Inclined Planes

5.4 Inclined Planes

Section Learning Objectives

By the end of this section, you will be able to do the following:

  • Distinguish between static friction and kinetic friction
  • Solve problems involving inclined planes

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Teacher Support

The learning objectives in this section will help your students master the following standards:

  • (4) Science concepts. The student knows and applies the laws governing motion in two dimensions for a variety of situations. The student is expected to:
    • (D) calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects.

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.

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Teacher Support

[BL][OL] Review the concept of friction.

[AL] Start a discussion about the two kinds of friction: static and kinetic. Ask students which one they think would be greater for two given surfaces. Explain the concept of coefficient of friction and what the number would imply in practical terms. Look at the table of static and kinetic friction and ask students to guess which other systems would have higher or lower coefficients.

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 5.33 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.

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.
Figure 5.33 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 fs is

f s ≤ μ s N , f s ≤ μ s N ,

where μ s 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 ≤ ≤ means less than or equal to, this equation says that static friction can have a maximum value of μ s N. μ s N. That is,

f s (max)= μ s N. 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 fk is given by

f k = μ k N. f k = μ k N.

where μ k μ k is the coefficient of kinetic friction.

Friction varies from surface to surface because different substances are rougher than others. Table 5.2 compares values of static and kinetic friction for different surfaces. The coefficient of the friction depends on the two surfaces that are in contact.

System Static Friction μ s μ s Kinetic Friction μ k 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
Table 5.2 Coefficients of Static and Kinetic Friction

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=mg=(100kg)(9.80 m/s 2 )=980 N, W=mg=(100kg)(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 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 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 5.34. The first step when setting up the problem is to break down the force of weight into components.

A diagram shows perpendicular and horizontal components of weight. The weight is on an inclined plane that slopes downward to the right.
Figure 5.34 The diagram shows perpendicular and horizontal components of weight on an inclined plane.

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Teacher Support

[BL] Review the concepts of mass, weight, gravitation and normal force.

[OL] Review vectors and components of vectors.

When an object rests on an incline that makes an angle θ 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 ⊥ w ⊥ , and a force acting parallel to the plane, w || w || . The perpendicular force of weight, w ⊥ w ⊥ , is typically equal in magnitude and opposite in direction to the normal force, N. N. The force acting parallel to the plane, w || w || , causes the object to accelerate down the incline. The force of friction, f 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 θ to the horizontal, then the magnitudes of the weight components are

w || =wsin(θ)=mgsin(θ) and w || =wsin(θ)=mgsin(θ) and
w ⊥ =wcos(θ)=mgcos(θ). w ⊥ =wcos(θ)=mgcos(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 w and w ⊥ w ⊥ . Knowing this property, you can use trigonometry to determine the magnitude of the weight components

cos(θ)= w ⊥ w w ⊥ =wcos(θ)=mgcos(θ) cos(θ)= w ⊥ w w ⊥ =wcos(θ)=mgcos(θ)
sin(θ)= w || w w || =wsin(θ)=mgsin(θ). sin(θ)= w || w w || =wsin(θ)=mgsin(θ).

Teacher Support

Teacher Support

[BL][OL][AL] Experiment with sliding different objects on inclined planes to understand static and kinetic friction. Which objects need a larger angle to slide down? What does this say about the coefficients of friction of those systems? Is a greater force required to start the motion of an object than to keep it in motion? What does this say about static and kinetic friction? When does an object slide down at constant velocity? What does this say about friction and normal force?

Watch Physics

Inclined Plane Force Components

This video shows how the weight of an object on an inclined plane is broken down into components perpendicular and parallel to the surface of the plane. It explains the geometry for finding the angle in more detail.