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5-6 - 5-7 Weight, Normal Force, & Inclined Surfaces

(A) Weight
We all live on the surface of the earth. As a result of this our everyday lives are partly governed by the fact that everything has weight. This weight is a direct result of Earth's gravitational pull; in fact, on Earth, a body's weight is the downward gravitational force exerted on the body by earth. You may recall that near the Earth's surface all bodies fall with the gravitational force on a body, its weight, equals the product of its mass and this gravitational acceleration

W = mg

The direction of the vectors W and g is toward the center of the earth.

The sensation of having weight is clear to us because of the force of contact between our feet and the ground beneath us. Earth's gravity presses us into the floor and the floor reacts back onto us. This reaction is what feels like our weight. However, if the object on which we are standing is accelerating this reaction force can trick us into feeling either heavier or lighter depending on the direction of the acceleration. This reaction force is called our apparent weight because it is the weight we perceive ourselves to have even when it differs from the actual force of gravity on us.

(B) Normal Force
Another "everyday" force occurs when two surfaces come into direct physical contact. The force of contact between the surfaces can be resolved into components that are parallel and perpendicular to the surfaces. The perpendicular component of this force is called the normal force (normal means perpendicular). The normal force that the floor exerts on your feet when you are standing is what we called the apparent weight just above. As you'll see in the next chapter, the normal force between two surfaces is an important factor in determining the friction between those surfaces.

Physlet Illustration: Riding Down in an Elevator
a = m/s2
A 50-kg box is riding in an elevator that accelerates downward at a constant rate. (Its velocity vector is shown in green.) The box rests on a digital scale that records its apparent weight in Newtons. Adjust the value of the acceleration (a < 9.8 m/s2) and see how it affects the apparent weight. How can the box become apparently "weightless"? Start
Hints Can you draw a free-body diagram for the box? What force is recorded on the scale? How does Newton's 3rd Law apply to this example? Apply Newton's 2nd Law to the box's motion, and determine the value of its acceleration that would make that force vanish.
Reference See Walker, Section 5-6

Example 5.3 Going Up in an Elevator: Low acceleration elevators, such as might be found in a hospital, typically accelerate at about 3.00 ft/s² when first starting upward. If a person knows her weight to be 125 lb, what would a scale read if she is standing on it when the elevator starts upward?

Picture the Problem The lefthand sketch shows a person standing on a scale in an upward accelerating elevator. The righthand scale is the free-body diagram.

Strategy The problem asks for the reading on the scale, which is determined by the force that the person exerts on the scale. By Newton's third law we know that the force on the scale equals the normal force N of the scale on the person (her apparent weight N = W_a). So, instead we draw a free-body diagram of the person because we know the person's true weight. Let's take up as +y.

Solution

1. Apply Newton's second law to the free-body diagram:
2. Solving for N gives the apparent weight:

Insight Notice that in the sum of forces in step 1 W_y = -W = -125 lb.

Physlet Illustration: The Hockey Puck
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F = Newtons  q = °
A 250-gm hockey puck, acted upon by a single force, is free to slide on the ice. The velocity components are given in m/s, and the times shown are in seconds. Adjust the angle and/or the magnitude of the applied force (0 < F < 12 N), and measure the puck's acceleration. How can you verify that it obeys Newton's Second Law?
Hints Measure the components of the acceleration by finding the change in the components of velocity over a specific time interval. How do the components of the acceleration depend on the force?
Reference See Walker, Section 5-3

Example 5.4 Rearranging: While rearranging the living room a student pushes a 27.5 kg sofa across the room at constant speed by applying a force that makes an angle of 35.0^o below the horizontal. If the floor opposes the sliding sofa (by friction) with a backward force of 160 N then, (a) what is the normal force of the floor on the sofa, and (b) what magnitude of force does the student apply to the sofa?

Solution Try to sketch a physical picture for this problem; the free-body diagram for the sofa is shown below.

Given: m = 27.5 kg,, f = 160N Find: (a) N, (b) F

The free-body diagram for the sofa is

Taking the +x direction to the right and the +y direction up we can apply Newton's second law to both the x- and y-directions. Since part (a) asks about the normal force, let's start with the y-direction and see where it leads us. The sofa does not accelerate vertically up nor down, so the net vertical force must be zero,

.

