Inclined Planes with Friction: Videos & Practice Problems

In problems involving objects on rough inclined planes, understanding the forces at play is crucial. The friction force can be static or kinetic, determined by comparing the gravitational component down the ramp, $\mathrm{mg}\sin \left(\theta \right)$, with the maximum static friction, $\mu$_s⁢mg⁢cos(θ). Critical angles for static and kinetic friction can be calculated using $\theta =\tan \mu \left(\theta \right)$, highlighting the relationship between friction coefficients and motion. Understanding these concepts is essential for analyzing motion on inclined planes.

Intro to Incline Plane with Friction

In this scenario, we are analyzing a 10-kilogram block resting on a rough inclined plane at an angle of 37 degrees. The presence of friction complicates the dynamics, requiring us to determine the frictional force acting on the block when it is released. To begin, we construct a free body diagram, identifying the gravitational force (mg) acting downward, the normal force (N) perpendicular to the ramp, and the frictional force (F) opposing the motion up the ramp.

To analyze the forces, we tilt our coordinate system so that the x-axis aligns with the ramp's incline. The gravitational force can be decomposed into two components: the component parallel to the incline (mg_x) and the component perpendicular to the incline (mg_y). The parallel component is calculated using the equation:

mg_x = mg * sin(θ)

Substituting the values, we find:

mg_x = 10 kg * 9.8 m/s² * sin(37°) ≈ 59 N

Next, we need to determine whether the friction is static or kinetic. This is done by comparing mg_x with the maximum static friction force (F_{s max}), which is given by:

F_{s max} = μ_s * N

To find the normal force (N), we use the equation:

N = mg * cos(θ)

Calculating this gives:

N = 10 kg * 9.8 m/s² * cos(37°) ≈ 47 N

Now, substituting into the static friction equation:

F_{s max} = 0.6 * 47 N ≈ 28.2 N

Since mg_x (59 N) exceeds F_{s max} (28.2 N), the block will start moving, and we will use kinetic friction instead. The kinetic friction force (F_k) is calculated as:

F_k = μ_k * N

Substituting the values, we find:

F_k = 0.4 * 47 N ≈ 18.8 N

Now, we can determine the block's acceleration (a_x) using Newton's second law:

ΣF_x = m * a_x

In the x-direction, the net force is the difference between mg_x and the kinetic friction force:

mg_x - F_k = m * a_x

Substituting the known values:

59 N - 18.8 N = 10 kg * a_x

Solving for a_x gives:

a_x = (59 N - 18.8 N) / 10 kg ≈ 4.02 m/s²

This positive acceleration indicates that the block accelerates down the ramp, confirming our calculations. Understanding the interplay between gravitational forces and friction on inclined planes is crucial for solving such problems effectively.

Pushing A Box Up A Ramp

In this scenario, we are analyzing the motion of a 20-kilogram block being pushed up an incline with a force of 110 Newtons. To determine the acceleration of the block, we first need to establish a free body diagram that includes the gravitational force acting downwards, the normal force acting perpendicular to the surface, and the applied force pushing the block up the ramp. Additionally, we must account for friction, which opposes the direction of motion.

The gravitational force can be broken down into two components: the component acting parallel to the incline, denoted as \( mg_x \), and the component acting perpendicular to the incline, denoted as \( mg_y \). The parallel component is calculated using the equation \( mg_x = mg \sin(\theta) \), where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), and \( \theta \) is the angle of the incline. For this problem, with \( m = 20 \) kg and \( \theta = 15^\circ \), we find \( mg_x = 20 \times 9.8 \times \sin(15^\circ) \approx 50.7 \) N.

Next, we compare the applied force (110 N) with the gravitational force component (50.7 N) to determine the direction of friction. Since the applied force is greater, friction will act down the ramp. To find the type of friction, we calculate the maximum static friction force using the equation \( f_s^{max} = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction (0.3) and \( N \) is the normal force, calculated as \( N = mg_y = mg \cos(\theta) \). Thus, \( N = 20 \times 9.8 \times \cos(15^\circ) \approx 56.8 \) N, leading to \( f_s^{max} = 0.3 \times 56.8 \approx 17.04 \) N.

