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Static and Kinetic Friction Forces

# Static and Kinetic Friction Forces

Friction is a force that opposes the relative motion between two surfaces that are in contact. It plays a critical role in everyday life, allowing us to walk without slipping, cars to move without skidding, and objects to stay put on inclined surfaces. There are two primary types of friction: static friction and kinetic (or dynamic) friction.

## Static Friction

Static friction is the force that keeps an object at rest when it is subjected to an external force. It needs to be overcome to start moving the object from rest. The force of static friction increases with the applied force until it reaches a maximum value, at which point the object starts to move. This maximum force of static friction is given by:

\[ F_{\text{static max}} = \mu_s N \]

where \( \mu_s \) is the coefficient of static friction, which depends on the types of surfaces in contact, and \( N \) is the normal force, the component of the contact force exerted perpendicular to the surfaces.

## Kinetic Friction

Kinetic friction, also known as sliding or dynamic friction, is the force that opposes the motion of two surfaces sliding against each other. Once an object is in motion, it experiences kinetic friction, which is usually less than the maximum static friction. The force of kinetic friction is given by:

\[ F_{\text{kinetic}} = \mu_k N \]

where \( \mu_k \) is the coefficient of kinetic friction.

The coefficients of friction are dimensionless and are determined empirically. They vary depending on the materials, texture, and even temperature of the surfaces in contact.

# Problems and Solutions on Static and Kinetic Friction Forces

1. **Problem:** A box weighing \(500 \text{ N}\) is at rest on a horizontal surface. If the coefficient of static friction is 0.4, what is the maximum horizontal force that can be applied without moving the box?

**Solution:**
The normal force (\(N\)) is equal to the weight of the box, which is \(500 \text{ N}\).

\( F_{\text{static max}} = \mu_s N = 0.4 \times 500 = 200 \text{ N} \).

The maximum force is \(200 \text{ N}\).

2. **Problem:** A sled with a mass of \(30 \text{ kg}\) is being pulled across a snowy surface with a coefficient of kinetic friction of 0.1. What force is required to keep the sled moving at a constant velocity?

**Solution:**
First, calculate the normal force:

\( N = m \times g = 30 \times 9.8 = 294 \text{ N}\).

Then, calculate the force of kinetic friction:

\( F_{\text{kinetic}} = \mu_k N = 0.1 \times 294 = 29.4 \text{ N}\).

A force of \(29.4 \text{ N}\) is required to keep the sled moving at a constant velocity.

3. **Problem:** If the coefficient of kinetic friction between a 10 kg block and the floor is 0.3, what is the acceleration of the block if a horizontal force of 50 N is applied?

**Solution:**
The normal force is equal to the weight (\(N = mg = 10 \times 9.8 = 98 \text{ N}\)).

The force of kinetic friction is \(F_{\text{kinetic}} = \mu_k N = 0.3 \times 98 = 29.4 \text{ N}\).

The net force is the applied force minus the friction force, which is \(50 – 29.4 = 20.6 \text{ N}\).

Use Newton’s second law to find the acceleration (\(a\)):

\(a = \frac{F_{\text{net}}}{m} = \frac{20.6}{10} = 2.06 \text{ m/s}^2\).

4. **Problem:** On an inclined plane with a 30° angle, a block with a mass of 5 kg is held in place by static friction. If the coefficient of static friction is 0.6, what is the maximum force parallel to the plane that can be applied to the block without causing it to slide?

**Solution:**
The normal force is \(N = mg \cos(\theta) = 5 \times 9.8 \times \cos(30°)\).

\(N \approx 5 \times 9.8 \times 0.866 \approx 42.41 \text{ N}\).

The maximum static friction force is \(F_{\text{static max}} = \mu_s N \approx 0.6 \times 42.41 \approx 25.45 \text{ N}\).

The maximum force parallel to the plane is the same as the maximum static friction, which is \(25.45 \text{ N}\).

5. **Problem:** A 20 kg box is being pushed across a horizontal surface with a constant speed. If the coefficient of kinetic friction is 0.5, what is the force of friction acting on the box?

**Solution:**
The normal force is \(N = mg = 20 \times 9.8 = 196 \text{ N}\).

The force of kinetic friction is \(F_{\text{kinetic}} = \mu_k N = 0.5 \times 196 = 98 \text{ N}\).

6. **Problem:** A book is at rest on a table that is inclined at an angle of 20° from the horizontal. The mass of the book is 2 kg, and the coefficient of static friction is 0.7. Determine if the book will start to slide down the incline.

**Solution:**
The force of gravity parallel to the incline is \(F_{\parallel} = mg \sin(\theta) = 2 \times 9.8 \times \sin(20°)\).

