Inclined Planes: Videos & Practice Problems

Inclined plane problems involve analyzing forces acting on an object on a slope. Begin with a free body diagram, identifying weight (mg) and normal force, which acts perpendicular to the incline. Decompose the weight into components: mg sin θ (down the incline) and mg cos θ (perpendicular to the incline). The acceleration (a) down the slope is given by $g^2\sin \left(\theta \right)$, where g is the acceleration due to gravity. The normal force is expressed as $F=\mathrm{mg}\cos \left(\theta \right)$.

Intro to Inclined Planes

Inclined plane problems are fundamental in physics, particularly when analyzing the motion of objects on ramps. To effectively solve these problems, it is essential to start with a free body diagram, which visually represents all forces acting on the object. For instance, consider a 5-kilogram block resting on a frictionless incline angled at 37 degrees. The first step is to identify the weight force, represented as \( F_g = mg \), acting straight down. Since the surface is inclined, the normal force, which acts perpendicular to the surface, will not point straight up but rather at an angle relative to the incline.

Next, to analyze the motion of the block, it is crucial to adjust the coordinate system to align with the incline. This involves tilting the x-y plane so that the new x-axis runs parallel to the slope of the incline. In this tilted coordinate system, the weight force \( mg \) must be decomposed into two components: one parallel to the incline (\( F_{gx} \)) and one perpendicular to it (\( F_{gy} \)). The decomposition is done using trigonometric functions, but it is important to note that the components of the weight force behave differently than those of typical forces. Specifically, for inclined planes, the component of weight acting down the incline is given by \( F_{gx} = mg \sin(\theta) \), while the component acting into the incline is \( F_{gy} = mg \cos(\theta) \).

To find the acceleration of the block down the incline, we apply Newton's second law, \( F = ma \). Focusing on the x-axis, the only force acting is \( F_{gx} \), leading to the equation \( mg \sin(\theta) = ma_x \). By canceling the mass \( m \) from both sides, we derive the acceleration down the incline as \( a_x = g \sin(\theta) \). For a 37-degree incline, this results in an acceleration of approximately \( 5.9 \, \text{m/s}^2 \).

In the y-direction, since there is no acceleration (the block does not move off the ramp), the forces must balance. This means that the normal force \( F_N \) must equal the component of the weight acting perpendicular to the incline, leading to the equation \( F_N = mg \cos(\theta) \). Thus, the normal force can be expressed as \( F_N = mg \cos(\theta) \), which is crucial for understanding the forces acting on the block.

In summary, when solving inclined plane problems, it is vital to draw a free body diagram, adjust the coordinate system to align with the incline, decompose the weight force into its components, and apply Newton's laws to find the acceleration and normal force. This systematic approach allows for a clear understanding of the dynamics involved in inclined plane scenarios.

A 4.0-kg box sits on a frictionless inclined plane that makes a 21° angle with the horizontal. It is held in place by a cord parallel to the plane. Calculate the tension in the cord.

• A. 14 N
• B. 37 N
• C. 33 N
• D. 21 N

Answer: (A)

Runaway Ramp

In this problem, we analyze the scenario of a truck moving at an initial velocity of 18 meters per second on a runaway ramp with a 20% grade. The goal is to determine the length of the ramp required to bring the truck to a complete stop. The truck's final velocity will be 0 meters per second when it comes to rest.

To solve for the length of the ramp, denoted as Δx, we can utilize kinematic equations. We start by identifying the known variables: the initial velocity (v₀ = 18 m/s) and the final velocity (v_f = 0 m/s). The acceleration and time are unknown, which means we need to find one of these to proceed.

On an inclined plane, the acceleration can be calculated using the formula:

a = g \cdot \sin(\theta)

where g is the acceleration due to gravity (approximately 9.8 m/s²) and θ is the angle of the incline. To find θ from the given 20% grade, we convert the percentage to a decimal and then use the inverse tangent function:

θ = \tan^{-1}\left(\frac{20}{100}\right) = \tan^{-1}(0.2)

This calculation yields an angle of approximately 11.3 degrees.

Next, we can find the acceleration:

a = 9.8 \cdot \sin(11.3^\circ) \approx 1.92 m/s²

Since the truck is decelerating as it moves up the ramp, we assign a negative sign to the acceleration, resulting in a = -1.92 m/s².

With three known variables now (initial velocity, final velocity, and acceleration), we can use the kinematic equation that does not involve time:

v_f² = v₀² + 2aΔx

Substituting the known values:

0 = (18)² + 2(-1.92)Δx

This simplifies to:

0 = 324 - 3.84Δx

Rearranging gives:

3.84Δx = 324

Solving for Δx results in:

Δx = \frac{324}{3.84} \approx 84.375 m

Thus, the length of the runaway ramp required for the truck to come to a stop is approximately 84 meters.

