A truck pulls boxes up an incline plane. The truck can exert a maximum force of 2,400 N. If the plane's incline is (2 pi )/3  and the coefficient of friction is 5/3 , what is the maximum mass that can be pulled up at one time?

May 10, 2018

The truck will be able to pull up to 144 kg.

Explanation:

This calculation is neglecting any friction or gravity acting on the truck, so it's net force is 2400 N up the slope. Also, I'll take theta as $\frac{\pi}{3}$ or 60˚. This is because $\frac{2 \pi}{3}$ is 120˚ and takes you beyond a vertical slope and back around to 60˚.

The parallel component of the gravitational force on the box is given by

${F}_{g | \setminus |} = m g \sin \left(\theta\right)$

This is the force that gravity exerts on the box down the slope.

The normal force acting on the box is given by the perpendicular component of the gravitational force

$N = m g \cos \left(\theta\right)$

The frictional force is then given by

${F}_{f} = {\mu}_{s} N = {\mu}_{s} m g \cos \left(\theta\right)$

${F}_{g | \setminus |}$ and ${F}_{f}$ are the opposing forces so if we set the sum of these equal to 2400 N, that will allow us to solve for the maximum mass

${F}_{g | \setminus |} + {F}_{f} = 2400$

$\Rightarrow m g \sin \left(\theta\right) + {\mu}_{s} m g \cos \left(\theta\right) = 2400$

$\Rightarrow m \left(g \sin \left(\theta\right) + {\mu}_{s} g \cos \left(\theta\right)\right) = 2400$

$\Rightarrow m = \frac{2400}{g \sin \left(\theta\right) + {\mu}_{s} g \cos \left(\theta\right)}$

$m = \frac{2400}{8.487 + 8.167} = 144 \text{ kg}$

Where

$g = 9.80 \text{ m/s}$

${\mu}_{s} = \frac{5}{3}$

$\theta = \frac{\pi}{3}$