The calculations for fluid friction are beautiful, but difficult. Instead, we are going to focus on the two types of dry friction: static and kinetic. Dry friction occurs when two solid surfaces are in contact.

speed

friction

Even seemingly smooth surfaces are rough at the microscopic level. Dry friction can occur because rough surfaces get caught on each other. This causes microscopic deformations to occur. Friction can also come from other sources like chemical bonding between surfaces.

Dry friction is difficult to model because surfaces can have a wide range of shapes and chemical compositions. These 5 rules are broken as often as they are followed.

Friction scales linearly with the normal force.

Friction is not affected by the area of contact between surfaces.

Stationary objects have more friction than sliding objects.

Sliding friction is not affected by sliding velocity.

You can look up the magnitude of friction for each pair of materials.

This friction model is only a "rough" approximation, so don't expect much precision or accuracy.

Static Friction

Static means not moving. Static friction is friction between solid objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface.

The static friction force balances applied forces to keep the object stationary. We can estimate the maximum static friction force.

$$ F_s \leq \mu_{s} F_{N}$$

$F_s$ = force of static friction [N, newtons]
direction and magnitude change to keep acceleration zero
but only up to the maximum value

$F_N$ = normal force [N, newtons]

$\mu_s$ = mu, coefficient of friction [no units]

The coefficients of friction(μ) are different for each pair of surfaces.

Friction Coefficient Data Table (wikipedia)

Materials		Static Friction		Kinetic Friction
Materials		Dry	Lubricated	Dry	Lubricated
Aluminium	Steel	0.61		0.47
Aluminum	Aluminum			1.5
Gold	Gold			2.5
Platinum	Platinum			3.0
Silver	Silver			1.5
Alumina ceramic	Silicon Nitride ceramic				0.004 (wet)
BAM (Ceramic alloy AlMgB₁₄)	Titanium boride (TiB₂)	0.04–0.05	0.02
Brass	Steel	0.35-0.51	0.19	0.44
Cast iron	Copper	1.05		0.29
Cast iron	Zinc	0.85		0.21
Concrete	Rubber	1.0	0.30 (wet)	0.6-0.85	0.45-0.75 (wet)
Concrete	Wood	0.62
Copper	Glass	0.68
Copper	Steel	0.53		0.36
Glass	Glass	0.9-1.0		0.4
Human synovial fluid	Cartilage		0.01		0.003
Ice	Ice	0.02-0.09
Polyethene	Steel	0.2	0.2
(Teflon)	PTFE (Teflon)	0.04	0.04		0.04
Steel	Ice	0.03
Steel	PTFE (Teflon)	0.04	0.04		0.04
Steel	Steel	0.74	0.16	0.42-0.62
Wood	Metal	0.2–0.6	0.2 (wet)
Wood	Wood	0.25–0.5	0.2 (wet)

Example: You place a 0.42 kg glass from IKEA called POKAL on a flat copper pan. How much horizontal force will you have to apply to get the glass to move?

solution

$$ F_s \leq \mu_{s} F_{N} $$ $$ F_{s\mathrm{\,max}} = \mu_{s} F_{N} $$ $$F_{N}=mg$$ $$F_{s\mathrm{\,max}}= \mu_{s}mg$$ $$F_{s\mathrm{\,max}}= (0.68)(0.42)(9.8)$$ $$F_{s\mathrm{\,max}}= 2.80 \, \mathrm{N}$$

Example: A 10 kg wood block is at rest on top of another 10 kg wood block which is resting on a concrete slab. How much force will it take to overcome the static friction between the ground and the lower box?

solution

Include both masses in the total mass.

$$F_g = mg$$ $$F_g = (20)(9.8)$$ $$F_g = 196\,\mathrm{N}$$ Since the block isn't accelerating in the vertical the normal force equals the force of gravity for both blocks. $$F_{s\mathrm{\,max}} = \mu_s F_N$$ $$F_{s\mathrm{\,max}} = \mu_s F_g$$ $$F_{s\mathrm{\,max}} = \mu_s 196$$ The coefficient of static friction for concrete and wood is 0.62. $$F_{s\mathrm{\,max}} = (0.62) (196)$$ $$F_{s\mathrm{\,max}} = 122 \, \mathrm{N}$$

μ_k =

μ_s =

m = kg

Example: Calculate the maximum value the hanging mass could have before the two masses begin to move. You can test your answer in the simulation.

