91影视

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\(a = 4\;{\rm{m/}}{{\rm{s}}^2}\).

The free body diagram of the skier is as follows:

The relation of static frictional force is given by:

\(\begin{aligned}{F_f} &= {\mu _s}N\\{F_f} &= {\mu _s}mg\end{aligned}\)

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

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

\(\begin{aligned}\Sigma F &= ma\\{F_f} &= ma\end{aligned}\)

On equating the above two relations, you get:

\(\begin{aligned}ma &= {\mu _s}mg\\{\mu _s} &= \left( {\frac{a}{g}} \right)\end{aligned}\)

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:

\(\begin{aligned}{\mu _s} &= \left( {\frac{{4\;{\rm{m/}}{{\rm{s}}^2}}}{{9.81\;{\rm{m/}}{{\rm{s}}^2}}}} \right)\\{\mu _s} &= 0.40\end{aligned}\)

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.