Since the net force acting on the block (110 N - 50.7 N = 59.3 N) exceeds the maximum static friction, the block experiences kinetic friction instead. The kinetic friction force is calculated using \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction (0.2). Therefore, \( f_k = 0.2 \times 56.8 \approx 11.36 \) N.

Now, we can apply Newton's second law, \( F = ma \), along the direction of motion. The equation becomes \( F - mg_x - f_k = ma \). Substituting the known values gives us:

110 N - 50.7 N - 11.36 N = 20 kg \cdot a.

This simplifies to:

48.94 N = 20 kg \cdot a.

Solving for acceleration \( a \) yields:

a = \frac{48.94}{20} \approx 2.447 \, \text{m/s}^2.

Thus, the block accelerates up the ramp at approximately 2.45 m/s², indicating that the applied force is sufficient to overcome both the gravitational pull and the kinetic friction acting against it.

Block Launched Up Ramp

In this scenario, we analyze a 2-kilogram block launched up a 40-degree ramp with an initial speed of 10 meters per second. The block ascends until it reaches a height of 3 meters above the base of the ramp. To determine the coefficient of kinetic friction, we start by constructing a free body diagram, identifying the forces acting on the block: the weight force (mg) acting downward, the normal force acting perpendicular to the ramp, and the kinetic friction force (f_k) opposing the motion.

We can express the weight force in terms of its components along the ramp. The component of the weight acting down the ramp is given by mg \sin(\theta), while the normal force is mg \cos(\theta). The equation of motion can be set up as follows:

-mg \sin(\theta) - f_k = ma

Substituting the expression for kinetic friction, we have:

-mg \sin(\theta) - \mu_k mg \cos(\theta) = ma

To find the acceleration (a), we need to determine the displacement (Δx) along the ramp. Given the vertical height (Δy) of 3 meters and the angle of the ramp, we can use the sine function:

Δx \sin(40^\circ) = Δy

Solving for Δx gives:

Δx = \frac{Δy}{\sin(40^\circ)} = \frac{3}{\sin(40^\circ)} \approx 4.67 \text{ meters}

Next, we apply the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

v_f^2 = v_i^2 + 2aΔx

Substituting the known values (final velocity v_f = 0, initial velocity v_i = 10 \text{ m/s}, and Δx \approx 4.67 \text{ m}), we can solve for acceleration:

0 = (10)^2 + 2a(4.67)

Rearranging gives:

a \approx -10.7 \text{ m/s}^2

Now, substituting the values back into the force equation to solve for the coefficient of kinetic friction:

-2(9.8) \sin(40^\circ) - \mu_k (2)(9.8) \cos(40^\circ) = 2(-10.7)

Calculating each term yields:

-12.6 - 15\mu_k = -21.4

Rearranging gives:

15\mu_k = 21.4 - 12.6 = 8.8

Finally, solving for the coefficient of kinetic friction:

\mu_k = \frac{8.8}{15} \approx 0.59

This value represents the coefficient of kinetic friction for the block sliding up the ramp, confirming that the calculations align with the expected physical behavior of the system.

Force Not Parallel To Ramp

When analyzing the motion of a mini fridge being pushed up an incline, it is essential to break down the forces acting on the fridge using a free body diagram. The applied force, which is 120 N at an angle of 30 degrees above the incline, must be resolved into its x (parallel to the incline) and y (perpendicular to the incline) components. The x-component of the applied force can be calculated using the cosine function:

\[ F_x = F \cdot \cos(30^\circ) = 120 \cdot \cos(30^\circ) \approx 104 \, \text{N} \]

In contrast, the gravitational force acting on the fridge can also be resolved into components. The component of gravitational force acting down the incline is given by:

\[ F_{mgx} = mg \cdot \sin(20^\circ) = 30 \cdot 9.8 \cdot \sin(20^\circ) \approx 100.5 \, \text{N} \]

Comparing these two forces, we find that the applied force component (104 N) is greater than the gravitational component (100.5 N). This indicates that, without friction, the fridge would move up the ramp. Consequently, the frictional force must act down the ramp to oppose this motion.

