation:

For our diagram, $m = \text{mass } k g$, $\theta = \frac{5 \pi}{12}$

If we apply Newton's Second Law up perpendicular to the plane we get:

$R - m g \cos \theta = 0$
$\therefore R = m g \cos \left(\frac{5 \pi}{12}\right) \setminus \setminus N$

With a maximum driving force of $3500 \setminus N$ upwards (along the plane) $D = 3500$. If we Apply Newton's Second Law up parallel to the plane we get:

$D + F - m g \sin \theta = 0$
$\therefore 3500 + F - m g \sin \left(\frac{5 \pi}{12}\right) = 0$
$\therefore F = m g \sin \left(\frac{5 \pi}{12}\right) - 3500$

And the friction is related to the Reaction (Normal) Force by

$F \le \mu R$
$\therefore m g \sin \left(\frac{5 \pi}{12}\right) - 3500 \le \frac{7}{12} \left(m g \cos \left(\frac{5 \pi}{12}\right)\right)$
$\therefore 12 m g \sin \left(\frac{5 \pi}{12}\right) - 42000 \le 7 m g \cos \left(\frac{5 \pi}{12}\right)$
$\therefore 12 m g \sin \left(\frac{5 \pi}{12}\right) - 7 m g \cos \left(\frac{5 \pi}{12}\right) \le 42000$
$\therefore m g \left(12 \sin \left(\frac{5 \pi}{12}\right) - 7 \cos \left(\frac{5 \pi}{12}\right)\right) \le 42000$
$\therefore m \le \frac{42000}{\left(12 \sin \left(\frac{5 \pi}{12}\right) - 7 \cos \left(\frac{5 \pi}{12}\right)\right) g}$
$\therefore m \le 438.2400291 \ldots$