(a) If the horizontal acceleration produced briefly by an earthquake is a, and if an object is going to “hold its place” on the ground, show that the coefficient of static friction with the ground must be at least${\mu }_s=\frac{a}{g}$. (b) The famous Loma Prieta earthquake that stopped the 1989 World Series produced ground accelerations of up to$4\mathrm{m}/{\mathrm{s}}^2$in the San Francisco Bay Area. Would a chair have started to slide on a floor with coefficient of static friction 0.25?

In this problem, consider that the earthquake is moving the chair to the right d.

The force used to accelerate the chair will be a static frictional force and, along with this force, include the normal force and weight of the chair in the free body diagram.

Given data:

The kinetic coefficient of friction between the skier and the water is ${\mu }_s=0.25$.

The acceleration produced by the earthquake is $a=4\mathrm{m}/{\mathrm{s}}^2$.

The free body diagram of the skier is as follows:

The relation of static frictional force is given by:

$\begin{array}{l}{F}_f={\mu }_{s}N\\ F_f={\mu }_{s}mg\end{array}$

Here, N is the normal force, m is the mass, and g is the gravitational acceleration.

The relation of force from Newton’s second law in the x-direction is given by:

$\begin{array}{c}\Sigma F=ma\\ F_f=ma\end{array}$

On equating the above two relations, you get:

$\begin{array}{c}ma={\mu }_{s}mg\\ {\mu }_s=\left(\frac{a}{g}\right)\end{array}$

Thus, ${\mu }_s=\left(\frac{a}{g}\right)$is the required relation.

The relation to calculate the static coefficient of friction is given by:

${\mu }_s=\left(\frac{a}{g}\right)$

On equating the above two relations, you get:

$\begin{array}{l}{\mu }_s=\left(\frac{4\mathrm{m}/{\mathrm{s}}^2}{9.81\mathrm{m}/{\mathrm{s}}^2}\right)\\ {\mu }_s=0.40\end{array}$

Since the given value of static coefficient friction is ${\mu }_s=0.25$, which is less than the obtained value ${\mu }_s=0.40$, so it concludes that the chair will slide.

Thus, the chair would slide.