Next, we need to determine whether the friction is static or kinetic. To do this, we calculate the net force acting on the fridge, which is the difference between the applied force component and the gravitational component:

\[ F_{\text{net}} = F_x - F_{mgx} = 104 - 100.5 = 3.5 \, \text{N} \]

To assess the type of friction, we compare this net force to the maximum static friction force, which is calculated using the normal force. The normal force is affected by both the weight of the fridge and the y-component of the applied force. The y-component of the applied force is given by:

\[ F_y = F \cdot \sin(30^\circ) = 120 \cdot \sin(30^\circ) = 60 \, \text{N} \]

Using the equilibrium condition in the y-direction, we can express the normal force as:

\[ N = mg \cdot \cos(20^\circ) - F_y = (30 \cdot 9.8 \cdot \cos(20^\circ)) - 60 \approx 216.3 \, \text{N} \]

The maximum static friction force is then calculated as:

\[ F_{s \, \text{max}} = \mu_s \cdot N = 0.3 \cdot 216.3 \approx 64.9 \, \text{N} \]

Since the net force (3.5 N) is less than the maximum static friction force (64.9 N), the frictional force acting on the fridge is equal to the net force, which means:

\[ F_{\text{friction}} = 3.5 \, \text{N} \, \text{(down the ramp)} \]

As a result, the fridge does not move, and its acceleration is zero. This analysis illustrates the importance of understanding the forces at play and how they interact to determine the motion of objects on inclined planes.

Incline with Friction & Critical Angle

In problems involving objects on rough inclined planes, two critical angles are often calculated: the angle at which an object begins to slide (critical angle for static friction) and the angle at which it slides down at a constant speed (critical angle for kinetic friction). Understanding these angles is essential for analyzing the forces acting on the object as the incline is adjusted.

When a block is placed on an adjustable ramp, it remains stationary until the ramp is tilted to a specific angle where the block starts to slide. This angle is determined by the balance of forces acting on the block, specifically the gravitational force components and the frictional force. The gravitational force can be broken down into two components: one acting parallel to the ramp (down the ramp) and the other acting perpendicular to the ramp (normal force). The force pulling the block down the ramp is given by mg \sin(\theta), while the maximum static frictional force opposing this motion is f_s^{max} = \mu_s N, where N = mg \cos(\theta).

At the critical angle where the block begins to slide, the forces are in equilibrium, leading to the equation:

mg \sin(\theta_{critical s}) = \mu_s mg \cos(\theta_{critical s})

By canceling out mg from both sides, we simplify this to:

\tan(\theta_{critical s}) = \mu_s

Thus, the critical angle for static friction can be calculated using:

\theta_{critical s} = \tan^{-1}(\mu_s)

For example, if the coefficient of static friction \mu_s is 0.75, the critical angle is:

\theta_{critical s} = \tan^{-1}(0.75) \approx 37^\circ

In the second scenario, when the block is sliding down the ramp at a constant speed, the forces are again in equilibrium, but now the frictional force is kinetic friction. The equation becomes:

mg \sin(\theta_{critical k}) = \mu_k mg \cos(\theta_{critical k})

Again, canceling mg gives:

\tan(\theta_{critical k}) = \mu_k

Thus, the critical angle for kinetic friction is calculated as:

\theta_{critical k} = \tan^{-1}(\mu_k)

For instance, if the coefficient of kinetic friction \mu_k is 0.31, the critical angle is:

\theta_{critical k} = \tan^{-1}(0.31) \approx 17^\circ

These two angles illustrate an important principle: it is generally harder to initiate motion (overcoming static friction) than to maintain it (overcoming kinetic friction). Notably, the critical angles depend solely on the coefficients of friction and are independent of the mass of the object, which may seem counterintuitive but is a result of the gravitational forces canceling out in the equations.