\(F_{\parallel} \approx 2 \times 9.8 \times 0.342 \approx 6.69 \text{ N}\).

The normal force is \(N = mg \cos(\theta) = 2 \times 9.8 \times \cos(20°)\).

\(N \approx 2 \times 9.8 \times 0.940 \approx 18.43 \text{ N}\).

The maximum static friction force is \(F_{\text{static max}} = \mu_s N = 0.7 \times 18.43 = 12.90 \text{ N}\).

Since \(F_{\text{static max}} \) is greater than \(F_{\parallel}\), the book will not slide down.

7. **Problem:** What is the coefficient of kinetic friction if a 15 kg crate initially at rest on a horizontal surface begins to move when a force of 60 N is applied, and subsequently moves with a constant speed when the force is changed to 45 N?

**Solution:**
When the crate is moving at a constant speed, the applied force is equal to the force of kinetic friction.

The normal force is \(N = mg = 15 \times 9.8 = 147 \text{ N}\).

The force of kinetic friction is \(F_{\text{kinetic}} = 45 \text{ N}\).

The coefficient of kinetic friction is \( \mu_k = \frac{F_{\text{kinetic}}}{N} = \frac{45}{147} \approx 0.306 \).

8. **Problem:** A 50 kg box sliding on a flat surface slows down at a uniform rate of \(3 \text{ m/s}^2\). Calculate the coefficient of kinetic friction between the box and the surface.

**Solution:**
The force causing the deceleration is the force of kinetic friction. Start by finding the friction force using Newton’s second law (\(F = ma\)):

\( F_{\text{kinetic}} = m \times a = 50 \times 3 = 150 \text{ N} \).

The normal force is \(N = mg = 50 \times 9.8 = 490 \text{ N}\).

The coefficient of kinetic friction is \( \mu_k = \frac{F_{\text{kinetic}}}{N} = \frac{150}{490} \approx 0.306 \).

9. **Problem:** A person is pushing a 40 kg crate on a flat surface with a force of 120 N, and the crate accelerates at \(2 \text{ m/s}^2\). What is the static frictional force?

**Solution:**
First, calculate the net force needed to produce the acceleration:

\( F_{\text{net}} = m \times a = 40 \times 2 = 80 \text{ N} \).

The force of static friction is the difference between the applied force and the net force: \( F_{\text{static}} = 120 – 80 = 40 \text{ N} \).

10. **Problem:** A hockey puck with a mass of 0.15 kg is sliding across the ice with a coefficient of kinetic friction of 0.02. If the puck’s initial velocity is \(10 \text{ m/s}\), how long will it take to come to a stop?

**Solution:**
The normal force is \(N = mg = 0.15 \times 9.8 = 1.47 \text{ N}\).

The force of kinetic friction is \(F_{\text{kinetic}} = \mu_k N = 0.02 \times 1.47 = 0.0294 \text{ N}\).

The deceleration due to friction is \(a = \frac{F_{\text{kinetic}}}{m} = \frac{0.0294}{0.15} \approx 0.196 \text{ m/s}^2\).

The time to stop is \(t = \frac{v_{\text{initial}}}{a} = \frac{10}{0.196} \approx 51.02 \text{ seconds}\).

11. **Problem:** A block of mass 3 kg is resting on a horizontal surface with a coefficient of static friction of 0.5. At what minimum force will the block start to slide?

**Solution:**
\( N = m \times g = 3 \times 9.8 = 29.4 \text{ N} \).

\( F_{\text{static max}} = \mu_s \times N = 0.5 \times 29.4 = 14.7 \text{ N} \).

The block will start to slide when the force applied exceeds \( 14.7 \text{ N} \).

12. **Problem:** A box sliding down a 45° inclined plane has a mass of 10 kg and a coefficient of kinetic friction of 0.2. Find the acceleration of the box.

**Solution:**
\( N = m \times g \times \cos(45°) = 10 \times 9.8 \times 0.707 \approx 69.29 \text{ N} \).

\( F_{\text{kinetic}} = \mu_k \times N = 0.2 \times 69.29 = 13.86 \text{ N} \).

\( F_{\parallel} = m \times g \times \sin(45°) = 10 \times 9.8 \times 0.707 \approx 69.29 \text{ N} \).

\( F_{\text{net}} = F_{\parallel} – F_{\text{kinetic}} = 69.29 – 13.86 = 55.43 \text{ N} \).

\( a = \frac{F_{\text{net}}}{m} = \frac{55.43}{10} = 5.54 \text{ m/s}^2 \).

13. **Problem:** Find the coefficient of static friction if a block at rest on an incline starts to slide when the incline angle is gradually increased to 25°. The mass of the block is 4 kg.

