Question

A 25.0-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 \(\alpha\) 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 m along the loading ramp?

Short answer

(a) The box starts to slip at \( \alpha \approx 19.3^\circ \). (b) The acceleration is approximately \( 1.32 \, \text{m/s}^2 \). (c) The box's speed after sliding 5 m is about \( 3.63 \, \text{m/s} \).

Step-by-step solution

Analyze Forces Acting on the Box

When the box is at the verge of slipping, the static friction force is at its maximum value. The forces acting are the gravitational force, normal force, and static friction. The gravitational force can be split into components parallel and perpendicular to the ramp. The force of static friction opposes the gravitational force component parallel to the ramp.

Calculate Static Friction and Normal Force

The force of static friction is given by \( f_{s} = \mu_{s} N \), where \( \mu_{s} = 0.35 \) is the coefficient of static friction, and \( N \) is the normal force. The normal force is \( N = mg \cos \alpha \). At the point of slipping, the static friction balances the downhill gravitational component: \( f_{s} = mg \sin \alpha \).

Determine the Minimum Angle for Sliding

Equate the maximum static friction force to the gravitational component parallel to the ramp: \( mg \sin \alpha = \mu_{s} mg \cos \alpha \). Simplifying gives \( \tan \alpha = \mu_{s} \). Calculate \( \alpha = \tan^{-1}(0.35) \). The minimum angle is \( \alpha \approx 19.3^\circ \).

Analyze Forces When Box is Moving

Once the box starts moving, the frictional force is kinetic. The net force causing acceleration is \( mg \sin \alpha - \mu_{k} N \), where \( \mu_{k} = 0.25 \).

Calculate the Acceleration

The net force can be written as \( F_{net} = ma = mg \sin \alpha - \mu_{k} mg \cos \alpha \). Solving for the acceleration \( a \), we use the earlier calculated angle: \( a = g(\sin \alpha - \mu_{k} \cos \alpha) \). Substituting \( g = 9.8 \, \text{m/s}^2 \), \( a \approx 1.32 \, \text{m/s}^2 \).

Calculate Final Velocity After Sliding 5 m

Use the kinematic equation \( v^2 = u^2 + 2as \), where initial velocity \( u = 0 \), acceleration \( a = 1.32 \, \text{m/s}^2 \), and displacement \( s = 5 \, \text{m} \). This gives \( v = \sqrt{0 + 2 \cdot 1.32 \cdot 5} \), which results in \( v \approx 3.63 \, \text{m/s} \).

Key concepts

Static Friction

Static friction is the force that keeps an object stationary even when there is a force trying to move it. Imagine a heavy box on a ramp. It doesn’t start sliding immediately because static friction is acting on it. This force is directly related to two things: the normal force (the perpendicular force the surface exerts on the object) and the coefficient of static friction, which describes how "sticky" the surface is.
The formula for static friction at its maximum is given by:

• \( f_{s} = \mu_{s} N \)
• Where \( \mu_{s} = 0.35 \) in our exercise
• \( N = mg \cos \alpha \), where \( \alpha \) is the angle of the incline

Static friction must counteract any forces trying to make the object slide downward. But once the necessary force to overcome this friction is applied, the object begins to move, transitioning to kinetic friction.

Kinetic Friction

Once an object begins moving, static friction switches to kinetic friction. Kinetic friction is usually less than static friction, which means that once an object starts sliding, it has less resistance to keep moving. In our example, once the box overcomes static friction on the inclined plane, it experiences kinetic friction instead.
The formula for kinetic friction is similar to static friction:

• \( f_{k} = \mu_{k} N \)
• Where \( \mu_{k} = 0.25 \) in the exercise

This means the box will continue to be affected by kinetic friction as it slides down the ramp. This force works against the motion of the box, slightly slowing it down but not enough to stop it. This is why kinetic friction impacts how quickly the box accelerates down the incline.

Inclined Plane

An inclined plane is a flat surface tilted at an angle, like a ramp. Such planes are often used to easily analyze forces. When an object, like our box, is placed on an inclined plane, gravity causes it to slide down. However, forces like friction are crucial in understanding its motion.
The angle of inclination determines the plane's steepness. Its effect on motion is described by breaking down gravitational force into two components:

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

When you adjust this angle \( \alpha \) in our exercise, it changes the balance of forces. At a specific angle, the component of gravity pulling the box downward becomes greater than the maximum static friction holding it in place, causing the box to start moving.

Acceleration

Acceleration refers to the rate at which an object speeds up or slows down. When the box begins to slide down the ramp, its acceleration is influenced by the forces acting on it. Primarily, the net force along the inclined plane surface determines the box's acceleration.
The net force accelerating the box downwards is the difference between the component of gravity pulling it downward and the kinetic friction opposing this pull. The formula is:

• \( a = g(\sin \alpha - \mu_{k} \cos \alpha) \)

For our box on a 19.3° inclined plane, this acceleration measures approximately 1.32 m/s². This value tells us how quickly the box picks up speed as it travels down the slope.

Kinematics

Kinematics deals with the motion of objects and how you calculate aspects like velocity and displacement over time. In this particular exercise, once we know the acceleration of the box as it slides, we can use kinematic equations to determine its final velocity after a certain distance.
For the box sliding 5 meters down the ramp, the initial velocity \( u \) is 0 since it starts from rest. Using the kinematic equation:

• \(v^2 = u^2 + 2as\)
• Where \( v \) is the final velocity
• \( a = 1.32 \, \text{m/s}^2 \)
• \( s = 5 \, \text{m} \)

Plugging in these values, the final velocity \( v \) comes out to approximately 3.63 m/s. This indicates how fast the box is moving after sliding the specified distance.