Chapter 4, Problem 39P

To determine

The coefficient of static friction between the drawer containing the silverware and the cabinet.

Answer to Problem 39P

Solution:

The coefficient of static friction between the drawer containing the silverware and the cabinet is 0.46.

Explanation of Solution

When force is applied on two objects in contact with each other and the objects are at rest, then, they experience a force of static friction. This force acts tangentially on the surfaces in contact so as to oppose the motion between the objects. The weight of the object acts vertically downwards. According to Newton’s third law, the object experiences a reaction called the normal force. The force of friction $F_{fr}$ is found to be proportional to the reaction force $F_N$. The constant of proportionality is called the coefficient of static friction ${\mu }_s$.

When the applied force F on the object increases, the objects continue to be at rest until the applied force reaches the value of the force of static friction between the two surfaces. The objects will start to move when the applied force exceeds the value of the force of static friction. The limiting case results when the applied force becomes equal to the frictional force.

Given:

Mass of the silverware drawer $m=2.0\text{ kg}$

The magnitude of the applied force $F=9.0\text{ N}$.

Formula used:

      ${\mu }_s=\frac{F_{fr}}{F_N}$

Here,

      $F_{fr}$ is the friction force

      $F_N$ is the normal force

Calculation:

At the instant when the drawer opens suddenly spilling its contents, the applied force F becomes equal to the force of friction.

      $F_{fr}=F=9.0\text{ N}$

The normal force when the drawer is at is equal to its weight W. The weight of the drawer is given by,

      $W=mg$

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

Therefore, the normal force is given by,

      $F_N=W=mg$

Calculate the value of the coefficient of static friction by substituting 9.0 N for $F_{fr}$, 2.0 kg and 9.8 m/s² for g in the expression.

${\mu }_s=\frac{F_{fr}}{F_N}$

    $=\frac{F_{fr}}{mg}$

    $=\frac{9.0\text{ N}}{\left(2.0\text{ kg}\right)\left(9.8{\text{ m s}}^{-2}\right)}$

    =0.4692

The coefficient of static friction between the drawer and the cabinet correct to 2 significant figures is found to be 0.46.