Inclined Planes, Motion in Two Dimensions | Physics

Learning objectives

Section Key Terms

kinetic friction

static friction

Static Friction and Kinetic Friction

Recall from the previous chapter that friction is a force that opposes relative motion parallel to the contact surface of the interacting objects and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.

There are different types of friction—kinetic and static. Kinetic friction acts on an object in relative motion, while static friction acts on an object or system at rest relative to each other. The maximum static friction is usually greater than the kinetic friction between the objects.

Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.

Figure_05_04_surface shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.

View original figure (external; image not reproduced in this text edition)

Original figure description: A box is shown sitting on a floor. A vector labeled f points to the left from the area where the bottom of the box meets the floor. A zoom-in diagram shows the bottom surface of the box and the surface of the floor as rough.

Frictional forces, such as f, always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view.

The magnitude of the frictional force has two forms: one for static friction, the other for kinetic friction. When there is no motion between the objects, the magnitude of static friction f_s is

f_{s} \leq μ_{s} N,

where $μ_{s}$ is the coefficient of static friction and N is the magnitude of the normal force. Recall that the normal force acts perpendicular to the surface and prevents the crate from falling through the floor. It opposes the force of gravity in this example, but that will not always be the case.

Since the symbol $\leq$ means less than or equal to, this equation says that static friction can have a maximum value of $μ_{s} N .$ That is,

f_{s} (max) = μ_{s} N.

Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds fs(max), the object will move. Once an object is moving, the magnitude of kinetic friction f_k is given by

f_{k} = μ_{k} N.

where $μ_{k}$ is the coefficient of kinetic friction.

Friction varies from surface to surface because different substances are rougher than others. Table_05_02 compares values of static and kinetic friction for different surfaces. The coefficient of the friction depends on the two surfaces that are in contact.

Coefficients of Static and Kinetic Friction
System	Static Friction $μ_{s}$	Kinetic Friction $μ_{k}$
Rubber on dry concrete	1.0	0.7
Rubber on wet concrete	0.7	0.5
Wood on wood	0.5	0.3
Waxed wood on wet snow	0.14	0.1
Metal on wood	0.5	0.3
Steel on steel (dry)	0.6	0.3
Steel on steel (oiled)	0.05	0.03
Teflon on steel	0.04	0.04
Bone lubricated by synovial fluid	0.016	0.015
Shoes on wood	0.9	0.7
Shoes on ice	0.1	0.05
Ice on ice	0.1	0.03
Steel on ice	0.4	0.02

Since friction always opposes relative motion, the direction of friction is upward along the plane if the object is at rest or sliding down the incline. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight

W = m g = (100 kg) (9.80 {m/s}^{2}) = 980 N,

perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than

f_{s} (max) = μ_{s} N = (0.45) (980 N) = 440 N

to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N

f_{k} = μ_{k} N = (0.30) (980 N) = 290 N

would keep it moving at a constant speed. If the floor were lubricated, both coefficients would be much smaller than they would be without lubrication. The coefficient of friction is unitless and is a number usually between 0 and 1.0, but there is no theoretical upper limit to its value.

Working with Inclined Planes

We discussed previously that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Up until now, we dealt only with normal force in one dimension, with gravity and normal force acting perpendicular to the surface in opposing directions (gravity downward, and normal force upward). Now that you have the skills to work with forces in two dimensions, we can explore what happens to weight and the normal force on a tilted surface such as an inclined plane. For inclined plane problems, it is easier breaking down the forces into their components if we rotate the coordinate system, as illustrated in Figure_05_04_weight. The first step when setting up the problem is to break down the force of weight into components.

View original figure (external; image not reproduced in this text edition)

Original figure description: A diagram shows perpendicular and horizontal components of weight. The weight is on an inclined plane that slopes downward to the right.

The diagram shows perpendicular and horizontal components of weight on an inclined plane.

When an object rests on an incline that makes an angle $θ$ with the horizontal, the force of gravity acting on the object is divided into two components: A force acting perpendicular to the plane, $w_{⊥}$ , and a force acting parallel to the plane, $w_{| |}$ . The perpendicular force of weight, $w_{⊥}$ , is typically equal in magnitude and opposite in direction to the normal force, $N .$ The force acting parallel to the plane, $w_{| |}$ , causes the object to accelerate down the incline. The force of friction, $f$ , opposes the motion of the object, so it acts upward along the plane.

It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle $θ$ to the horizontal, then the magnitudes of the weight components are

w_{| |} = w s i n (θ) = m g s i n (θ) and

w_{⊥} = w c o s (θ) = m g c o s (θ) .