The 5 kg block is made of aluminum and the table is made of steel.

Friction Coefficient Data Table (wikipedia)

Materials		Static Friction		Kinetic Friction
Materials		Dry	Lubricated	Dry	Lubricated
Aluminium	Steel	0.61		0.47
Aluminum	Aluminum			1.5
Gold	Gold			2.5
Platinum	Platinum			3.0
Silver	Silver			1.5
Alumina ceramic	Silicon Nitride ceramic				0.004 (wet)
BAM (Ceramic alloy AlMgB₁₄)	Titanium boride (TiB₂)	0.04–0.05	0.02
Brass	Steel	0.35-0.51	0.19	0.44
Cast iron	Copper	1.05		0.29
Cast iron	Zinc	0.85		0.21
Concrete	Rubber	1.0	0.30 (wet)	0.6-0.85	0.45-0.75 (wet)
Concrete	Wood	0.62
Copper	Glass	0.68
Copper	Steel	0.53		0.36
Glass	Glass	0.9-1.0		0.4
Human synovial fluid	Cartilage		0.01		0.003
Ice	Ice	0.02-0.09
Polyethene	Steel	0.2	0.2
(Teflon)	PTFE (Teflon)	0.04	0.04		0.04
Steel	Ice	0.03
Steel	PTFE (Teflon)	0.04	0.04		0.04
Steel	Steel	0.74	0.16	0.42-0.62
Wood	Metal	0.2–0.6	0.2 (wet)
Wood	Wood	0.25–0.5	0.2 (wet)

hint

First, draw a free body diagram. Then calculate the max static friction force on the 5 kg block.

Since the blocks aren't moving, the opposing forces are equal. The friction force equals the tension force, which also equals the force of gravity for the right block. This means we can set the max static friction equal to the force of gravity for the right block.

Replace the force of gravity with "mg", and solve for the mass.

solution

$$\text{aluminum on steel}$$ $$\mu_s = 0.61$$

$$F_N = F_g$$ $$F_g = mg$$ $$F_N = (5) (9.8)$$ $$F_N = 49 \, \mathrm{N}$$

$$ F_{s\mathrm{\,max}} = \mu_{s} F_{N} $$ $$ F_{s\mathrm{\,max}} = (0.61)(49 \, \mathrm{N}) $$ $$ F_{s\mathrm{\,max}} = 29.89 \, \mathrm{N} $$

Since the system isn't moving the acceleration is zero. This means that opposing forces need to be equal. The friction force equals the tension force, which also equals the force of gravity on the right block.

$$F_{s\mathrm{\,max}} = T$$ $$T = F_g$$ $$F_g = mg$$ $$F_{s\mathrm{\,max}} = mg$$ $$29.89 = m(9.8)$$ $$m = 3.05 \, \mathrm{kg}$$

Kinetic Friction

Kinetic means motion. Kinetic friction is a force that occurs when two surfaces in contact slide against each other. The kinetic friction force remains constant over a wide range of speeds.

$$F_{k}=\mu_{k} F_{N}$$

$F_k$ = force of kinetic friction [N,newtons]
pointed opposite the direction of motion

$F_N$ = normal force [N,newtons]

$\mu _k$ = mu, coefficient of friction [no units]

In most situations, the friction force doesn't depend on the amount of contact between surfaces. This is because a larger contact area spreads out the normal force.

Example: You slide the 0.42 kg POKAL glass cup on a glass table at 3.0 m/s.
Find the force of kinetic friction, and the time for the cup to come to a stop.