We cannot solve this immediately for N because we don't yet know F_y. Notice, however, that since we know the direction of F, we can determine F_y provided we can figure out F_x. So we next apply Newton's second law to the x-direction. The sofa is being pushed at constant speed, therefore the horizontal acceleration is zero which means that the net horizontal force must also be zero,

.

Now that we know F_x we determine F_y through the relation . Therefore,

.

The minus sign just means that it's along the -y direction. Having F_y, we are now ready to calculate the normal force on the sofa.

,

which is our final result for part (a).

For part (b) we seek the magnitude of F. Since we have already calculated the two components of F, we can determine its magnitude from the Pythagorean theorem.

.

Insights An important point here is that Newton's second law must be satisfied in every direction and that often a result from an equation for one direction is useful in solving an equation for a different direction.

Practice Quiz

An object is at rest on a horizontal table. The normal force exerted by the table on the object... points vertically up points vertically down is parallel to the table is zero [none of the above]

An object sits undisturbed on a horizontal table. The normal force exerted by the table on the object... equals its weight is greater than its weight is less than its weight is zero [none of the above]

An object has a mass of 3.25 kg. Its weight is... 3.25 3.25 N 3.02 N 7.15 N [none of the above]

A woman of mass 55.0 kg sits in a chair of mass 62.0 kg. The normal force of the chair on the woman is... 540 N 608 N 1150 N 117 N 7.00 N

A man stands in an elevator that is moving upward at constant speed. Compared to his actual weight on Earth's surface, his apparent weight is... greater less the same zero [none of the above]

By pushing on the floor, a person accelerates a 120 lb box across a room at 1.3 m/2. The floor opposes the motion of the box with a frictional force of 110 N. What horizontal force does the person apply to the box? 230 N 71 N 180 N 110 N 423 N

(C) Inclined Surfaces

Life isn't conveniently arranged to take place only on smooth, flat, horizontal surfaces. Sometimes you just have to go uphill or downhill. In these situations, it is best to choose a coordinate system with axes that are parallel and perpendicular to the surface. Generally, I'll take the x-axis to be parallel to the surface and the y-axis to be perpendicular to the surface as shown below.

In the above coordinate system the weight, which always acts vertically downward, can be resolved into x- and y-components. If is the angle that the surface makes with the horizontal then

With this recognition that the weight now has components in both directions, the rest of the analysis on inclined surfaces follow precisely as it does on horizontal surfaces.

Example 5.5 Finding a Better Way: A heavy suitcase that weighs 450 N, and has wheels, needs to be placed onto the back of a truck 0.65 m above the ground. Instead of lifting it straight up, you decided to get a thick piece of wood to use as a ramp and roll it up onto the truck. If the extended ramp makes an angle of 60^o with the horizontal, (a) determine the minimum force needed to roll the suitcase up the ramp, and (b) compare this to the minimum force needed to lift it up onto the truck.

Picture the Problem The uppermost picture shows the truck with the wooden ramp and the suitcase. The lowermost picture is a free-body diagram of the suitcase where F is the applied force to pull it up the ramp.

Strategy For part (a) we first recognize that the minimum force to roll it up must be parallel to the ramp (so all of the force is used) and just enough to balance the suitcase's tendency to roll down the ramp (due to gravity). Therefore, suitcase rolls up at constant speed. Likewise, for part (b) the minimum force just balances the weight of the suitcase.

Solution

Part (a)

1. Apply Newton's second law to the x-direction:
2. Solve this equation for Fmin:

Part (b)

3. Compare the result of (a) to the weight of the suitcase:
4. Calculate the percent difference:

Insights In the free-body diagram for this problem I drew in the components of the weight directly on the diagram instead of drawing the vertically downward weight vector. This is sometimes more convenient; either way you resolve the vectors into components, as part of the recommended strategy for force analysis, whether before or after drawing the free-body diagram.

Practice Quiz

An object is held at rest on a surface inclined at an angle with the horizontal. If the force holding it in place is parallel to the surface, the normal force exerted by the surface on the object... equals its weight is greater than its weight is less than its weight is zero [none of the above]

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