# An object with a mass of 7 kg is on a ramp at an incline of pi/4 . If the object is being pushed up the ramp with a force of  8 N, what is the minimum coefficient of static friction needed for the object to remain put?

Aug 13, 2018

${\mu}_{s} \approx 0.835$

#### Explanation:

As seen in the free-body diagram, there are four forces acting on the object.

• Gravitational Pull $W = m \cdot g = 68.67 \textcolor{w h i t e}{l} \text{N}$;
• Normal Force $N$;
• Static Friction;
• The externally-applied force, $\textcolor{p u r p \le}{F}$ acting on the object upwards along the slope.

Consider forces acting on the object in two perpendicular directions: normal to the slope $y$ and along the slope $x$.

Forces acting normal to the slope include

• the normal force, ${N}_{y} = \text{N}$
• a portion of the gravitation pull, ${W}_{y} = W \cdot \cos \left(\frac{\pi}{4}\right) = 48.56 \textcolor{w h i t e}{l} \text{N}$

The object experiences no acceleration on the direction normal to the slope as it stays in contact with the ramp. Thus the net force the object experiences on the direction normal to the slope shall equal zero.

$\Sigma {F}_{y} = {N}_{y} - {W}_{y} = 0$
${N}_{y} = {W}_{y} = 48.56 \textcolor{w h i t e}{l} \text{N}$

Similarly, forces acting along the slope include

• the static friction, ${f}_{s}$ ($f$ in the right-hand diagram)
• a portion of the gravitational pull, ${W}_{x} = W \cdot \sin \left(\frac{\pi}{4}\right) = 48.56 \textcolor{w h i t e}{l} \text{N}$
• the force seeking to push the object up along the ramp, $\textcolor{p u r p \le}{F} = 8 \textcolor{w h i t e}{l} \text{N}$

Let the direction up along the slope be the positive direction of ${F}_{x}$, the sign in front of ${f}_{s}$ would be positive as it acts in the same direction of $\textcolor{p u r p \le}{F}$; otherwise the sign shall be negative.

The net force along the slope would also be zero since the box stays still and experiences no movement in that direction. Thus

$\Sigma F = \textcolor{p u r p \le}{{F}_{x}} \boldsymbol{\textcolor{n a v y}{+}} {f}_{s} - {W}_{x} = 0$

$8 \textcolor{w h i t e}{l} \text{N" + f_s = 48.56 color(white)(l) "N}$
${f}_{s} = \left(48.56 - 8\right) \textcolor{w h i t e}{l} \text{N}$

The static friction coefficient ${\mu}_{s}$ determines the maximum static friction the contact surface can possibly deliver under normal force $N$:

${f}_{s} \le {f}_{\max} = {\mu}_{s} \cdot N$

Rearranging gives

${\mu}_{s} = {f}_{\max} / N \ge {f}_{s} / N$

${\mu}_{s} \ge {f}_{s} / N$

$\textcolor{w h i t e}{{\mu}_{s}} = \left(\left(48.56 - 8\right) \textcolor{w h i t e}{l} \text{N") / (48.56 color(white)(l) "N}\right)$

$\textcolor{w h i t e}{{\mu}_{s}} \approx 0.835$

Therefore the minimum value of the static frictional coefficient would be

${\mu}_{s} \approx 0.835$