Chapter 5.2, Problem 31SSC

To determine

To Write: Conclusion about coefficients of kinetic and static friction.

Explanation of Solution

Given:

Mass of the table, $m=13\text{kg}$

Applied horizontal force when table does not move, $F_s=20\text{N}$

Applied horizontal force applied when table accelerates, $F=25\text{N}$

Acceleration of the table, $a=0.26{\text{m/s}}^2$

Formula Used:

By Newton’s second law:

$F=ma$

Here, m is the mass and a is the acceleration.

If f is the frictional force, then

$f=\mu N$

Here, $\mu$ is the coefficient of friction and N is the normal force.

$N=mg$

Calculations:

If the applied force is greater than frictional force then the body will move else it will not.

20N is less than f

$\begin{array}{l}20<f\\ 20<{\mu }_s\times 13\times 9.8\\ 0.15<{\mu }_s\end{array}$

${\mu }_s$ is the coefficient of static friction.

$\begin{array}{l}25>f\\ 25>{\mu }_{k}13\times 9.8\\ 0.19>{\mu }_k\end{array}$

  ${\mu }_s>{\mu }_k$ always

So, from above analysis

$0.19>{\mu }_k$

$0.15<{\mu }_s$

Conclusion:

Thus, static friction lies between 0.15 to 0.19 and kinetic friction is less than that.