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

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

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. $$F_{\mathrm{max}}= \mu_{s}F_{N}$$ $$F_{\mathrm{max}}= \mu_{s}mg$$ $$\frac{F_{\mathrm{max}}}{mg} = \mu_{s}$$ $$\frac{125}{(50)(9.8)} = \mu_{s}$$ $$0.255 = \mu_{s}$$

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)

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

$$F_{k}=\mu_{k} F_{N}$$ $$\frac{F_{k}}{F_{N}}=\mu_{k}$$ $$\frac{94\, \mathrm{N}}{490\, \mathrm{N}}=\mu_{k}$$ $$0.1918 =\mu_{k}$$

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.

More Practice

Friction problems usually start with one decision: are the surfaces stuck, or are they sliding? Find the normal force first, then decide whether static friction is only matching the applied force or kinetic friction is doing a fixed-size job against the motion.

For the mixed problems, keep the force picture in your head before doing algebra. A good setup should make the direction of friction, the size of the normal force, and the reason an object starts moving or stays at rest clear.

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

solution

$$F_k=\mu_kF_N$$ $$F_k=(0.32)(75\,\mathrm{N})$$ $$F_k=24\,\mathrm{N}$$

The friction force is 24 N opposite the sliding motion.

Example: A 12 000 g shipping box slides across a level warehouse floor. The label says it contains books, but the coefficient of kinetic friction is the important surface detail: 0.25. What is the kinetic friction force?

solution

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

$$12\,000\,\mathrm{g}=12\,\mathrm{kg}$$ $$F_N=mg$$ $$F_N=(12\,\mathrm{kg})(9.8\,\mathrm{m/s^2})$$ $$F_N=117.6\,\mathrm{N}$$ $$F_k=\mu_kF_N$$ $$F_k=(0.25)(117.6\,\mathrm{N})$$ $$F_k=29.4\,\mathrm{N}$$

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

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

The push is less than the maximum static friction, so the crate does not move.

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

Static friction only uses as much force as it needs, so here it is 70 N opposite the push.

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

solution

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

The 100 N push is larger than the maximum static friction, so the crate starts moving.

$$F_k=\mu_kF_N$$ $$F_k=(0.30)(196\,\mathrm{N})$$ $$F_k=58.8\,\mathrm{N}$$

Once it is sliding, use kinetic friction, not the maximum static friction.

Example: A 10 kg box slides on a level floor and the friction force is 24.5 N. What is the coefficient of kinetic friction?

solution

$$F_N=mg$$ $$F_N=(10)(9.8)$$ $$F_N=98\,\mathrm{N}$$ $$F_k=\mu_kF_N$$ $$\mu_k=\frac{F_k}{F_N}$$ $$\mu_k=\frac{24.5\,\mathrm{N}}{98\,\mathrm{N}}$$ $$\mu_k=0.25$$

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

Example: A box slides on a level surface with coefficient of kinetic friction 0.40. The friction force is 0.0392 kN. What is the mass of the box?

solution

$$0.0392\,\mathrm{kN}=39.2\,\mathrm{N}$$ $$F_k=\mu_kF_N$$ $$F_N=\frac{F_k}{\mu_k}$$ $$F_N=\frac{39.2\,\mathrm{N}}{0.40}$$ $$F_N=98\,\mathrm{N}$$

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

$$F_N=mg$$ $$m=\frac{F_N}{g}$$ $$m=\frac{98\,\mathrm{N}}{9.8\,\mathrm{m/s^2}}$$ $$m=10\,\mathrm{kg}$$

Example: An 8 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

$$F_N=mg$$ $$F_N=(8)(9.8)$$ $$F_N=78.4\,\mathrm{N}$$ $$F_k=\mu_kF_N$$ $$F_k=(0.20)(78.4)$$ $$F_k=15.68\,\mathrm{N}$$ $$\sum F=F_{\mathrm{pull}}-F_k$$ $$\sum F=40\,\mathrm{N}-15.68\,\mathrm{N}$$ $$\sum F=24.32\,\mathrm{N}$$ $$\sum F=ma$$ $$a=\frac{24.32\,\mathrm{N}}{8\,\mathrm{kg}}$$ $$a=3.04\,\mathrm{m/s^2}$$

