# A horizontal force of 26.3 N is necessary to push a 99 N wooden box across a tile floor at constant velocity. How do you find the coefficient of sliding friction between the two surfaces?

May 30, 2017

The answer I got was $0.27 = \mu$

#### Explanation:

There is some force of $26.3 \text{ N}$ acting on the object in the $\textcolor{b l u e}{x}$ direction and a frictional force that directly opposes the motion of the box, ${f}_{k}$. Since the box is not accelerating (constant velocity), there is no net force.

Since we know that the net force is $0$, then we also know that the two opposing forces are equal to each other.

$\sum {\vec{F}}_{x} = m a = 0$

therefore,

${F}_{x} = {f}_{k}$

$26.3 N = {f}_{k}$

We also know that frictional force is calculated as follows

${f}_{k} = \mu \cdot N$

Where

$\text{mu = coefficient of kinetic friction}$
$\text{N = normal force (N)}$

Knowing this we can rewrite the equation for the forces acting on the box in the $\textcolor{b l u e}{x}$ direction as

color(white)(aaaaaaaaaaaaaaaaaa)color(magenta)(26.3 N = mu*N

in order to solve for the coefficient of kinetic friction.

But how do we figure out the Normal force?

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See the diagram? Look at the $\textcolor{red}{y}$ components. The forces acting on the object in the $\textcolor{red}{y}$ direction are the force of gravity acting downwards and the Normal force acting upwards. Since the object is not accelerating in the $\textcolor{red}{y}$ direction, we too can set the force of gravity equal to the Normal force.

${F}_{\text{gravity") = F_("Normal}}$

And since the weight of the object was given to us, we have figured out the Normal force.

${F}_{\text{gravity") = F_("Normal}}$

$99 N = {F}_{\text{Normal}}$

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Now, back to our original equation that we were working on

color(white)(aaaaaaaaaaaaaaaaaa)color(magenta)(26.3 N = mu*N

We can plugin $99 N$ for the Normal force and solve for the coefficient of kinetic friction.

color(magenta)(26.3 N = mu*N

$\frac{26.3 N}{99 N} = \mu$

$0.27 = \mu \left(\text{the coefficient of kinetic friction is unitless}\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Answer: 0.27 = mu}}$