Paul Peter Urone; Roger Hinrichs

Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The most important of these categories are discussed in this section, together with some interesting applications. Further examples of forces are discussed later in this text.

Normal Force

Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure 4.12(a). But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in Figure 4.12(b)? When the bag of dog food is placed on the table, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card table, but even rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or trampoline or diving board). The greater the deformation, the greater the restoring force. So when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. At this point the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it.

A person is holding a bag of dog food at some height from a table. He is exerting a force F sub hand, shown by a vector arrow in upward direction, and the weight W of the bag is acting downward, shown by a vector arrow having the same length as vector F sub hand. In a free-body diagram two forces are acting on the red point; one is F sub hand, shown by a vector arrow upward, and another is the weight W, shown by a vector arrow having the same length as vector F sub hand but pointing downward. (b) The bag of dog food is on the table, which deforms due to the weight W, shown by a vector arrow downward; the normal force N is shown by a vector arrow pointing upward having the same length as W. In the free-body diagram, vector W is shown by an arrow downward and vector N is shown by an arrow having the same length as vector W but pointing upward.

Figure 4.12 (a) The person holding the bag of dog food must supply an upward force

F_{hand}

equal in magnitude and opposite in direction to the weight of the food

w

. (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force

N

equal in magnitude and opposite in direction to the weight of the load.

We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force and here is given the symbol $N$ . (This is not the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object’s weight if the object is on an incline, as you will see in the next example.

Common Misconception: Normal Force (N) vs. Newton (N)

In this section we have introduced the quantity normal force, which is represented by the variable $N$ . This should not be confused with the symbol for the newton, which is also represented by the letter N. These symbols are particularly important to distinguish because the units of a normal force ( $N$ ) happen to be newtons (N). For example, the normal force $N$ that the floor exerts on a chair might be $N = 100 N$ . One important difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! You will encounter more similarities among variables and units as you proceed in physics. Another example of this is the quantity work ( $W$ ) and the unit watts (W).

Example 4.5

Weight on an Incline, a Two-Dimensional Problem

Consider the skier on a slope shown in Figure 4.13. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?

A skier is skiing down the slope and the slope makes a twenty-five degree angle with the horizontal. Her weight W, shown by a vector vertically downward, breaks into two components—one is W parallel, which is shown by a vector arrow parallel to the slope, and the other is W perpendicular, shown by a vector arrow perpendicular to the slope in the downward direction. Vector N is represented by an arrow pointing upward and perpendicular to the slope, having the same length as W perpendicular. Friction vector f is represented by an arrow along the slope in the uphill direction. IIn a free-body diagram, the vector arrow W for weight is acting downward, the vector arrow for f is shown along the direction of the slope, and the vector arrow for N is shown perpendicular to the slope.

Figure 4.13 Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier).

N

is perpendicular to the slope and f is parallel to the slope, but

w

has components along both axes, namely

w_{⊥}

and

w_{∥}

N

is equal in magnitude to

w_{⊥}

, so that there is no motion perpendicular to the slope, but

f

is less than

w_{∥}

, so that there is a downslope acceleration (along the parallel axis).

Strategy

This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one-dimensional problems to solve. 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 mutually perpendicular axes are independent.) We use the symbols $⊥$ and $∥$ to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled $w$ , $f$ , and $N$ in Figure 4.13. $N$ is always perpendicular to the slope, and $f$ is parallel to it. But $w$ is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining $w_{∥}$ 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 $w_{∥} = w sin (25º) = mg sin (25º)$ , and the magnitude of the component of the weight perpendicular to the slope is $w_{⊥} = w cos (25º) = mg cos (25º)$ .

(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only 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 $f$ . Using Newton’s second law, with subscripts to denote quantities parallel to the slope,

a_{∥} = \frac{F_{net ∥}}{m}

4.30

where $F_{net ∥} = w_{∥} = mg sin (25º)$ , assuming no friction for this part, so that

a_{∥} = \frac{F_{net ∥}}{m} = \frac{mg sin (25º)}{m} = g sin (25º)

4.31

(9.80 {m/s}^{2}) (0.4226) = 4.14 {m/s}^{2}

4.32

is the acceleration.

