Question

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches \(28.0^{\circ}\), the box starts to slip and slides \(2.53 \mathrm{~m}\) down the plank in \(3.92 \mathrm{~s}\). Find the coefficients of friction.

Short answer

The coefficients of static and kinetic friction are approximately 0.53 and 0.49 respectively.

Step-by-step solution

Determination of static friction

First, find the coefficient of static friction. This can be determined by equating the gravitational force component acting parallel to the incline to the force of static friction. The force due to gravity acting down the incline is \( mg\sin{\theta} \) and the maximum static friction force is \( \mu_{s} mg\cos{\theta} \). At the point of slip, these two forces are equal. Solving \( \mu_{s} = \tan{\theta} \) for \(\mu_{s}\) and substituting \(\theta = 28.0^{\circ}\), we get \(\mu_{s} \approx 0.53\).

Determination of kinetic friction

Next, find the coefficient of kinetic friction. When the box is sliding, the force of kinetic friction balances the component of gravitational force along the incline. These must be equal for constant velocity. We can use the equation of motion that relates distance, time, initial velocity and acceleration to find the acceleration: \( x = ut + \frac{1}{2} a t^2 \), where \( u \) is initial velocity (zero in this case), \( x \) is distance slid (2.53m), and \( t \) is time (3.92s). Solving for acceleration, we get \( a \approx 0.327 m/s^2 \). Equating the net force \( mg\sin{\theta} - \mu_{k}mg\cos{\theta} \) to \( ma \), and solving for \( \mu_{k} \), we find that \( \mu_{k} \approx 0.49 \).

Solution

The coefficients of static and kinetic friction between the box and the plank are approximately \( \mu_{s} = 0.53 \) and \( \mu_{k} = 0.49 \), respectively.

Key concepts

Static Friction

Static friction is the force that keeps an object at rest when it is placed on a surface. Think of it as a kind of "glue" that holds the object in place until you apply enough force to break this bond. When the student tilted the plank, she was testing the static friction between the box and the plank. The angle at which the box began to slide, 28 degrees in this scenario, is critical. This angle tells us that the component of the box's weight acting down the plank matched the maximum static friction force.

• The formula used is: \( \mu_{s} = \tan{\theta} \).
• \( \theta \) is the angle of inclination.
• For the box to start sliding, \( \theta = 28^{\circ} \).
• Calculating with \( \tan{28^{\circ}} \) gives us the static friction coefficient, approximately 0.53.

Static friction varies depending on the materials and their surface roughness. It is usually higher than kinetic friction because it takes more force to start moving an object than to keep it moving.

Kinetic Friction

Once the box started moving, it was under the influence of kinetic friction. Kinetic friction acts on moving objects, opposing their motion. It's generally less than static friction because overcoming initial resistance (static) requires more effort than keeping something in motion.

• In this scenario, kinetic friction worked against the gravity component along the plank.
• To find the kinetic friction, you first need the acceleration: \( a = \frac{2x}{t^2} = 0.327 \text{ m/s}^2 \), where \( x = 2.53 \text{ m} \) and \( t = 3.92 \text{ s} \).
• The forces in balance are: \( mg\sin{\theta} = \mu_{k}mg\cos{\theta} + ma \).
• Rearranging and solving gives \( \mu_{k} \approx 0.49 \).

The coefficient of kinetic friction, which is about 0.49, tells us how the box's motion was resisted once it started sliding down the plank.

Inclined Plane

An inclined plane is just a flat surface tilted at an angle relative to the horizontal, like the plank in this exercise. It makes it easier to lift heavy loads by dispersing the force needed over a longer distance.

• For objects on an incline: part of their weight causes a downward force along the plane, defined by \( mg\sin{\theta} \).
• The normal force, acting perpendicular to the surface, is \( mg\cos{\theta} \).
• This scenario’s inclined plane demonstrates how different forces interact: gravity pulls the box down, while friction fights against this motion.

The angle of the inclined plane, 28 degrees here, played a crucial role in overcoming static friction and determining the moment the box began to move. Inclined planes are simple machines that help us understand fundamental physics forces.

Newton's Laws of Motion

Newton's Laws of Motion are fundamental to understanding mechanics. They describe how objects move and interact with forces.

• The first law, inertia, explains that the box remains still until the tilted angle supplies enough force to overcome static friction.
• According to the second law, \( F = ma \), the net force acting on the box once it started moving produced an acceleration, calculated as 0.327 m/s².
• The third law states that every action has an equal and opposite reaction, so the force of friction opposing motion plays a critical role in determining motion.

These laws explain why static and kinetic frictions behave differently. They also help establish the relationships between forces acting on the inclined plane, allowing us to calculate the coefficients of friction accurately. Understanding Newton's laws gives us insight into why the box initially stayed put and how it eventually moved.