Example: A 5 kg sled is already sliding at 21.6 km/h on level snow. The coefficient of kinetic friction is 0.30, and the hill behind it is no longer affecting the motion. If no one pushes the sled, how long does it take to stop?

solution

$$21.6\,\mathrm{km/h}=6.0\,\mathrm{m/s}$$ $$F_N=mg$$ $$F_N=(5)(9.8)$$ $$F_N=49\,\mathrm{N}$$ $$F_k=\mu_kF_N$$ $$F_k=(0.30)(49)$$ $$F_k=14.7\,\mathrm{N}$$

Friction is the only horizontal force, so it causes acceleration opposite the motion.

$$a=\frac{-14.7\,\mathrm{N}}{5\,\mathrm{kg}}$$ $$a=-2.94\,\mathrm{m/s^2}$$ $$v=v_0+at$$ $$0=6.0+(-2.94)t$$ $$t=2.04\,\mathrm{s}$$

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

answer

This cannot be solved from the information given. The mass lets you find the normal force on a level floor, but kinetic friction also needs the coefficient of kinetic friction.

Example: A 20 kg crate is sliding at 28.8 km/h on a level floor. The coefficient of kinetic friction is 0.25. How far does it slide before stopping?

solution

$$28.8\,\mathrm{km/h}=8.0\,\mathrm{m/s}$$ $$F_N=mg$$ $$F_N=(20)(9.8)$$ $$F_N=196\,\mathrm{N}$$ $$F_k=\mu_kF_N$$ $$F_k=(0.25)(196)$$ $$F_k=49\,\mathrm{N}$$ $$a=\frac{-49\,\mathrm{N}}{20\,\mathrm{kg}}$$ $$a=-2.45\,\mathrm{m/s^2}$$ $$v^2=v_0^2+2a\Delta x$$ $$0^2=(8.0)^2+2(-2.45)\Delta x$$ $$\Delta x=13.1\,\mathrm{m}$$

Example: A 5 kg block rests on a level table and is attached by a light string over a pulley to a hanging mass. The coefficient of static friction between the block and table is 0.60. What is the largest hanging mass that can be held at rest?

solution

The table block can only provide static friction up to its maximum value.

$$F_N=mg$$ $$F_N=(5)(9.8)$$ $$F_N=49\,\mathrm{N}$$ $$F_{s\mathrm{max}}=\mu_sF_N$$ $$F_{s\mathrm{max}}=(0.60)(49)$$ $$F_{s\mathrm{max}}=29.4\,\mathrm{N}$$

At the largest hanging mass that can stay at rest, the hanging weight equals the maximum static friction.

$$m_h g=29.4\,\mathrm{N}$$ $$m_h=\frac{29.4\,\mathrm{N}}{9.8\,\mathrm{m/s^2}}$$ $$m_h=3.0\,\mathrm{kg}$$

Example: A 6 kg block sits on a level table with coefficient of static friction 0.40. It is attached over a pulley to a 2 kg hanging mass. Will the system stay at rest? If so, what is the static friction force on the table block?

solution

$$F_N=(6)(9.8)$$ $$F_N=58.8\,\mathrm{N}$$ $$F_{s\mathrm{max}}=\mu_sF_N$$ $$F_{s\mathrm{max}}=(0.40)(58.8)$$ $$F_{s\mathrm{max}}=23.52\,\mathrm{N}$$ $$F_{g,h}=m_hg$$ $$F_{g,h}=(2)(9.8)$$ $$F_{g,h}=19.6\,\mathrm{N}$$

The hanging weight is less than the maximum static friction, so the system can stay at rest.