Friction Coefficient Data Table (wikipedia)

Materials		Static Friction		Kinetic Friction
Materials		Dry	Lubricated	Dry	Lubricated
Aluminium	Steel	0.61		0.47
Aluminum	Aluminum			1.5
Gold	Gold			2.5
Platinum	Platinum			3.0
Silver	Silver			1.5
Alumina ceramic	Silicon Nitride ceramic				0.004 (wet)
BAM (Ceramic alloy AlMgB₁₄)	Titanium boride (TiB₂)	0.04–0.05	0.02
Brass	Steel	0.35-0.51	0.19	0.44
Cast iron	Copper	1.05		0.29
Cast iron	Zinc	0.85		0.21
Concrete	Rubber	1.0	0.30 (wet)	0.6-0.85	0.45-0.75 (wet)
Concrete	Wood	0.62
Copper	Glass	0.68
Copper	Steel	0.53		0.36
Glass	Glass	0.9-1.0		0.4
Human synovial fluid	Cartilage		0.01		0.003
Ice	Ice	0.02-0.09
Polyethene	Steel	0.2	0.2
(Teflon)	PTFE (Teflon)	0.04	0.04		0.04
Steel	Ice	0.03
Steel	PTFE (Teflon)	0.04	0.04		0.04
Steel	Steel	0.74	0.16	0.42-0.62
Wood	Metal	0.2–0.6	0.2 (wet)
Wood	Wood	0.25–0.5	0.2 (wet)

solution

$$F_{k}= \mu_{k}F_{N}$$ $$F_{N}=mg$$ $$F_{k}= \mu_{k}mg$$ $$F_{k}= (0.4)(0.42)(9.8)$$ $$F_{k}= 1.64 \, \mathrm{N}$$
$$F=ma$$ $$1.64=(0.42)a$$ $$\frac{1.64}{0.42}=a$$ $$3.90 \mathrm{\tfrac{m}{s^{2}}} = a$$
$$a=-3.90 \mathrm{\tfrac{m}{s^{2}}}$$ $$\Delta t=?$$ $$v_{i}=3.0 \mathrm{\tfrac{m}{s}}$$ $$v_{f}=0$$ $$v_{f} = v_{i}+a \Delta t$$ $$0=3+(-3.90)\Delta t$$ $$3.90\Delta t=3$$ $$\frac{3}{3.90}=\Delta t$$ $$0.77 \mathrm{s}=\Delta t$$

Static friction will match an applied force until the applied force exceeds the maximum value of static friction. Forces above that point will cause motion.

Once the body is moving, friction transitions to kinetic. Kinetic friction is lower and less precise. This transition often causes a jerky motion as the friction force quickly drops to the lower value.

Example: Imagine the graph above is for a 20 kg box. Calculate the coefficients of kinetic and static friction.

kinetic friction solution

$$F_N = F_g$$ $$F_g = mg$$ $$F_N = (20 \, \mathrm{kg})(9.8 \, \mathrm{ \tfrac{m}{s^2}}) $$ $$F_N = 196 \, \mathrm{N}$$

We can tell from the graph that the force of kinetic friction is about 70 N.

$$F_k = \mu_{k} F_{N} $$ $$70 \, \mathrm{N}= \mu_{k} 196 \, \mathrm{N}$$ $$\frac{70 \, \mathrm{N}}{196 \, \mathrm{N}} = \mu_{k} $$ $$0.36 = \mu_{k}$$

static friction solution

$$F_N = F_g$$ $$F_g = mg$$ $$F_N = (20 \, \mathrm{kg})(9.8 \, \mathrm{ \tfrac{m}{s^2}}) $$ $$F_N = 196 \, \mathrm{N}$$

We can tell from the dotted line that the max value of static friction is 100 N.

$$ F_s \leq \mu_{s} F_{N} $$ $$ F_{s\mathrm{\,max}} = \mu_{s} F_{N} $$ $$ 100 \, \mathrm{N} = \mu_{s} 196 \, \mathrm{N}$$ $$\frac{100 \, \mathrm{N}}{196 \, \mathrm{N}} = \mu_{s} $$ $$0.51 = \mu_{s}$$

Click to Run

Simulation: Calculate the coefficients of kinetic and static friction? Use the default settings on the friction mode for the simulation above.

