Problem 31 - College Physics

Question

\(\bullet\) A box of bananas weighing 40.0 \(\mathrm{N}\) rests on a horizontal sur- face. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is \(0.20 .\) (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a hor- izontal force of 6.0 \(\mathrm{N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

Step-by-Step Solution

Answer

a) 0 N, b) 6.0 N, c) 16.0 N, d) 8.0 N.

Step 1: Understanding Static Friction

Static friction acts to counteract any applied force up to its maximum limit before the box begins to move. The maximum static friction force is calculated using the formula \( f_{s, ext{max}} = \mu_s imes N \), where \( \mu_s = 0.40 \) is the coefficient of static friction and \( N = 40.0 \text{ N} \) is the normal force (equal to the weight of the box). Therefore, \( f_{s, ext{max}} = 0.40 \times 40.0 \text{ N} = 16.0 \text{ N} \).

Step 2: Calculating Friction Force Without Applied Force

When no horizontal force is applied, the box remains at rest, and the static friction force is zero, as no external force is attempting to move the box. Thus, the friction force is \( 0 \text{ N} \).

Step 3: Calculating Friction Force With 6.0 N Applied Force

When a force of 6.0 N is applied horizontally, the static friction force will match this force (as long as it doesn't exceed the maximum static friction) to keep the box stationary. Since \( 6.0 \text{ N} \leq 16.0 \text{ N} \), the static friction force exerted is \( 6.0 \text{ N} \).

Step 4: Minimum Force to Initiate Motion

To start moving the box, an applied force must exceed the maximum static friction force. Therefore, the minimum force required to initiate motion is equal to the maximum static friction: \( 16.0 \text{ N} \).

Step 5: Minimum Force to Maintain Constant Velocity

Once the box is moving, kinetic friction takes over. The kinetic friction force is given by \( f_k = \mu_k \times N \), where \( \mu_k = 0.20 \). Thus, \( f_k = 0.20 \times 40.0 \text{ N} = 8.0 \text{ N} \). This is the force needed to maintain constant velocity.

Key Concepts

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces that are in contact and at rest. It acts in response to an applied force and works to keep the object stationary. This force has a maximum limit before motion occurs, known as the maximum static friction force, or \( f_{s, \text{max}} \). The formula to calculate this maximum force is:

• \( f_{s, \text{max}} = \mu_s \times N \)

Here, \( \mu_s \) is the coefficient of static friction, which is a measure of how much static friction exists between the two surfaces, and \( N \) is the normal force, typically equal to the object's weight when on a horizontal surface. If you apply a force less than \( f_{s, \text{max}} \), static friction will simply balance it, preventing motion.
In the case of the box of bananas, the maximum static friction force is \( 16.0 \) N. As long as any applied force is below this value, the box will not move.

Kinetic Friction

Kinetic friction comes into play once an object has overcome static friction and is in motion. This type of friction acts in the opposite direction of motion, resisting the movement of the object. It is typically less than static friction, which is why it is easier to keep an object moving than to start it. The kinetic friction force can be calculated with the formula:

• \( f_k = \mu_k \times N \)

In this equation, \( \mu_k \) is the coefficient of kinetic friction. For the box of bananas, the kinetic friction force is \( 8.0 \) N, calculated using \( 0.20 \times 40.0 \) N. This implies that once the box is in motion, a continuous force of 8.0 N will maintain its constant velocity; any less force will result in deceleration, while more will cause acceleration.

Normal Force

Normal force is a crucial concept when analyzing problems involving friction. It is the perpendicular force exerted by a surface to support the weight of an object resting on it. In many cases, especially simple horizontal surfaces, the normal force is equal to the gravitational force acting on the object, which means it equals the object's weight.The significance of this force is that it directly relates to the frictional forces: static and kinetic friction. For the box of bananas resting on a horizontal floor, the normal force is \( 40.0 \) N, which is the weight of the box. Thus, every calculation involving friction coefficients depends on this normal force value.

Coefficient of Friction

The coefficient of friction is a dimensionless number that represents the degree of interaction between two surfaces. It's derived from experiments and helps quantify how "slippery" or "grippy" a surface combination is. Each type — static or kinetic — has its own coefficient value, denoted as \( \mu_s \) for static friction and \( \mu_k \) for kinetic friction.

• For static friction, the coefficient determines the extent of force needed to initiate motion.
• For kinetic friction, it dictates the force needed to maintain motion at a constant speed.

In our example with the box, \( \mu_s = 0.40 \) and \( \mu_k = 0.20 \). These values mean that more force is needed to start moving the box (static) than to keep it moving (kinetic). The coefficient values are integral to solving and understanding friction-dominated problems.