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle of the incline is the same as the angle formed between $w$ and $w_{⊥}$ . Knowing this property, you can use trigonometry to determine the magnitude of the weight components

\begin{matrix} c o s (θ) = \frac{w_{⊥}}{w} \\ w_{⊥} = w c o s (θ) = m g c o s (θ) \end{matrix}

\begin{matrix} s i n (θ) = \frac{w_{| |}}{w} \\ w_{| |} = w s i n (θ) = m g s i n (θ) . \end{matrix}

To review, the process for solving inclined plane problems is as follows:

Draw a sketch of the problem.
Identify known and unknown quantities, and identify the system of interest.
Draw a free-body diagram (which is a sketch showing all of the forces acting on an object) with the coordinate system rotated at the same angle as the inclined plane. Resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x -direction) then Fnet x = 0. If the object does accelerate in that direction, Fnet x = ma.
Check your answer. Is the answer reasonable? Are the units correct?

Finding the Coefficient of Kinetic Friction on an Inclined Plane

A skier, illustrated in Figure_05_04_skiera (a), with a mass of 62 kg is sliding down a snowy slope at an angle of 25 degrees. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.

View original figure (external; image not reproduced in this text edition)

Original figure description: A skier is moving down a slope with an angle measure of twenty-five degrees to the ground. Force vectors are shown: N points upward from the skier, f points to the right of the skis, w points at a downward left angle from the skiers' feet, forming an angle with vector w that has a measure of twenty-five degrees. A free-body diagram illustrates the right triangle formed by these vectors.

Use the diagram to help find the coefficient of kinetic friction for the skier.

Strategy

The magnitude of kinetic friction was given as 45.0 N. Kinetic friction is related to the normal force N as $f_{k} = μ_{k} N$ . Therefore, we can find the coefficient of kinetic friction by first finding the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier’s weight perpendicular to the slope.

That is,

N = w_{⊥} = w cos (25^{\circ}) = m g cos (25^{\circ}) .

Substituting this into our expression for kinetic friction, we get

f_{k} = μ_{k} m g {cos 25}^{\circ},

which can now be solved for the coefficient of kinetic friction μ_k.

Solution

Solving for $μ_{k}$ gives

$μ_{k} = \frac{f_{k}}{w {cos 25}^{\circ}} = \frac{f_{k}}{m g {cos 25}^{\circ}} .$

Substituting known values on the right-hand side of the equation,

$μ_{k} = \frac{45.0 N}{(62 kg)(9 {.80 m/s}^{2}) (0.906)} = 0.082 .$

Discussion

This result is a little smaller than the coefficient listed in Table_05_02 for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly. In situations like this, where an object of mass m slides down a slope that makes an angle θ with the horizontal, friction is given by $f_{k} = μ_{k} m g cos θ .$

Weight on an Incline, a Two-Dimensional Problem

The skier’s mass, including equipment, is 60.0 kg. (See Figure_05_04_skierb (b).) (a) What is her acceleration if friction is negligible? (b) What is her acceleration if the frictional force is 45.0 N?

View original figure (external; image not reproduced in this text edition)

Now use the diagram to help find the skier's acceleration if friction is negligible and if the frictional force is 45.0 N.

Strategy

The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. Remember that motions along perpendicular axes are independent. We use the symbol $⊥$ to mean perpendicular, and $| |$ to mean parallel.

The only external forces acting on the system are the skier’s weight, friction, and the normal force exerted by the ski slope, labeled $w$ , $f$ , and $N$ in the free-body diagram. $N$ is always perpendicular to the slope and $f$ is parallel to it. But $w$ is not in the direction of either axis, so we must break it down into components along the chosen axes. We define $w_{| |}$ to be the component of weight parallel to the slope and $w_{⊥}$ the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.

Solution

The magnitude of the component of the weight parallel to the slope is $w_{| |} = w sin (25 °) = m g sin (25 °)$ , and the magnitude of the component of the weight perpendicular to the slope is $w_{⊥} = w cos (25 °) = m g cos (25 °).$

(a) Neglecting friction: Since the acceleration is parallel to the slope, we only need to consider forces parallel to the slope. Forces perpendicular to the slope add to zero, since there is no acceleration in that direction. The forces parallel to the slope are the amount of the skier’s weight parallel to the slope $w_{| |}$ and friction $f$ . Assuming no friction, by Newton’s second law the acceleration parallel to the slope is

a_{| |} = \frac{F_{net ||}}{m},

Where the net force parallel to the slope $F_{net ||} = w_{| |} = m g sin (25 °)$ , so that

\begin{matrix} a_{| |} & = & \frac{F_{net ||}}{m} = \frac{m g sin (25 °)}{m} = g sin (25 °) \\ = & (9.80 {m/s}^{2}) (0.423) = 4.14 {m/s}^{2} \end{matrix}

is the acceleration.