(b) Including friction. 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,

4.33

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

a_{∥} = \frac{F_{net ∣ ∣}}{m} = \frac{w_{∥} - f}{m} = \frac{mg sin (25º) - f}{m} .

4.34

We substitute known values to obtain

a_{∥} = \frac{(60 . 0 kg) (9 . {80 m/s}^{2}) (0.4226) - 45 . 0 N}{60 . 0 kg},

4.35

which yields

a_{∥} = 3 . {39 m/s}^{2},

4.36

which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.

Discussion

Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is $a = g sin θ$ , regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).

Resolving Weight into Components

Vector arrow W for weight is acting downward. It is resolved into components that are parallel and perpendicular to a surface that has a slope at angle theta to the horizontal. The coordinate direction x is labeled parallel to the sloped surface, with positive x pointing uphill. The coordinate direction y is labeled perpendicular to the sloped surface, with positive y pointing up from the surface. The components of w are w parallel, represented by an arrow pointing downhill along the sloped surface, and w perpendicular, represented by an arrow pointing into the sloped surface. W parallel is equal to w sine theta, which is equal to m g sine theta. W perpendicular is equal to w cosine theta, which is equal to m g cosine theta.

Figure 4.14 An object rests on an incline that makes an angle θ with the horizontal.

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 sin (θ) = mg sin (θ)

4.37

and

w_{⊥} = w cos (θ) = mg cos (θ) .

4.38

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 $θ$ 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{array}{l} cos (θ) & = & \frac{w_{⊥}}{w} \\ w_{⊥} & = & w cos (θ) = mg cos (θ) \end{array}

4.39

\begin{array}{l} sin (θ) & = & \frac{w_{∥}}{w} \\ w_{∥} & = & w sin (θ) = mg sin (θ) \end{array}

4.40

Take-Home Experiment: Force Parallel

To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of the rubber band, and a board you can position at different angles. How much does the rubber band stretch when you hang the object from the end of the board? Now place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show?

Tension

A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons. Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The tension force pulls outward along the two ends of a rope.

Consider a person holding a mass on a rope as shown in Figure 4.15.

An object of mass m is attached to a rope and a person is holding the rope. A weight vector W points downward starting from the lower point of the mass. A tension vector T is shown by an arrow pointing upward initiating from the hook where the mass and rope are joined, and a third vector, also T, is shown by an arrow pointing downward initiating from the hand of the person.

Figure 4.15 When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force

T

, that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope.

Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus $F_{net} = 0$ . The only external forces acting on the mass are its weight $w$ and the tension $T$ supplied by the rope. Thus,

F_{net} = T - w = 0,

4.41

where $T$ and $w$ are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass:

T = w = mg .

4.42

For a 5.00-kg mass, then (neglecting the mass of the rope) we see that

T = mg = (5.00 kg) (9 . {80 m/s}^{2}) = 49.0 N .

4.43

If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope.

Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in Figure 4.16 (a) and (b).

The internal structure of a finger with tendon, extensor muscle, and flexor muscle is shown. The force in the muscles is shown by arrows pointing along the tendon. In the second figure, part of a bicycle with a brake cable is shown. Three tension vectors are shown by the arrows along the brake cable, starting from the handle to the wheels. The tensions have the same magnitude but different directions.

Figure 4.16 (a) Tendons in the finger carry force

T

from the muscles to other parts of the finger, usually changing the force’s direction, but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension

T

from the handlebars to the brake mechanism. Again, the direction but not the magnitude of

T

is changed.

Example 4.6

What Is the Tension in a Tightrope?

Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 4.17.

A tightrope walker is walking on a wire. His weight W is acting downward, shown by a vector arrow. The wire sags and makes a five-degree angle with the horizontal at both ends. T sub R, shown by a vector arrow, is toward the right along the wire. T sub L is shown by an arrow toward the left along the wire. All three vectors W, T sub L, and T sub R start from the foot of the person on the wire. In a free-body diagram, W is acting downward, T sub R is acting toward the right with a small inclination, and T sub L is acting toward the left with a small inclination.

Figure 4.17 The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the tightrope walker is standing.