Connected Block on a Ramp

In this scenario, we analyze a system of two connected objects on a frictionless incline. The first object, labeled as object A, is positioned on the inclined plane, while the second object, object B, hangs vertically. To understand the forces acting on these objects, we begin by constructing free body diagrams for each.

For object A, the forces include its weight, represented as \( m_A g \), acting straight down, and a normal force acting perpendicular to the inclined surface. Additionally, there is a tension force from the cable connected to object B. For object B, the forces are simpler, consisting of its weight \( m_B g \) acting downward and the tension force acting upward. Since both objects are connected by the same cable, the tension in the cable is the same for both objects.

To analyze the motion, we tilt the coordinate system so that the new y-axis aligns with the normal force acting on object A. We can choose the direction down the incline as positive. When released, object B will accelerate downward, causing object A to move upward along the incline.

Next, we decompose the gravitational force acting on object A into components parallel and perpendicular to the incline. The component of the gravitational force acting down the incline is given by \( m_A g \sin(\theta) \), where \( \theta \) is the angle of the incline.

We then apply Newton's second law, \( F = ma \), to both objects. For object B, the equation can be expressed as:

\[ m_B g - T = m_B a \]

Substituting the known values, we find that \( m_B = 5 \) kg leads to \( 49 - T = 5a \).

For object A, the equation becomes:

\[ T - m_A g \sin(\theta) = m_A a \]

With \( m_A = 2 \) kg and \( \theta = 53^\circ \), we calculate the gravitational component as \( m_A g \sin(53^\circ) \). This results in:

\[ T - 15.7 = 2a \]

Now, we have two equations with two unknowns (tension \( T \) and acceleration \( a \)). We can solve these equations simultaneously. By rearranging and substituting, we eliminate \( T \) and combine the equations:

\[ 49 - 15.7 = 7a \]

Solving for \( a \), we find:

\[ 33.3 = 7a \]

Thus, the acceleration \( a \) is:

\[ a = 4.8 \, \text{m/s}^2 \]

This value represents the magnitude of the acceleration of the system, confirming that the analysis of forces and motion in connected systems on inclined planes leads to a clear understanding of the dynamics involved.

Here’s what students ask on this topic:

What is the formula for calculating the acceleration of an object on an inclined plane?

The acceleration of an object on an inclined plane can be calculated using the formula:

$a=g\sin \theta$

where $a$ is the acceleration down the slope, $g$ is the acceleration due to gravity (approximately 9.8 m/s²), and $\theta$ is the angle of the incline relative to the horizontal.

How do you draw a free body diagram for an object on an inclined plane?

To draw a free body diagram for an object on an inclined plane, follow these steps:

1. Draw the object on the inclined plane.
2. Identify and draw the weight force ($\mathrm{mg}$), which acts straight down.
3. Draw the normal force, which acts perpendicular to the surface of the incline.
4. If there is friction, draw the frictional force parallel to the surface of the incline, opposing the motion.

Label all the forces and their directions. This diagram helps in analyzing the forces and solving for unknowns.

What is the normal force on an inclined plane and how is it calculated?

The normal force on an inclined plane is the force exerted by the surface perpendicular to the object. It can be calculated using the formula:

$F=\mathrm{mg}\cos \theta$

where $F$ is the normal force, $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $\theta$ is the angle of the incline.

Why do we decompose the weight force in inclined plane problems?

We decompose the weight force in inclined plane problems to simplify the analysis of forces acting on the object. The weight force ($\mathrm{mg}$) is decomposed into two components:

• $\mathrm{mg}\sin \theta$ (parallel to the incline) - responsible for the object's acceleration down the slope.
• $\mathrm{mg}\cos \theta$ (perpendicular to the incline) - counteracted by the normal force.

This decomposition allows us to apply Newton's second law separately in the directions parallel and perpendicular to the incline.

How does the angle of the incline affect the acceleration of an object on an inclined plane?

The angle of the incline ($\theta$) directly affects the acceleration of an object on an inclined plane. The acceleration is given by:

$a=g\sin \theta$

As the angle $\theta$ increases, the sine of the angle increases, leading to a higher acceleration. Conversely, as the angle decreases, the acceleration decreases. At $\theta$ = 0°, the acceleration is zero, and at $\theta$ = 90°, the acceleration is maximum (equal to $g$).