coefficient of static friction solution

Click the reset icon to make sure the simulation is at the default friction.
Check the "Masses" box.
Slowly increase the force until the mass moves.
This is the maximum force of static friction. I got 125 N. $$F_{\mathrm{max}}= \mu_{s}F_{N}$$ $$F_{\mathrm{max}}= \mu_{s}mg$$ $$125=\mu_s(50)(9.8)$$ $$125=490\mu_s$$ $$\frac{125}{490}=\mu_s$$ $$0.255 = \mu_{s}$$

coefficient of kinetic friction solution

Click the reset icon to make sure the simulation is at the default friction.
Check the "Forces" and "Values" boxes.
Set the applied force to be enough to keep the box moving.
You should be able to see the value of the friction force. (94 N)

$$F_N = mg$$ $$F_N = (50)(9.8)$$ $$F_N = 490 \, \mathrm{N}$$

$$F_{k}=\mu_{k} F_{N}$$ $$94=\mu_k(490)$$ $$\frac{94}{490}=\mu_k$$ $$0.1918 =\mu_{k}$$

Investigation: As we increase the angle of an incline, a stationary mass on the incline will begin to slide.

What variables determine the coefficient of static friction?

solution

Gravity isn't in the same directions as the other forces so we can't use Newton's second law. We need to separate the gravity vector into components parallel and perpendicular to the ground.

$$F_{g}=mg$$

$$\text{perpendicular to ground}$$
$$F_{g\perp}=F_{g}\cos(\theta)$$ $$F_{g\perp}=mg\cos(\theta)$$
$$\text{parallel to ground}$$
$$F_{g\parallel}=F_{g}\sin(\theta)$$ $$F_{g\parallel}=mg\sin(\theta)$$

Since the acceleration is zero all opposite forces must be the same.

This tells us the normal force and the static friction force.

$$F_N = F_{g\perp}$$ $$F_{g\perp}=mg \cos{\theta}$$ $$F_s =F_{g\parallel}$$ $$F_{g\parallel}= mg \sin{\theta}$$

For the maximum angle, right at the point where the mass will start to slide, we can use the static friction equation.

$$F_{\mathrm{s\,max}} = \mu_s F_N$$

The maximum static friction will equal the parallel component of gravity at the largest possible angle before the mass will start to slide.

$$F_{g\parallel} = F_{\mathrm{s\,max}}$$ $$F_{g\parallel} = \mu_s F_N$$ $$mg \sin{\theta}= \mu_s mg \cos{\theta}$$ $$ \sin{\theta}= \mu_s \cos{\theta}$$ $$ \frac{\sin{\theta}}{\cos{\theta}}= \mu_s $$ $$ \boxed{\tan{\theta}= \mu_s} $$

The angle at which an object begins to slide depends on only the coefficient of static friction, not the mass or the acceleration of gravity!

What variables determine the coefficient of kinetic friction?

solution

Gravity isn't in the same directions as the other forces so we can't use Newton's second law. We need to separate the gravity vector into components parallel and perpendicular to the ground.

$$F_{g}=mg$$

$$\text{perpendicular to ground}$$
$$F_{g\perp}=F_{g}\cos(\theta)$$ $$F_{g\perp}=mg\cos(\theta)$$
$$\text{parallel to ground}$$
$$F_{g\parallel}=F_{g}\sin(\theta)$$ $$F_{g\parallel}=mg\sin(\theta)$$

The acceleration perpendicular to the ground is zero. This means the normal force equals the gravity component in that direction.

$$\sum F = ma$$ $$-F_k + mg \sin{\theta} = ma$$ $$-\mu_kF_N + mg \sin{\theta} = ma$$ $$-\mu_k m g \cos{\theta} + mg \sin{\theta} = ma$$ $$-\mu_k g \cos{\theta} + g \sin{\theta} = a$$ $$-\mu_k g \cos{\theta} = a - g \sin{\theta}$$ $$\mu_k g \cos{\theta} = g \sin{\theta}-a$$ $$ \boxed{\mu_k = \frac{g \sin{\theta}-a}{g \cos{\theta}}}$$

The coefficient of kinetic friction is dependent on the acceleration of the body, the acceleration of gravity, and the angle of the incline.

practice problems (20)

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

printout.pdf show topics and mark harder problems

Example: A cafeteria tray slides across a counter. The coefficient of kinetic friction is 0.32, and the normal force on the tray is 75 N. What is the kinetic friction force?

solution

$$\mu_k = 0.32$$ $$F_N = 75\,\mathrm{N}$$ $$F_k = \,?$$
$$F_k = \mu_k F_N$$ $$F_k = (0.32)(75)$$ $$F_k = 24\,\mathrm{N}$$

The friction force points opposite the tray's sliding motion.

