A truck pulls boxes up an incline plane. The truck can exert a maximum force of 4,200 N. If the plane's incline is (5 pi )/8  and the coefficient of friction is 14/3 , what is the maximum mass that can be pulled up at one time?

Feb 26, 2018

The mass is $= 497.2 k g$

Explanation:

Resolving in the direction parallel to the plane ↗^+

Let the force exerted by the truck be $F = 4200 N$

Let the frictional force be $= {F}_{r} N$

The coefficient of friction $\mu = {F}_{r} / N = \frac{14}{3}$

The normal force is $N = m g \cos \theta$

The angle of the plane is $\theta = \frac{5}{8} \pi$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Therefore,

$F = {F}_{r} + m g \sin \theta$

$= \mu N + m g \sin \theta$

$= \mu m g \cos \theta + m g \sin \theta$

$m = \frac{F}{g \left(\mu \cos \theta + \sin \theta\right)}$

=4200/(9.8(14/3*cos(5/8pi)+sin(5/8pi))

$= - 497.2 k g$