**Solution:**
When the block starts to slide, \( F_{\text{static max}} \) is equal to the component of the gravitational force along the incline.

\( F_{\parallel} = m \times g \times \sin(25°) = 4 \times 9.8 \times \sin(25°) \).

\( F_{\parallel} \approx 4 \times 9.8 \times 0.4226 \approx 16.54 \text{ N} \).

\( N = m \times g \times \cos(25°) = 4 \times 9.8 \times \cos(25°) \).

\( N \approx 4 \times 9.8 \times 0.9063 \approx 35.45 \text{ N} \).

\( \mu_s = \frac{F_{\text{static max}}}{N} = \frac{16.54}{35.45} \approx 0.466 \).

14. **Problem:** What is the force of kinetic friction if a 20 kg box initially moving at \(5 \text{ m/s}\) comes to a stop after traveling \(10 \text{ m}\) along a horizontal surface?

**Solution:**
Use the work-energy principle, where the work done by friction equals the change in kinetic energy.

\( W = F_{\text{kinetic}} \times d = -\Delta KE = -\frac{1}{2} m v^2 \).

\( F_{\text{kinetic}} \times 10 = -\frac{1}{2} \times 20 \times 5^2 \).

\( F_{\text{kinetic}} = -\frac{1}{2} \times 20 \times 25 / 10 \).

\( F_{\text{kinetic}} = -\frac{250}{10} = -25 \text{ N} \).

The force of kinetic friction is \(25 \text{ N}\) (the negative sign indicates that friction acts in the opposite direction of motion).

15. **Problem:** A car weighing 1500 kg is parked on a hill inclined at an angle of 10°. If the static friction coefficient is 0.4, can the car remain stationary without any external support?

**Solution:**
\( N = m \times g \times \cos(10°) = 1500 \times 9.8 \times \cos(10°) \).

\( F_{\text{static max}} = \mu_s \times N = 0.4 \times 1500 \times 9.8 \times \cos(10°) \).

\( F_{\parallel} = m \times g \times \sin(10°) = 1500 \times 9.8 \times \sin(10°) \).

If \( F_{\text{static max}} \geq F_{\parallel} \), the car will not slide.

16. **Problem:** Calculate the static friction force between a 10 kg block and the surface if the block does not move when a horizontal force of 20 N is applied. The coefficient of static friction is 0.6.

**Solution:**
The static friction force is equal to the applied force until it reaches the maximum value, after which the block will move.

\( F_{\text{static}} = 20 \text{ N} \) (since the block didn’t move, this is the friction force).

17. **Problem:** A box is traveling down an incline with a 30° slope at a constant velocity. The box has a mass of 8 kg, and the coefficient of kinetic friction is 0.4. Find the force parallel to the incline.

**Solution:**
When the velocity is constant, \( F_{\parallel} = F_{\text{kinetic}} \).

\( N = m \times g \times \cos(30°) = 8 \times 9.8 \times \cos(30°) \).

\( F_{\text{kinetic}} = \mu_k \times N = 0.4 \times 8 \times 9.8 \times \cos(30°) \).

\( F_{\parallel} = F_{\text{kinetic}} \).

18. **Problem:** A child pulls a toy car with a mass of 0.5 kg across a rough floor producing an acceleration of \(1 \text{ m/s}^2\). If the coefficient of kinetic friction is 0.1, what is the pulling force?

**Solution:**
\( N = m \times g = 0.5 \times 9.8 \).

\( F_{\text{kinetic}} = \mu_k \times N = 0.1 \times 0.5 \times 9.8 \).

The net force equals the mass times acceleration: \( F_{\text{net}} = m \times a = 0.5 \times 1 \).

The pulling force is the sum of the force of kinetic friction and the net force: \( F_{\text{pull}} = F_{\text{kinetic}} + F_{\text{net}} \).

19. **Problem:** On a surface with a coefficient of kinetic friction of 0.25, how much work is done by friction on a 3 kg block that slides 4 meters after being pushed?

**Solution:**
\( N = m \times g = 3 \times 9.8 \).

\( F_{\text{kinetic}} = \mu_k \times N = 0.25 \times 3 \times 9.8 \).

The work done by friction is the force of friction times the distance: \( W = F_{\text{kinetic}} \times d \).

20. **Problem:** A block of mass 5 kg is being pushed with a force of 25 N on a surface where the coefficient of kinetic friction is 0.3. Calculate the resulting acceleration of the block.

**Solution:**
\( N = m \times g = 5 \times 9.8 \).

\( F_{\text{kinetic}} = \mu_k \times N = 0.3 \times 5 \times 9.8 \).

The net force

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