Example: A 12 kg box of books slides across a level warehouse floor. The coefficient of kinetic friction between the box and the floor is 0.25. What is the kinetic friction force?

solution

On a level floor with no vertical acceleration, the normal force equals the weight.

$$F_N = mg$$ $$F_N = (12)(9.8)$$ $$F_N = 117.6\,\mathrm{N}$$
$$F_k = \mu_k F_N$$ $$F_k = (0.25)(117.6)$$ $$F_k = 29.4\,\mathrm{N}$$

Example: A 20 kg crate sits on a level floor. The coefficient of static friction is 0.45. If you push horizontally with 70 N, does the crate move? What is the static friction force?

solution

$$F_N = mg$$ $$F_N = (20)(9.8)$$ $$F_N = 196\,\mathrm{N}$$
$$F_{s\mathrm{\,max}} = \mu_s F_N$$ $$F_{s\mathrm{\,max}} = (0.45)(196)$$ $$F_{s\mathrm{\,max}} = 88.2\,\mathrm{N}$$

The 70 N push is less than the 88.2 N maximum, so the crate doesn't move.

$$F_s = 70\,\mathrm{N}$$

Static friction only pushes back as hard as it needs to. Here it is 70 N opposite the push, not 88.2 N. The 88.2 N is only the maximum possible value.

Example: The same 20 kg crate has a coefficient of static friction of 0.45 and a coefficient of kinetic friction of 0.30. If you push horizontally with 100 N, will it move? If it moves, what is the kinetic friction force, and what is the crate's acceleration?

solution

From the previous problem, the maximum static friction is 88.2 N. The 100 N push is larger, so the crate starts to move.

$$F_k = \mu_k F_N$$ $$F_k = (0.30)(196)$$ $$F_k = 58.8\,\mathrm{N}$$
$$\sum F = ma$$ $$100 - 58.8 = 20a$$ $$41.2 = 20a$$ $$a = 2.06\,\mathrm{\tfrac{m}{s^{2}}}$$

Once the crate is sliding, use kinetic friction, not the maximum static friction. That's why it's easier to keep something sliding than to get it started.

Example: A 10 kg box slides on a level floor, and the friction force on it is 34.3 N. What is the coefficient of kinetic friction?

solution

$$F_N = mg$$ $$F_N = (10)(9.8)$$ $$F_N = 98\,\mathrm{N}$$
$$F_k = \mu_k F_N$$ $$34.3 = \mu_k(98)$$ $$\mu_k = \frac{34.3}{98}$$ $$\mu_k = 0.35$$

The coefficient has no units because it is a ratio of two forces.

Example: A student uses a spring scale to pull a wooden block across a level lab table at a constant speed. The scale reads 3.92 N. The coefficient of kinetic friction is 0.40. What is the mass of the block?

solution

Constant speed means zero acceleration, so the pull is balanced by kinetic friction.

$$F_k = 3.92\,\mathrm{N}$$
$$F_k = \mu_k F_N$$ $$3.92 = (0.40)F_N$$ $$F_N = \frac{3.92}{0.40}$$ $$F_N = 9.8\,\mathrm{N}$$

On a level table, the normal force equals the weight.

$$F_N = mg$$ $$9.8 = m(9.8)$$ $$m = 1.0\,\mathrm{kg}$$

This is how a coefficient of friction is measured in a lab. Pulling at a constant speed makes the scale reading equal to the friction force.

Example: An 8.0 kg crate is pulled across a level floor with a 40 N horizontal force. The coefficient of kinetic friction is 0.20. What is the crate's acceleration?

solution

$$F_N = mg$$ $$F_N = (8.0)(9.8)$$ $$F_N = 78.4\,\mathrm{N}$$
$$F_k = \mu_k F_N$$ $$F_k = (0.20)(78.4)$$ $$F_k = 15.7\,\mathrm{N}$$
$$\sum F = ma$$ $$40 - 15.7 = 8.0a$$ $$24.3 = 8.0a$$ $$a = \frac{24.3}{8.0}$$ $$a = 3.04\,\mathrm{\tfrac{m}{s^{2}}}$$

