A truck pulls boxes up an incline plane. The truck can exert a maximum force of 3,200 N. If the plane's incline is (5 pi )/8  and the coefficient of friction is 3/7 , what is the maximum mass that can be pulled up at one time?

Jul 21, 2017

${m}_{\text{max}} = 322$ $\text{kg}$

Explanation:

We're asked to find the maximum mass a truck can pull up at one time, with some given information.

We'll treat the boxes as one body, and our goal is finding the mass of this body, $m$.

The maximum mass the truck can pull is the largest so that the net force is not down the incline (then the boxes would not be going up, intuitively). Thus, at this condition, the boxes are in equilibrium (no net forces acting, pulls upward with constant velocity).

Thus, treating the positive $x$-axis as up the incline:

$\sum {F}_{x} = \overbrace{{F}_{\text{truck")^(3200color(white)(l)"N}}} - {\overbrace{{f}_{k}}}^{= {\mu}_{k} n = \frac{3}{7} m g \cos \theta} - m g \sin \theta = 0$

$3200$ $\text{N}$ $- \frac{3}{7} m g \cos \theta - m g \sin \theta = 0$

$3200$ $\text{N}$ $= \frac{3}{7} m g \cos \theta + m g \sin \theta$

Divide all terms by the mass, $m$:

$\frac{3200 \textcolor{w h i t e}{l} \text{N}}{m} = \frac{3}{7} g \cos \theta + g \sin \theta$

Therefore,

$\frac{3200 \textcolor{w h i t e}{l} \text{N}}{\frac{3}{7} g \cos \theta + g \sin \theta} = m$

= (3200color(white)(l)"N")/(3/7(9.81color(white)(l)"m/s"^2)sin((5pi)/8) + (9.81color(white)(l)"m/s"^2)cos((5pi)/8))

= color(red)(322 color(red)("kg"