| The Definition and Interpretation of the Normal Force |
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In many situations, an object is in contact with a surface, such as a tabletop. Because of the contact, there is a force acting on the object. The present section discusses only one component of this force, the component that acts perpendicular to the surface. The next section discusses the component that acts parallel to the surface. The perpendicular component is called the
normal force.
| Definition of the Normal Force |
| The normal force is one component of the force that a surface exerts on an object with which it is in contact—namely, the component that is perpendicular to the surface.
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Newton’s third law plays an important role in connection with the normal force. In Figure 4-14, for instance, the block exerts a force on the table by pressing down on it. Consistent with the third law, the table exerts an oppositely directed force of equal magnitude on the block. This reaction force is the normal force. The magnitude of the normal force indicates how hard the two objects press against each other.
If an object is resting on a horizontal surface and there are no vertically acting forces except the object’s weight and the normal force, the magnitudes of these two forces are equal; that is,
. This is the situation in Figure 4-14. The weight must be balanced by the normal force for the object to remain at rest on the table. If the magnitudes of these forces were not equal, there would be a net force acting on the block, and the block would accelerate either upward or downward, in accord with Newton’s second law.
If other forces in addition to
and
act in the vertical direction, the magnitudes of the normal force and the weight are no longer equal. In Figure 4-15
a, for instance, a box whose weight is 15 N is being pushed downward against a table. The pushing force has a magnitude of 11 N. Thus, the total downward force exerted on the box is 26 N, and this must be balanced by the upward-acting normal force if the box is to remain at rest. In this situation, then, the normal force is 26 N, which is considerably larger than the weight of the box.
| | | | Figure 4-15 (a) The normal force is greater than the weight of the box, because the box is being pressed downward with an 11-N force. (b) The normal force is smaller than the weight, because the rope supplies an upward force of 11 N that partially supports the box.
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Figure 4-15b illustrates a different situation. Here, the box is being pulled upward by a rope that applies a force of 11 N. The net force acting on the box due to its weight and the rope is only 4 N, downward. To balance this force, the normal force needs to be only 4 N. It is not hard to imagine what would happen if the force applied by the rope were increased to 15 N—exactly equal to the weight of the box. In this situation, the normal force would become zero. In fact, the table could be removed, since the block would be supported entirely by the rope. The situations in Figure 4-15 are consistent with the idea that the magnitude of the normal force indicates how hard two objects press against each other. Clearly, the box and the table press against each other harder in part a of the picture than in part b.
Like the box and the table in Figure 4-15, various parts of the human body press against one another and exert normal forces. Example 8 illustrates the remarkable ability of the human skeleton to withstand a wide range of normal forces.
| Example 8 | | | A Balancing Act |
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Reasoning To begin, we draw a free-body diagram for the neck and head of the standing performer. Before the act, there are only two forces, the weight of the standing performer’s head and neck, and the normal force. During the act, an additional force is present due to the woman’s weight. In both cases, the upward and downward forces must balance for the head and neck to remain at rest. This condition of balance will lead us to values for the normal force.
The physics of the human skeleton. |
Solution
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a. | Figure 4-16 b shows the free-body diagram for the standing performer’s head and neck before the act. The only forces acting are the normal force and the 50-N weight. These two forces must balance for the standing performer’s head and neck to remain at rest. Therefore, the seventh cervical vertebra exerts a normal force of . | |
b. | Figure 4-16 c shows the free-body diagram that applies during the act. Now, the total downward force exerted on the standing performer’s head and neck is , which must be balanced by the upward normal force, so that . | | |
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In summary, the normal force does not necessarily have the same magnitude as the weight of the object. The value of the normal force depends on what other forces are present. It also depends on whether the objects in contact are accelerating. In one situation that involves accelerating objects, the magnitude of the normal force can be regarded as a kind of “apparent weight,” as we will now see.
| Apparent Weight |
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Usually, the weight of an object can be determined with the aid of a scale. However, even though a scale is working properly, there are situations in which it does not give the correct weight. In such situations, the reading on the scale gives only the “apparent” weight, rather than the gravitational force or “true” weight. The apparent weight is the force that the object exerts on the scale with which it is in contact.
If the elevator is accelerating, the apparent weight and the true weight are not equal. When the elevator accelerates upward, the apparent weight is greater than the true weight, as Figure 4-17b shows. Conversely, if the elevator accelerates downward, as in part c, the apparent weight is less than the true weight. In fact, if the elevator falls freely, so its acceleration is equal to the acceleration due to gravity, the apparent weight becomes zero, as part d indicates. In a situation such as this, where the apparent weight is zero, the person is said to be “weightless.” The apparent weight, then, does not equal the true weight if the scale and the person on it are accelerating.
The discrepancies between true weight and apparent weight can be understood with the aid of Newton’s second law. Figure 4-18 shows a free-body diagram of the person in the elevator. The two forces that act on him are the true weight
and the normal force
exerted by the platform of the scale. Applying Newton’s second law in the vertical direction gives
where
a is the acceleration of the elevator and person. In this result, the symbol
g stands for the magnitude of the acceleration due to gravity and can never be a negative quantity. However, the acceleration
a may be either positive or negative, depending on whether the elevator is accelerating upward (+) or downward (−). Solving for the normal force
FN shows that
In Equation 4.6,
FN is the magnitude of the normal force exerted on the person by the scale. But in accord with Newton’s third law,
FN is also the magnitude of the downward force that the person exerts on the scale—namely, the apparent weight.
Equation 4.6 contains all the features shown in Figure 4-17. If the elevator is not accelerating,
a = 0 m/s
2, and the apparent weight equals the true weight. If the elevator accelerates upward,
a is positive, and the equation shows that the apparent weight is greater than the true weight. If the elevator accelerates downward,
a is negative, and the apparent weight is less than the true weight. If the elevator falls freely,
a = −
g, and the apparent weight is zero. The apparent weight is zero because when both the person and the scale fall freely, they cannot push against one another. In this text, when the weight is given, it is assumed to be the true weight, unless stated otherwise.
| | | | Figure 4-18 A free-body diagram showing the forces acting on the person riding in the elevator of Figure 4-17. is the true weight, and is the normal force exerted on the person by the platform of the scale.
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Copyright © 2007 John Wiley & Sons, Inc. All rights reserved. |