# 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 (3 pi )/8  and the coefficient of friction is 7/4 , what is the maximum mass that can be pulled up at one time?

Apr 19, 2017

The mass is $= 224.1 k g$

#### Explanation:

Let the mass be $m k g$

Force of truck is $F = 3500 N$

The coefficient of friction is $\mu$

$\mu = {F}_{r} / N$

$N = m g \cos \theta$

${F}_{r} = \mu \cdot N = \mu \cdot m g \cos \theta$

Resolving in the direction patrallel to the plane ↗^+

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

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

$= \frac{3500}{9.8 \left(\frac{7}{4} \cdot \cos \left(\frac{3}{8} \pi\right) + \sin \left(\frac{3}{8} \pi\right)\right)}$

$= 224.1 k g$

Apr 19, 2017

$m = 224.4 \text{ } k g$

#### Explanation:

$\text{Firstly,look into the animation several times to catch the situations that is given below.}$

• the vector of weight is perpendicular to the horizontal surface.
• The vector of weight can be splitted into two component
• The component that is perpendicular to the inclined plane causes a Frictional force.
• The yellow vector represent the react force that it has applied inclined plane.
• The component that is parallel to the inclined plane causes sliding down..
• A Tension occurs on rope (T)

$\textcolor{red}{\text{Weight of object (the red vector):} G = m \cdot g}$
$\textcolor{b l u e}{\text{the vertical component that is perpendicular to the inclined plane(the blue vector) :} {G}_{y} = m g \cdot \cos \theta}$
$\textcolor{g r e e n}{\text{the horizontal component that is parallel to the inclined plane(the green vector)} {F}_{x} = m g \cdot \sin \theta}$

$\text{The Frictional Force: } {F}_{f} = {\mu}_{k} \cdot {G}_{y}$
${F}_{f} = {\mu}_{k} \cdot m g \cdot \cos \theta$

$T \ge {F}_{f} + {F}_{x}$

$T \ge {\mu}_{k} \cdot m g \cdot \cos \theta + m g \cdot \sin \theta$

$3500 = m g \left({\mu}_{k} \cdot \cos \theta + \sin \theta\right)$

$m = \frac{3500}{g \left({\mu}_{k} \cdot \cos \theta + \sin \theta\right)}$

$m = \frac{3500}{9.81 \left(\frac{7}{4} \cdot 0.383 + 0.924\right)}$

$m = \frac{3500}{9.81 \cdot 1.59}$

$m = \frac{3500}{15.5970}$

$m = 224.4 \text{ } k g$