In this lesson students will learn about friction, the difference between static and kinetic friction and how to solve problems involving friction.

### Key Equations

\begin{align*}\text{Static and Kinetic Friction} \begin{cases} f_s \le \mu_s | F_N| & \text{Opposes potential motion of surfaces in contact}\\ f_k = \mu_k | F_N| & \text{Opposes motion of surfaces in contact} \end{cases}\end{align*}

- The force of
**friction**can actually be described in terms of the*coefficient of friction,*\begin{align*}\mu\end{align*}. It is determined experimentally and varies depending upon the two surfaces in contact. -
**Static friction**(\begin{align*}\mu_s\end{align*}) acts between two surfaces that are in contact but not in motion with respect to each other. This force prevents objects from sliding. It always opposes potential motion, and it rises in magnitude to a maximum value given by the formula below.\begin{align*}^2\end{align*} -
**Kinetic friction**(\begin{align*}\mu_k\end{align*}) acts between two surfaces that are in contact and in motion with respect to each other. This force reduces the acceleration and it always opposes the direction of motion.\begin{align*}^2\end{align*}

#### Example 1

Calculate the force necessary to slide a 4.7-kg chair across a room at a constant speed if the coefficient of kinetic friction between the chair and the floor is 0.68.

*Question:* \begin{align*}F = ? [N]\end{align*}

*Given:* \begin{align*}m = 4.7\ kg\end{align*}

\begin{align*}{\;}\qquad \quad \mu_k = 0.68\end{align*}

\begin{align*}{\;} \qquad \quad g = 10.0\ m/s^2\end{align*}

*Equations:* \begin{align*}\sum F = ma\end{align*}

\begin{align*}{\;}\qquad \ \qquad \sum F_y = N - mg = 0, \text{so}\ N = mg\end{align*}

\begin{align*}{\;}\qquad \ \qquad \sum F_x = ma_x\end{align*}

\begin{align*}{\;}\qquad \ \qquad F_{\text{pull}} - f_k = 0 \ (\text{because the chair is moving at}\ constant \ speed,\ \text{so}\ a=0)\end{align*}

\begin{align*}{\;}\qquad \ \qquad F_{\text{pull}} = \mu_k N\end{align*}

*Plug n’ Chug:* The force necessary to move the chair at a constant speed is equal to the frictional force between the chair and the floor. However in order to calculate the frictional force you must first determine the normal force which is (in this case) equal to the weight (i.e. \begin{align*}F_g\end{align*}) of the chair.

\begin{align*}N & = mg = (4.7 \ kg)(10 \ m/s^2) = 47 \ N \\ F_{\text{pull}} &= \mu_k \ N = (0.68)(47 \ N) = 32 \ N\end{align*}

*Answer:* \begin{align*}\boxed{\mathbf{32 \ N}}\end{align*}

#### Example 2

### Watch this Explanation

### Simulation

Ramp Forces and Motion (PhET Simulation)

### Explore More

- Nathan pulls his little brother in a wagon, using a rope inclined at \begin{align*}30^\circ\end{align*} above the horizontal. The wagon and brother have a total mass of 80 kg, the average coefficient of friction between the wagon wheels and the floor is 0.08, and Nathan pulls on the rope with a force of 100 N.
- Draw a force diagram for the wagon, labeling all forces.
- Calculate the horizontal and vertical components of Nathan’s pull. Label them on your diagram (use dotted lines for components so as not to confuse them with other forces).
- Calculate the normal force acting on the wagon. (HINT: It is NOT equal to the weight! Use your FBD above).
- Calculate the force of friction on the wagon.
- Calculate the horizontal acceleration of the wagon.

- When the 20 kg box to the right is pulled with a force of 100 N, it just starts to move (i.e. the maximum value of static friction is overcome with a force of 100 N). What is the value of the coefficient of static friction, \begin{align*}\mu_S\end{align*}?
- A different box, this time 5 kg in mass, is being pulled with a force of 20 N and is sliding with an acceleration of \begin{align*}2 \ m/s^2\end{align*}. Find the coefficient of kinetic friction, \begin{align*}\mu_K\end{align*}.
- The large box on the table is \begin{align*}30 \;\mathrm{kg}\end{align*} and is connected via a rope and pulley to a smaller \begin{align*}10 \;\mathrm{kg}\end{align*} box, which is hanging. The \begin{align*}10 \;\mathrm{kg}\end{align*} mass is the highest mass you can hang without moving the box on the table. Find the coefficient of static friction \begin{align*}\mu_S\end{align*}.
- A block has a little block hanging out to its side, as shown: As you know, if the situation is left like this, the little block will just fall. But if we accelerate the leftmost block to the right, this will create a normal force between the little block and the big block, and if there is a coefficient of friction between them, then the little block won’t slide down! Clever, eh?
- The mass of the little block is \begin{align*}0.15 \;\mathrm{kg}\end{align*}. What frictional force is required to keep it from falling? (State a magnitude and direction.)
- If both blocks are accelerating to the right with an acceleration \begin{align*}a = 14.0 \;\mathrm{m/s}^2\end{align*}, what is the normal force on the little block provided by the big block?
- What is the minimum coefficient of static friction required?

**Answers**

- b. \begin{align*}F_x = 87 \ N, F_y = 50 \ N\end{align*} c. \begin{align*}N = 750 \ N\end{align*} d. \begin{align*}f = 60 \ N\end{align*} e. \begin{align*}0.34 \ m/s^2\end{align*}
- 0.5
- 0.2
- 0.33
- a. \begin{align*}1.5 \;\mathrm{N}; 2.1 \;\mathrm{N}; 0.71\end{align*}