Example: A kid on a sled has a total mass of 30 kg. They reach the flat bottom of a hill moving at 18 km/h. The coefficient of kinetic friction between the sled and the snow is 0.10. How long does it take the sled to stop?

solution

$$18\,\mathrm{\tfrac{\textcolor{DeepPink}{km}}{\textcolor{DodgerBlue}{h}}}\left(\frac{1000\,\mathrm{m}}{1\,\textcolor{DeepPink}{\mathrm{km}}}\right)\left(\frac{1\,\textcolor{DodgerBlue}{\mathrm{h}}}{3600\,\mathrm{s}}\right)$$ $$5.0\,\mathrm{\tfrac{m}{s}}$$
$$F_N = mg$$ $$F_N = (30)(9.8)$$ $$F_N = 294\,\mathrm{N}$$
$$F_k = \mu_k F_N$$ $$F_k = (0.10)(294)$$ $$F_k = 29.4\,\mathrm{N}$$

Friction is the only horizontal force, and it points backward.

$$\sum F = ma$$ $$-29.4 = 30a$$ $$a = -0.98\,\mathrm{\tfrac{m}{s^{2}}}$$
$$u = 5.0\,\mathrm{\tfrac{m}{s}}$$ $$v = 0$$ $$a = -0.98\,\mathrm{\tfrac{m}{s^{2}}}$$ $$\Delta t = \,?$$
$$v = u + a\Delta t$$ $$0 = 5.0 + (-0.98)\Delta t$$ $$0.98\Delta t = 5.0$$ $$\Delta t = 5.1\,\mathrm{s}$$

Snow has a low coefficient of friction, so the sled glides for a while before it stops.

Example: A 10 kg box slides across a warehouse floor at 4.0 m/s. What is the kinetic friction force on the box?

solution

This cannot be solved from the information given. The mass gives the normal force on a level floor, 98 N, but kinetic friction also needs the coefficient of kinetic friction. The speed doesn't help, because in this model kinetic friction doesn't depend on speed.

Example: A 1400 kg car is driving at 20 m/s on a dry concrete road when the driver slams on the brakes and the tires lock up and skid. Use a coefficient of kinetic friction of 0.70 for rubber on concrete (from the table on this page). How far does the car skid?

solution

Friction is the only horizontal force. Let forward be positive.

$$F_k = \mu_k F_N$$ $$F_k = \mu_k mg$$ $$\sum F = ma$$ $$-\mu_k mg = ma$$ $$a = -\mu_k g$$ $$a = -(0.70)(9.8)$$ $$a = -6.86\,\mathrm{\tfrac{m}{s^{2}}}$$
$$u = 20\,\mathrm{\tfrac{m}{s}}$$ $$v = 0$$ $$a = -6.86\,\mathrm{\tfrac{m}{s^{2}}}$$ $$\Delta x = \,?$$
$$v^{2} = u^{2} + 2a\Delta x$$ $$0^{2} = (20)^{2} + 2(-6.86)\Delta x$$ $$0 = 400 - 13.7\Delta x$$ $$\Delta x = \frac{400}{13.7}$$ $$\Delta x = 29\,\mathrm{m}$$

The car's mass cancels, so a heavy truck with the same tires would skid the same distance. Police use skid marks and this calculation to estimate how fast a car was going.

Example: A 6.0 kg block sits on a level table. The coefficient of static friction is 0.40. The block is attached by a string over a pulley to a 2.0 kg hanging mass. Will the system stay at rest? If so, what is the static friction force on the table block?

solution

$$\text{table block}$$ $$F_N = (6.0)(9.8)$$ $$F_N = 58.8\,\mathrm{N}$$ $$F_{s\mathrm{\,max}} = \mu_s F_N$$ $$F_{s\mathrm{\,max}} = (0.40)(58.8)$$ $$F_{s\mathrm{\,max}} = 23.5\,\mathrm{N}$$

$$\text{hanging mass}$$ $$F_g = mg$$ $$F_g = (2.0)(9.8)$$ $$F_g = 19.6\,\mathrm{N}$$

The hanging weight, 19.6 N, is less than the maximum static friction, 23.5 N, so the system stays at rest.

$$F_s = 19.6\,\mathrm{N}$$

Static friction only needs to balance the tension from the hanging mass, so it's 19.6 N, not 23.5 N.

