Chapter 12, Problem 74P

To determine

ToCalculate: Whether there are any values for the coefficients of static friction.

Answer to Problem 74P

There are values for the coefficients of static friction as below.

  ${\mu }_{s,beam-floor}=0.58$

  ${\mu }_{s,beam-floor}=0.14$

Explanation of Solution

Given information :

Length of the beam $=20\text{cm}$

Radius of the cylinder $=4.0\text{cm}$

Mass of the beam $=5.0\text{kg}$

Mass of the cylinder $=8.0\text{kg}$

The coefficient of static friction between beam and cylinder $=0$

The coefficients of static friction between the cylinder and the floor, and between the beam and the floor, are not zero.

Formula used :

Weight of a body can be obtained by:

  $F=mg$

Where, m is the mass and g is the acceleration due to gravity.

Friction force is:

  $f_s={\mu }_s{F}_n$

Where, ${\mu }_s$ is the coefficient of static friction and $F_n$ is the normal force.

Calculation:

The forces that act on the beam are its weight $=mg$

The force of the cylinder acting along the radius of the cylinder $=F_c$

The normal force of the ground $=F_n$

Friction force $f_s={\mu }_s{F}_n$

The forces acting on the cylinder are its weight $=Mg$

The force of the beam on the cylinder acting radially inward $F_{cb}=F_c$

The normal force of the ground on the cylinder $=F_{nc}$

The force of friction $f_{sc}={\mu }_{sc}{F}_{nc}$

Choose the coordinate system as in the below given figure and apply the conditions for rotational and translational equilibrium.

  

Express ${\mu }_{s,beam-floor}$ in terms of $f_s$ and $F_n$ :

  ${\mu }_{s,beam-floor}=\frac{f_s}{F_n}$

Express ${\mu }_{s,cylinder-floor}$ in terms of $f_{sc}$ and $F_{nc}$ :

  ${\mu }_{s,cylinder-floor}=\frac{f_{sc}}{F_{nc}}$

Apply ${\displaystyle \sum \overrightarrow{\tau }}=0$ about an axis through the right end of the beam.

  $\left[\left(10\text{cm}\right)\cos \theta \right]mg-\left(15\text{cm}\right)F_c=0$

Solve for $F_c$ to obtain,

  $F_c=\frac{\left[\left(10\text{cm}\right)\cos \theta \right]mg}{15\text{cm}}$

  $\begin{array}{l}{F}_c=\frac{\left[10\cos {30}^{\circ }\right]\left(5.0\text{kg}\right)\left(9.81\text{m}\cdot {\text{s}}^{\text{-2}}\right)}{15}\\ F_c=28.3\text{N}\end{array}$

Apply ${\displaystyle \sum F_y=0}$ to the beam,

  $F_n+F_{c}cos\theta -mg=0$

  $F_n=mg-F_c\cos \theta$

  $\begin{array}{l}{F}_n=\left(5.0\text{kg}\right)\left(9.81\text{m}\cdot {\text{s}}^{\text{-2}}\right)-\left(28.3\text{N}\right)\cos {30}^{\circ }\\ F_n=24.5\text{N}\end{array}$

Apply ${\displaystyle \sum F_x=0}$ to the beam

  $\begin{array}{l}{f}_s+F_{c}cos\left(90°-\theta \right)=0\\ f_s\text{ = }{F}_{c}cos\left(90°-\theta \right)\end{array}$

  $\begin{array}{l}{f}_s=F_c\cos \left({90}^{\circ }-\theta \right)=\left(28.3\text{N}\right)\cos {60}^{\circ }\\ f_s=14.2\text{N}\end{array}$

  ${\overrightarrow{F}}_{cb}$ is the reaction force to ${\overrightarrow{F}}_c$

  $F_{cb}=F_c=28.3\text{N}$ radially inward.

Apply ${\displaystyle \sum F_y=0}$ to the cylinder:

  $F_{nc}-F_{cb}\cos \theta -Mg=0$

Solve for $F_{nc}$ to obtain,

  $F_{nc}=F_{cb}\cos \theta +Mg$

  $\begin{array}{l}{F}_{nc}=\left(28.3\text{N}\right)\cos {30}^{\circ }+\left(8.0\text{kg}\right)\left(9.81{\text{ms}}^{\text{-2}}\right)\\ F_{nc}=103\text{N}\end{array}$

Apply ${\displaystyle \sum F_x=0}$ to the cylinder,

  $f_{sc}-F_{cb}\cos \left({90}^{\circ }-\theta \right)=0$

  $\begin{array}{l}{f}_{sc}=F_{cb}\cos \left({90}^{\circ }-\theta \right)\\ f_{sc}=28.3{\text{N cos 60}}^{\circ }\\ f_{sc}=14.2\text{N}\end{array}$

  $\begin{array}{l}{\mu }_{s,beam-floor}=\frac{14.2\text{N}}{24.5\text{N}}\\ {\mu }_{s,beam-floor}=0.58\end{array}$

And

  $\begin{array}{l}{\mu }_{s,cylinder-floor}=\frac{14.2\text{N}}{103\text{N}}\\ {\mu }_{s,beam-floor}=0.14\end{array}$

Conclusion:

There are values for the coefficients of static friction. They are given as below.

  ${\mu }_{s,beam-floor}=0.58$

  ${\mu }_{s,beam-floor}=0.14$