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

Static friction matches the tension needed to hold the hanging mass, so it is 19.6 N on the table block.

Example: A 15 kg crate sits on a level floor. The coefficient of static friction is 0.50. A person pushes horizontally with 0.060 kN. Does the crate move, and what is the friction force?

solution

$$0.060\,\mathrm{kN}=60\,\mathrm{N}$$ $$F_N=mg$$ $$F_N=(15)(9.8)$$ $$F_N=147\,\mathrm{N}$$ $$F_{s\mathrm{max}}=\mu_sF_N$$ $$F_{s\mathrm{max}}=(0.50)(147)$$ $$F_{s\mathrm{max}}=73.5\,\mathrm{N}$$

The push is less than the maximum static friction, so the crate does not move.

$$F_s=60\,\mathrm{N}$$

This is a common static friction mistake: the friction force is not 73.5 N here. That is only the maximum possible value.

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

solution

The normal force equals the perpendicular component of gravity.

$$F_N=mg\cos(\theta)$$ $$F_N=(10)(9.8)\cos(15^\circ)$$ $$F_N=94.7\,\mathrm{N}$$

The static friction force must match the parallel component of gravity.

$$F_s=mg\sin(\theta)$$ $$F_s=(10)(9.8)\sin(15^\circ)$$ $$F_s=25.4\,\mathrm{N}$$

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 degree ramp. The coefficient of static friction is 0.35. Will the box stay at rest or slide?

solution

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

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

Now find the maximum static friction available.

$$F_N=mg\cos(\theta)$$ $$F_N=(10)(9.8)\cos(25^\circ)$$ $$F_N=88.8\,\mathrm{N}$$ $$F_{s\mathrm{max}}=\mu_sF_N$$ $$F_{s\mathrm{max}}=(0.35)(88.8)$$ $$F_{s\mathrm{max}}=31.1\,\mathrm{N}$$

The required 41.4 N is greater than the maximum available 31.1 N, so the box slides down the ramp.

Example: A small block begins to slide when a board reaches an angle of 28.8 degrees. What is the coefficient of static friction between the block and the board?

solution

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

$$\mu_s=\tan(\theta)$$ $$\mu_s=\tan(28.8^\circ)$$ $$\mu_s=0.55$$

Example: A 12 kg box is sliding down a 20 degree ramp. The coefficient of kinetic friction is 0.20. What is the 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.

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

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

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

solution

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

$$\mu_k=\frac{g\sin(\theta)-a}{g\cos(\theta)}$$ $$\mu_k=\frac{(9.8)\sin(18^\circ)-1.2}{(9.8)\cos(18^\circ)}$$ $$\mu_k=0.196$$

The coefficient of kinetic friction is about 0.20.

Example: A 25 kg crate starts at rest on a level floor in a storage room. 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 0.050 min, and the crate's handle is 12 cm wide. How far does the crate move?

solution

First convert the time, then check whether the crate starts moving.

$$0.050\,\mathrm{min}=3.0\,\mathrm{s}$$ $$F_N=mg$$ $$F_N=(25)(9.8)$$ $$F_N=245\,\mathrm{N}$$ $$F_{s\mathrm{max}}=\mu_sF_N$$ $$F_{s\mathrm{max}}=(0.50)(245)$$ $$F_{s\mathrm{max}}=122.5\,\mathrm{N}$$

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

$$F_k=\mu_kF_N$$ $$F_k=(0.35)(245)$$ $$F_k=85.75\,\mathrm{N}$$ $$\sum F=150-85.75$$ $$\sum F=64.25\,\mathrm{N}$$ $$a=\frac{64.25}{25}$$ $$a=2.57\,\mathrm{m/s^2}$$ $$\Delta x=v_0t+\frac{1}{2}at^2$$ $$\Delta x=0+\frac{1}{2}(2.57)(3.0)^2$$ $$\Delta x=11.6\,\mathrm{m}$$