Question

A package rests on the back seat of your car. The coefficient of friction between the seat and the package is 0.24. Assuming you drive on a level road, what is the maximum acceleration your car can have if the package is to remain in place relative to your back seat?

Answer 1

In car's frame of referance (non-inertial), the second Newton's law in horizontal direction has the following view:

$$ma = F_{fr}-mA$$

where $a$ is the acceleration of the box and $A$ is the acceleration of the car, and $m$ is the mass of the box. $F_{fr}$ is the frictional force. While $F_{fr}\ge mA$ the acceleration of the box is equal to and the box does not move. Thus, the maximum acceleration of the car is:

$$A = \dfrac{F_{fr}}{m}$$

By definition, the frictional force is:

$$F_{fr} = \mu N$$

where $\mu = 0.24$ is the coefficient of friction between the seat and the package, and $N$ is the normal force on the package. Since there is no vertical acceleration, $N = mg$ ($g = 9.81m/s^2$ ). Thus, obtain

$$F_{fr} = \mu mg$$

and

$$A = \dfrac{F_{fr}}{m} = \dfrac{\mu mg}{m} = \mu g$$

Finally, obtain:

$$A = 0.24\cdot 9.81m/s^2 \approx 2.35m/s^2$$

Answer. 2.35 m/s^2.