# A truck pulls boxes up an incline plane. The truck can exert a maximum force of 5,700 N. If the plane's incline is (2 pi )/3  and the coefficient of friction is 8/5 , what is the maximum mass that can be pulled up at one time?

Nov 8, 2017

$\text{8130.2 kg}$

#### Explanation:

If $m$ is the mass of the boxes and $M$ is the mass of the truck:

$N =$ the normal force

So, we want to know the maximum mass of the boxes.

$m g \sin \left(\theta\right) + M g \sin \left(\theta\right) + {F}_{f} = 5700$

This is because when the boxes reach the maximum mass, the truck at this stage cannot pull more boxes. Thus, the force that pulls everything up the incline must be equal to the forces against the motion.

$m g \sin \left(\theta\right) \to$ The component of the weight of the boxes along the plane.

$M g \sin \left(\theta\right) \to$ The component of truck's weight along the plane.

Now, ${F}_{f}$ or static friction is

${F}_{f} = \text{the coefficient of friction" xx "the normal force}$

Here

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

The problem is that $M$ is not given. I will make another equation;

$5700 - M g \sin \left(\theta\right) - \mu \times M g \cos \left(\theta\right) = 0$

Here

$\mu =$ the coefficient of friction

Rearrange the equation to get $M$

$M = \frac{5700}{g \sin \left(\theta\right) + \mu \times g \cos \left(\theta\right)}$

Now plug in the numbers to get

$M = \text{8800.2 kg}$

Now use that equation

$m g \sin \left(\theta\right) + M g \sin \left(\theta\right) + {F}_{f} = 5700$

Solve for $m$:

$m = \frac{5700 - {F}_{f}}{g \sin \left(\theta\right)} - M$

After you plug in the numbers you will get :

$m = \text{8130.2 kg}$