Problem 39 - College Physics

Question

A \(25.0 \mathrm{~kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .\) (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid \(5.0 \mathrm{~m}\) along the loading ramp?

Step-by-Step Solution

Answer

(a) The box starts to slip at \( 19.29^\circ \). (b) The acceleration is \( 1.01 \ m/s^2 \). (c) The box's speed after sliding 5 m is \( 3.19 \ m/s \).

Step 1: Understanding the forces involved

The box is acted upon by three main forces: gravitational force (directed vertically downward), normal force (perpendicular to the ramp), and frictional force (opposing the motion along the ramp). The gravitational force can be broken down into components parallel and perpendicular to the ramp's surface.

Step 2: Setup the equations for static friction

The force of static friction can be described by \( f_s = \mu_s N \), where \( \mu_s = 0.35 \) is the coefficient of static friction, and \( N \) is the normal force. Normal force is given by \( N = mg \cos(\alpha) \). To find the angle where the box starts to slip, set the maximum static friction equal to the component of weight parallel to the ramp: \( mg \sin(\alpha) = \mu_s mg \cos(\alpha) \).

Step 3: Calculate the minimum angle for slipping

Dividing both sides of the friction equation by \(mg\) and rearranging, we get \( \tan(\alpha) = \mu_s \). Now solve for \( \alpha \): \( \alpha = \tan^{-1}(0.35) \). Calculating gives \( \alpha \approx 19.29^\circ \).

Step 4: Setup the equations for kinetic friction

Once the box starts moving, kinetic friction comes into play. The force of kinetic friction is given by \( f_k = \mu_k N \), with \( \mu_k = 0.25 \). Substitute \( N = mg \cos(\alpha) \) to get \( f_k = 0.25 \cdot mg \cos(\alpha) \).

Step 5: Calculate the net force and acceleration

The net force acting on the box down the ramp when it is moving can be described as \( F_{net} = F_{gravity} - f_k = mg \sin(\alpha) - \mu_k mg \cos(\alpha) \). Therefore, the acceleration is \( a = \frac{F_{net}}{m} = g (\sin(\alpha) - \mu_k \cos(\alpha)) \). Substitute \( \alpha = 19.29^\circ \) and \( g = 9.81 \ m/s^2 \) to find \( a \approx 1.01 \ m/s^2 \).

Step 6: Calculate the final velocity after sliding 5.0 m

Use the kinematic equation \( v^2 = u^2 + 2a s \), where initial velocity \( u = 0 \), \( s = 5.0 \ m \), and \( a \approx 1.01 \ m/s^2 \). Solving for final velocity \( v \), we find \( v = \sqrt{2 \cdot 1.01 \cdot 5.0} \approx 3.19 \ m/s \).

Key Concepts

Static Friction

Static friction is the force that keeps an object at rest when it is placed on a surface. It prevents the object from sliding down an inclined plane. This frictional force adjusts itself up to a maximum limit to counteract any external forces trying to move the object.
Understanding static friction is crucial when solving problems involving objects that might start slipping or moving. The equation for static friction is given by:

• \( f_s = \mu_s N \)

Here, \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force.
The static frictional force is what holds the box in place when placed on an inclined plane. As the inclination angle \( \alpha \) increases, this force has to overcome the component of gravitational force trying to pull the box down the incline.
Once this static threshold is surpassed, the object will begin to slide. In our example, the static friction helps determine the exact angle at which slipping begins, calculated using \( \tan(\alpha) = \mu_s \).

Normal Force

The normal force is a key concept in understanding motion on an inclined plane. It is the force exerted by a surface that supports the weight of an object resting on it. This force acts perpendicular (normal) to the contact surface, hence the name.

• The normal force can be calculated using the equation: \( N = mg \cos(\alpha) \)

Where \( m \) is the mass of the object, \( g \) is the gravitational acceleration, and \( \alpha \) is the angle of the incline.
In our context, the normal force influences both static and kinetic friction calculations. It affects how much friction is experienced because the frictional force is directly proportional to the normal force.
Understanding the normal force helps us analyze how the pressure from the weight of an object gets distributed on an inclined surface. Without a clearly defined normal force, we wouldn't be able to accurately calculate frictional forces acting on the box of textbooks.

Gravitational Force

Gravitational force, often simply referred to as weight, is the force exerted by the Earth to pull objects down towards its center. For an object on an inclined plane, this force can be split into two components:

• Parallel to the plane: \( mg \sin(\alpha) \)
• Perpendicular to the plane: \( mg \cos(\alpha) \)

These components are essential when analyzing motion on inclines.
The parallel component \( mg \sin(\alpha) \) causes the object to slide down the plane, while the perpendicular component affects the normal force exerted by the surface.
In the context of our example, determining the balance between these gravitational components and static friction is crucial for finding the angle \( \alpha \) at which the box will start to slip. It's also fundamental in calculating the net force and thus the resulting acceleration once the box begins to move.

Inclined Plane

An inclined plane is a flat surface tilted at an angle, different from being horizontal. It is a classic physics problem due to the multiple forces interacting when an object is placed on it.

• The angle of the incline \( \alpha \) determines how much of the gravitational force acts to slide the object down, opposed by friction.

Understanding inclined planes helps break down the effects of gravitational force into manageable components.
When dealing with problems on inclined planes, it is important to calculate angles precisely, especially when assessing conditions for static and kinetic friction. Inclined planes provide a practical example of how these forces interplay to determine whether an object will remain static or start sliding.
In our specific problem, the inclined plane made it necessary to analyze how the forces change as the angle \( \alpha \) increases and to interpret how these conditions affect the motion and speed of the box.