(b) Including friction: Here we now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now

F_{net | |} = w_{| |} - f,

and substituting this into Newton’s second law, $a_{| |} = \frac{F_{net ||}}{m}$ gives

a_{| |} = \frac{F_{net ||}}{m} = \frac{w_{| |} - f}{m} = \frac{m g sin (25 °) - f}{m} .

We substitute known values to get

a_{| |} = \frac{(60 .0 kg)(9 {.80 m/s}^{2})(0 .423) - 45 .0 N}{60 .0 kg},

a_{| |} = 3 {.39 m/s}^{2},

which is the acceleration parallel to the incline when there is 45 N opposing friction.

Discussion

Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is not.

Practice Problems

When an object sits on an inclined plane that makes an angle θ with the horizontal, what is the expression for the component of the objects weight force that is parallel to the incline?

$w_{||} = w cos θ$
$w_{||} = w sin θ$
$w_{||} = w sin θ - cos θ$
$w_{||} = w cos θ - sin θ$

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Friction at an Angle: Sliding a Coin

An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in the first Worked Example, the kinetic friction on a slope $f_{k} = μ_{k} m g cos θ$ , and the component of the weight down the slope is equal to $m g sin θ$ . These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out

\begin{matrix} f_{k} & = & F g_{x} \\ μ_{k} m g cos θ & = & m g sin θ. \end{matrix}

Solving for $μ_{k}$ , since $tan θ = sin θ /cos θ$ we find that

$μ_{k} = \frac{m g sin θ}{m g cos θ} = tan θ.$

1 coin
1 book
1 protractor
1. Put a coin flat on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move.
2. Measure the angle of tilt relative to the horizontal and find $μ_{k}$ .

True or FalseIf only the angles of two vectors are known, we can find the angle of their resultant addition vector.

True
False

Section Summary

Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. A normal force is always perpendicular to the contact surface between systems. Friction depends on both of the materials involved.
µ_s is the coefficient of static friction, which depends on both of the materials.
µ_k is the coefficient of kinetic friction, which also depends on both materials.
When objects rest on an inclined plane that makes an angle $θ$ with the horizontal surface, the weight of the object can be broken into components that act perpendicular $(w_{⊥})$ and parallel ( $w_{| |}$ ) to the surface of the plane.

Key Equations

force of static friction	$f_{s} \leq μ_{s} N$
force of kinetic friction	$f_{k} = μ_{k} N$
perpendicular component of weight on an inclined plane	$w_{⊥} = w cos (θ) = m g cos (θ)$
parallel component of weight on an inclined plane	$w_{\| \|} = w sin (θ) = m g sin (θ)$

Check Your Understanding

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Concept Items

True or FalseKinetic friction is less than the limiting static friction because once an object is moving, there are fewer points of contact, and the friction is reduced. For this reason, more force is needed to start moving an object than to keep it in motion.

True
False

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Problems

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A skier with a mass of 55 kg is skiing down a snowy slope that has an incline of 30°. Find the coefficient of kinetic friction for the skier if friction is known to be 25 N .

$μ k = 0$
$μ k = 0.05$
$μ k = 0.09$
$μ k = ∞$

Critical Thinking

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Test Prep Multiple Choice

For objects of identical masses but made of different materials, which of the following experiences the most static friction?

Shoes on ice
Metal on wood
Teflon on steel

If an object sits on an inclined plane and no other object makes contact with the object, what is typically equal in magnitude to the component of the weight perpendicular to the plane?

The normal force
The total weight
The parallel force of weight

A 5 kg box is at rest on the floor. The coefficient of static friction between the box and the floor is 0.4. A horizontal force of 50 N is applied to the box. Will it move?

No, because the applied force is less than the maximum limiting static friction.
No, because the applied force is more than the maximum limiting static friction.
Yes, because the applied force is less than the maximum limiting static friction.
Yes, because the applied force is more than the maximum limiting static friction.

A skier with a mass of 67 kg is skiing down a snowy slope with an incline of 37°. Find the friction if the coefficient of kinetic friction is 0.07.

27.66 N
34.70 N
36.71 N
45.96 N

Test Prep Short Answer

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Test Prep Extended Response

True or False—Friction varies from surface to surface because different substances have different degrees of roughness or smoothness.

True
False

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Glossary

kinetic friction: a force that opposes the motion of two systems that are in contact and moving relative to one another
static friction: a force that prevents the relative motion of two systems that are in contact and are not moving relative to one another

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