Example: A 20 kg sled is pulled across level snow by a rope with 60 N of tension at 30° above the horizontal. The coefficient of kinetic friction is 0.10. What is the sled's acceleration?

solution

$$T_x = (60)\cos(30\degree)$$ $$T_x = 52.0\,\mathrm{N}$$

$$T_y = (60)\sin(30\degree)$$ $$T_y = 30.0\,\mathrm{N}$$

The rope lifts on the sled, so the normal force is less than the weight.

$$F_N + T_y - mg = 0$$ $$F_N + 30.0 - (20)(9.8) = 0$$ $$F_N + 30.0 - 196 = 0$$ $$F_N = 166\,\mathrm{N}$$
$$F_k = \mu_k F_N$$ $$F_k = (0.10)(166)$$ $$F_k = 16.6\,\mathrm{N}$$
$$\sum F = ma$$ $$52.0 - 16.6 = 20a$$ $$35.4 = 20a$$ $$a = 1.77\,\mathrm{\tfrac{m}{s^{2}}}$$

The normal force isn't always equal to the weight. Pulling up at an angle reduces the normal force, which also reduces friction.

Question: A brick slides across a floor lying on its largest face. Then it is turned on its narrow side, so much less of it touches the floor. How does the kinetic friction force change?

answer

In our friction model, it doesn't change. Friction depends on the normal force and the coefficient of friction, not the area of contact.

On its narrow side, less area touches the floor, but the brick's weight presses down harder on each part of that area. The two effects cancel. Real surfaces don't always follow this rule, but it works well for most dry, hard surfaces.

Example: A 10 kg box rests on a 15° ramp and does not slide. What are the normal force and the static friction force holding it in place?

solution

Make sure your calculator is in degree mode. The box isn't accelerating, so the forces balance in both directions.

$$\text{perpendicular to ramp}$$ $$F_N = mg\cos(15\degree)$$ $$F_N = (10)(9.8)\cos(15\degree)$$ $$F_N = 94.7\,\mathrm{N}$$

$$\text{parallel to ramp}$$ $$F_s = mg\sin(15\degree)$$ $$F_s = (10)(9.8)\sin(15\degree)$$ $$F_s = 25.4\,\mathrm{N}$$

Static friction points up the ramp, because gravity tries to pull the box down the ramp.

Example: A 10 kg box is placed on a 25° ramp. The coefficient of static friction is 0.35. Will the box stay at rest or slide?

solution

Find how much static friction would be needed to hold the box.

$$F_{g\parallel} = mg\sin(25\degree)$$ $$F_{g\parallel} = (10)(9.8)\sin(25\degree)$$ $$F_{g\parallel} = 41.4\,\mathrm{N}$$

Now find the maximum static friction available.

$$F_N = mg\cos(25\degree)$$ $$F_N = (10)(9.8)\cos(25\degree)$$ $$F_N = 88.8\,\mathrm{N}$$ $$F_{s\mathrm{\,max}} = \mu_s F_N$$ $$F_{s\mathrm{\,max}} = (0.35)(88.8)$$ $$F_{s\mathrm{\,max}} = 31.1\,\mathrm{N}$$

The box needs 41.4 N to stay put, but friction can only provide 31.1 N, so the box slides. You can also check with tan(25°) = 0.47, which is more than 0.35.

Example: To measure friction, a student slowly tilts a wooden board with a small block on it. The block starts to slide when the board reaches 31°. What is the coefficient of static friction between the block and the board?

solution

At the angle where the block just begins to slip, we derived that the coefficient of static friction equals the tangent of the angle.

$$\mu_s = \tan(\theta)$$ $$\mu_s = \tan(31\degree)$$ $$\mu_s = 0.60$$

The block's mass isn't needed. It cancels out of the derivation.

