Problem 66 - College Physics

Question

A packing crate is placed on a \(20^{\circ}\) inclined plane. If the coefficient of static friction between the crate and the plane is \(0.65,\) will the crate slide down the plane if released from rest? Justify your answer.

Step-by-Step Solution

Answer

The crate will not slide down the plane.

Step 1: Understanding the Problem

The problem involves a crate on an inclined plane and we need to determine if it will slide down. The incline is at a 20-degree angle, and the coefficient of static friction is 0.65. We will compare the gravitational force component down the plane with the maximum static friction force to solve this.

Step 2: Calculate Force of Gravity Along the Incline

The force of gravity causing the crate to slide is given by the formula: \[ F_{gravity} = mg \sin(\theta) \]where \( m \) is the mass of the crate, \( g \) is the acceleration due to gravity \(9.8 \text{ m/s}^2 \), and \( \theta = 20^{\circ} \).

Step 3: Calculate Maximum Static Friction Force

The maximum static friction force that can act on the crate is given by:\[ F_{friction} = \mu_s \cdot N \]where \( \mu_s = 0.65 \) is the coefficient of static friction and \( N = mg \cos(\theta) \) is the normal force. So,\[ F_{friction} = 0.65 \cdot mg \cos(20^{\circ}) \].

Step 4: Set Forces in Balance

For the crate to not slide, the static friction force should be equal to or greater than the gravitational component:\[ 0.65 \cdot mg \cos(20^{\circ}) \geq mg \sin(20^{\circ}) \].

Step 5: Solve the Inequality

Dividing both sides of the inequality by \( mg \) gives:\[ 0.65 \cdot \cos(20^{\circ}) \geq \sin(20^{\circ}) \].Calculate both sides to see if the inequality holds. Using a calculator, we get:\( 0.65 \cdot \cos(20^{\circ}) \approx 0.6101 \)and \( \sin(20^{\circ}) \approx 0.3420 \).Since \( 0.6101 \geq 0.3420 \), the crate will not slide.

Key Concepts

Static Friction

Static friction is the force that keeps an object at rest when it is on a surface. It is characterized by the coefficient of static friction, represented by \( \mu_s \). This coefficient is a unitless number that quantifies how much frictional force exists between two static surfaces. In our exercise, \( \mu_s \) is 0.65 for the crate and the inclined plane.
A higher \( \mu_s \) means greater resistance to the start of motion. The static friction can adjust up to a maximum limit to counteract forces attempting to make the object move. This maximum is the product of \( \mu_s \) and the normal force \( N \), given by:

• \( F_{\text{friction max}} = \mu_s \cdot N \)

In this particular problem, we calculated the maximum static friction to determine if the crate would stay put or slide down the plane. Understanding static friction can help predict and control the motion of objects on inclined surfaces.

Inclined Plane

An inclined plane is a flat surface tilted at an angle relative to the horizontal. It allows objects to be lifted or lowered more easily compared to lifting them vertically. In our scenario, the inclined plane is tilted at a 20-degree angle.
The main forces acting on an object on an inclined plane are gravity, friction, and the normal force. The angle of the inclined plane affects both the gravitational force's components parallel and perpendicular to the surface. Here, the angle is used to calculate these components through trigonometric functions:

• Parallel to the plane: \( F_{\text{gravity parallel}} = mg \sin(\theta) \)
• Perpendicular to the plane: \( F_{\text{gravity perpendicular}} = mg \cos(\theta) \)

These components directly determine whether an object will remain static or start sliding down the incline. The inclined plane makes it crucial to understand how angles impact forces in physics problems.

Force of Gravity

The force of gravity is a fundamental force that pulls objects towards the Earth. It acts on the crate on the inclined plane, causing it to slide down.
For objects on an incline, gravity is split into two components: one parallel and one perpendicular to the plane. The parallel component, \( F_{\text{gravity parallel}} = mg \sin(\theta) \), tries to pull the object down the slope. This component is counteracted by static friction when the object is at rest.
The perpendicular component, \( F_{\text{gravity perpendicular}} = mg \cos(\theta) \), contributes to the normal force but doesn't move the object along the plane. Understanding these components is key because they help evaluate whether static friction will be sufficient to prevent sliding.

Normal Force

The normal force is a support force exerted upon an object in contact with another stable surface. It's always perpendicular to the contact surface.
For the crate on an inclined plane, the normal force counters the force of gravity's perpendicular component. It is crucial for calculating static friction's maximum value as it relies on the normal force:

• \( N = mg \cos(\theta) \)

In our exercise, the normal force is a result of gravity acting on the crate and can be used to calculate the force that resists the downward motion, which is static friction. By understanding normal force, it becomes evident how the plane’s angle impacts the frictional interaction between surfaces.