What force is needed to slide a 250-kg crate across the floor at constant velocity if the coefficient of sliding friction between a crate and a...

Sunday, November 23, 2014

What force is needed to slide a 250-kg crate across the floor at constant velocity if the coefficient of sliding friction between a crate and a...

Hello!

Newton's Second law will help us. It states that the (vector) sum of all forces acting on a body is the same as its mass multiplied by its acceleration, $\displaystyle{\vec{{F}}}={m}{\vec{{a}}}.$

There are four forces here: gravitational force $\displaystyle{m}{g{}}$ downwards, reaction force $\displaystyle{N}$ upwards, traction force $\displaystyle{F}_{{T}}$ horizontally and friction $\displaystyle{F}_{{f{}}}$ also horizontally but in reverse direction.

The acceleration is zero, because velocity is constant. So we obtain

$\displaystyle{m}{\vec{{{g}}}}+{\vec{{N}}}+{\vec{{{F}_{{T}}}}}+{\vec{{{F}_{{f}}}}}={\vec{{0}}}.$

Vertical forces must balance each other and horizontal, too, so in magnitudes we obtain

$\displaystyle{N}={m}{g{}}$ and $\displaystyle{F}_{{T}}={F}_{{f{.}}}$

Also we know that $\displaystyle{F}_{{f{=}}}\mu{N},$ where $\displaystyle\mu$ is the coefficient of sliding friction. Hence $\displaystyle{F}_{{T}}={F}_{{f{=}}}\mu{N}=\mu{m}{g{.}}$ This is the final formula, and numerically $\displaystyle{F}_{{T}}\approx{0.25}\cdot{250}\cdot{9.8}={612.5}{\left({N}\right)}.$