Example: A 12 kg box slides down a 20° ramp. The coefficient of kinetic friction is 0.20. What is the box's acceleration down the ramp?

solution

Choose down the ramp as positive. Gravity's parallel component points down the ramp, and kinetic friction points up the ramp.

$$\sum F = ma$$ $$mg\sin(\theta) - F_k = ma$$ $$mg\sin(\theta) - \mu_k mg\cos(\theta) = ma$$ $$g\sin(\theta) - \mu_k g\cos(\theta) = a$$ $$(9.8)\sin(20\degree) - (0.20)(9.8)\cos(20\degree) = a$$ $$3.35 - 1.84 = a$$ $$a = 1.51\,\mathrm{\tfrac{m}{s^{2}}}$$

The mass cancels, so a heavier box would have the same acceleration.

Example: A block slides down an 18° ramp with an acceleration of 1.2 m/s². What is the coefficient of kinetic friction?

solution

We derived this relationship for a block sliding down an incline.

$$\mu_k = \frac{g\sin(\theta) - a}{g\cos(\theta)}$$ $$\mu_k = \frac{(9.8)\sin(18\degree) - 1.2}{(9.8)\cos(18\degree)}$$ $$\mu_k = \frac{3.03 - 1.2}{9.32}$$ $$\mu_k = 0.20$$

Example: A 25 kg crate starts at rest on a level storage room floor. The coefficient of static friction is 0.50 and the coefficient of kinetic friction is 0.35. A person pulls horizontally with 150 N for 3.0 s. How far does the crate move?

solution

$$F_N = mg$$ $$F_N = (25)(9.8)$$ $$F_N = 245\,\mathrm{N}$$
$$F_{s\mathrm{\,max}} = (0.50)(245)$$ $$F_{s\mathrm{\,max}} = 122.5\,\mathrm{N}$$

The 150 N pull is larger than the maximum static friction, so the crate moves. Once it's sliding, use kinetic friction.

$$F_k = (0.35)(245)$$ $$F_k = 85.8\,\mathrm{N}$$
$$\sum F = ma$$ $$150 - 85.8 = 25a$$ $$64.2 = 25a$$ $$a = 2.57\,\mathrm{\tfrac{m}{s^{2}}}$$
$$u = 0$$ $$a = 2.57\,\mathrm{\tfrac{m}{s^{2}}}$$ $$\Delta t = 3.0\,\mathrm{s}$$ $$\Delta x = \,?$$
$$\Delta x = u\Delta t + \tfrac{1}{2}a\Delta t^{2}$$ $$\Delta x = (0)(3.0) + \tfrac{1}{2}(2.57)(3.0)^{2}$$ $$\Delta x = 11.6\,\mathrm{m}$$

Example: A force sensor slowly pulls harder and harder on a 2.0 kg block resting on a level table. The table shows the friction force at each pull. Find the coefficients of static and kinetic friction.

applied force (N)	friction force (N)	block
0	0	at rest
4.0	4.0	at rest
8.0	8.0	at rest
11.8	11.8	just starts to slide
12.0	7.8	sliding

solution

$$F_N = mg$$ $$F_N = (2.0)(9.8)$$ $$F_N = 19.6\,\mathrm{N}$$

While the block is at rest, static friction matches the pull. The largest static friction is 11.8 N, right before it slides.

$$\text{static}$$ $$F_{s\mathrm{\,max}} = \mu_s F_N$$ $$11.8 = \mu_s(19.6)$$ $$\mu_s = 0.60$$

$$\text{kinetic}$$ $$F_k = \mu_k F_N$$ $$7.8 = \mu_k(19.6)$$ $$\mu_k = 0.40$$

The friction force drops once the block starts sliding. That drop is the jerk you feel when a heavy object suddenly breaks loose.

Reading (10 minutes): Read What Even Is Friction, Anyway? by Rhett Allain from WIRED. Then answer these questions.

A book is pushed across a table but does not move. What does the article say about the static friction force compared with the push?

answer

Static friction matches the push in the opposite direction while the book remains at rest. It can increase up to its maximum value instead of always having one fixed size.

A coin slides down one side of a bowl and does not rise to its starting height on the other side. Where did the missing mechanical energy go?

answer

Friction transfers some of the coin's mechanical energy to thermal energy in the coin and bowl. The energy is not destroyed, but less remains as kinetic and gravitational potential energy.

Why would walking become difficult on nearly frictionless ice even though friction often slows objects down?

answer

Walking needs static friction from the ground to push a foot forward. With too little friction, the foot slips backward instead of helping the person accelerate forward.