Critical Angle

In this scenario, we are analyzing a car parked on an inclined road at an angle of 35 degrees, which becomes slippery due to a snowstorm. The objective is to determine the coefficient of static friction (μ_s) between the car and the icy road as the car begins to slide downhill. This situation is characterized as a critical angle problem, where the angle at which the car starts to slide indicates the maximum static friction before motion occurs.

The relationship between the coefficient of static friction and the critical angle can be expressed using the formula:

μ_s = tan(θ_critical)

In this case, the critical angle (θ_critical) is given as 35 degrees. To find the coefficient of static friction, we simply need to calculate the tangent of this angle:

μ_s = tan(35°)

When you compute this using a calculator set to degree mode, you will find:

μ_s ≈ 0.7

This value indicates the coefficient of static friction required to prevent the car from sliding down the icy road. Understanding this concept is crucial for evaluating the effects of incline and friction in real-world scenarios, particularly in conditions that affect vehicle stability.

A 3-kg block is at rest on an adjustable ramp. When the ramp is tilted to a 20° angle, the block slides with constant velocity. What is the coefficient of kinetic friction between the ramp and the block?

• A. μk = 0.24
• B. μk = 0.36
• C. μk = 0.52
• D. Not enough information given

Answer: (B)

Here’s what students ask on this topic:

How do you calculate the friction force on an inclined plane with friction?

To calculate the friction force on an inclined plane with friction, you first need to determine whether the friction is static or kinetic. Draw a free body diagram and resolve the gravitational force into components parallel and perpendicular to the plane. The friction force can be calculated using the equations:

For static friction: $f_s\le {\mu }_s{N}_{}$

For kinetic friction: $f_k={\mu }_k{N}_{}$

where ${\mu }_s$ and ${\mu }_k$ are the coefficients of static and kinetic friction, respectively, and $N_{}$ is the normal force, which is ${\mathrm{mg}}_{}cos\theta$.

What are critical angles in inclined plane problems with friction?

Critical angles in inclined plane problems with friction are specific angles at which an object starts to slide or slides at constant speed. These angles are calculated using the coefficients of static and kinetic friction. The critical angle for static friction, ${\theta }_c$, is given by:

${\theta }_c=tan⁻¹{\mu }_s$

For kinetic friction, the critical angle is:

${\theta }_k=tan⁻¹{\mu }_k$

These angles help determine the point at which an object overcomes static friction and begins to move, or continues to move at a constant velocity.

How do you determine if friction is static or kinetic on an inclined plane?

To determine if friction is static or kinetic on an inclined plane, compare the gravitational component down the ramp, ${\mathrm{mg}}_{}sin\theta$, with the maximum static friction, ${\mu }_s{\mathrm{mg}}_{}cos\theta$. If ${\mathrm{mg}}_{}sin\theta$ is less than or equal to ${\mu }_s{\mathrm{mg}}_{}cos\theta$, the friction is static. If it exceeds this value, the friction becomes kinetic, and the object starts to move.

What is the relationship between the coefficients of friction and the critical angles on an inclined plane?

The relationship between the coefficients of friction and the critical angles on an inclined plane is given by the tangent function. The critical angle for static friction, ${\theta }_c$, is calculated using:

${\theta }_c=tan⁻¹{\mu }_s$

For kinetic friction, the critical angle is:

${\theta }_k=tan⁻¹{\mu }_k$

These equations show that the critical angles depend solely on the coefficients of static and kinetic friction, and not on other factors like mass.

How do you calculate the acceleration of a block on an inclined plane with friction?

To calculate the acceleration of a block on an inclined plane with friction, use Newton's second law in the direction parallel to the plane. The net force is the difference between the gravitational component down the ramp and the friction force. The equation is:

$F_x={\mathrm{mg}}_{}sin\theta -f_k=m_{}{a}_x$

where $f_k={\mu }_k{N}_{}$ and $N_{}={\mathrm{mg}}_{}cos\theta$. Solve for $a_x$:

$a_x=\; ({\mathrm{mg}}_{}sin\theta -{\mu }_k{\mathrm{mg}}_{}cos\theta )\; /m_{}$

This simplifies to:

$a_x=g(sin\theta -{\mu }_{k}cos\theta )$