Applying Newton's second law

The work of a curious fellow
   

This law strictly enforced (for the large and slow)...

In the previous lesson we introduced the free body diagram as a tool for analyzing the motion of objects using Newton's second law. The one that says that the acceleration of a particle is equal to the force applied to it divided by its mass. In this lesson we will work a few examples using that tool.

Remember that so far we have restricted ourselves to forces that are constant within the limits of space and time where our observations take place. This restriction is not as severe as you might think because we can choose our observation place and time, much as we choose the orientation of our reference frame, so as to minimize the complications in our life. For example in the tossed ball situation, we did not start observing the ball until the instance it was released. The details of what happened to it to give it that initial velocity were ignored.

If you need to review the basic setup for that problem you may revisit that experiment. Click on the Ball Toss Experiment.

Now suppose that the ball is tossed in the same way from the top of a large building so that it is released from a height of 30 meters. What is the maximum height to which the ball rises? How far from the building does it hit the ground? At what time would an observer at the base of the building see the ball at a 45-degree elevation? Run the Ball Toss from Height Experiment to get the answers, which are all contained in the image at the right by the way.

ball tossed from height
Martian ball toss on a windy day

Now suppose the same questions were to be answered for a planet a bit like Mars where the force of gravity on a 1kg object was 5N rather than 9.8N as here on Earth. Run the Extra-terrestrial Ball Toss Experiment for this situation.

As one last exercise using this tossed ball, suppose that there was a strong wind blowing on this new planet such that there was a force on the ball of 3N directly back towards the building from which the ball was tossed. At what height above the ground will the ball strike the building? To find out, run the Ball Toss With Wind Experiment. The free-body model output for this situation is shown at the left.

You perhaps are beginning to recognize the power of this free body notion. Let's set up a different kind of problem. Consider an inclined plane without friction, set up so that it is at an angle of 30 degrees from the horizontal and passing through the origin of our reference frame. Don't worry for now how long the plane is. Now design a free body model for a 1kg box sitting on the inclined plane at 10 meters from the origin. To do this you need to know what the height of the plane is at x=10m so you can set in the correct initial position. The tangent of 30 degrees is .577. The altitude is 10*tan(30) or 5.77m.

Now what are the forces acting on the box. Well it is interacting with only 2 objects. The Earth and the inclined plane. The interaction with the Earth produces the force known as the weight of the box. The weight of an object is its mass times the acceleration due to gravity (m*g). For this object it is 1kg*9.8m/s2 or 9.8 Newtons directed at an angle of -90 degrees. Since the plane is frictionless, the only direction in which it can produce a force on the box is in a direction perpendicular to the plane so we know the direction of the plane's force on the box is at 120 degrees.

We know from Newton's third law that the magnitude of the force of the plane on the box is the same as the magnitude of the force of the box on the plane. It is the weight of the box that is the source of its force on the plane, but not the entire weight. If the plane were at an angle of 90 degrees, the force of the box on the plane would be zero since the weight vector would be parallel to the plane. If the plane were at an angle of zero degrees, the entire weight of the box would be pushing on the plane. In fact the force of the box on the plane is the weight of the box times the cosine of the plane's angle. In this case that is .866*9.8N or 8.49N, which also is the magnitude of the force of the plane on the box.

Now we have everything we need to set up the free body diagram for the box. Run the Sliding Box Experiment.

box sliding down inclined plane
box sliding up inclined plane

Anyone can calculate the motion of a box sliding down a ramp. How about a box sliding up a ramp. Run the Box Sliding Up Experiment to see this situation.

At this point you might want to go to some of the problems in your physics text and try to solve them using this free body diagram tool.

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Next we will add in another level of reality in the problems we are studying. This requires a new kind of force be considered, the force of friction. In fact there are three types of friction we will work with. There is the frictional force between two objects in contact but not moving relative to one another. This is called a static friction force. If the objects have a relative velocity there is a different frictional force, called the kinetic friction force. Both the static and kinetic frictional forces involve objects sliding over one another, sort of sharing a common surface. The third type of friction is that between an object and any fluid it finds itself in, air or water for example. We will stick to sliding friction for now.

The reason sliding friction happens is that surfaces are not perfectly smooth. Even highly polished surfaces have microscopic hills and valleys. When hills on one surface lie in valleys on the other, the objects cannot slide without enough force applied tangent to the surface to either break off the high spots of slip them over the hills into the next valley. As long as the objects are kept moving relative to one another, the high spots never really nestle down into the lows. When the objects have no relative motion, the surfaces sort of settle together and at the points of contact some atoms from one object actually share some electrons with atoms from the other, welding the objects together at numerous tiny sites. This is why it takes more force to start objects sliding over each other than it does to keep them sliding.

box sliding down with friction
box sliding up by friction

It should be evident that the harder the two objects are pressed together, the more interference there will be between the surfaces. In fact the force of sliding friction is proportional to the normal force holding the objects together. The ratio of the force required to slide an object to the force pressing the object against another is called the coefficient of friction. The Greek letter mu () is used to designate sliding friction coefficients. The symbol s stands for the static friction coefficient and k stands for the kinetic friction coefficient if there is any chance for confusion.

The magnitude of the frictional force then is |F| = * n , where n is the normal force vector. The direction of the frictional force is always directed opposite the applied force in the static friction case and opposite the relative velocity in the kinetic friction case. With this much information we can include a kinetic friction force in out free body analysis where appropriate. Remember that we can only deal with forces that are constant over the time of observation in this model so we have to limit ourselves to situations where the object we are analyzing does not stop or change directions. In that event the friction force is not constant.

Let's look again at the box on the inclined plane. Run the Sliding Box with Friction Experiment

Using the free body diagram requires you to do some calculations, for example to adjust the weight when the mass is changed or to find the normal force for an inclined plane. If you want a program that handles all those details for you, try our Physics-1 program which has more power and flexibility than I can build into a Java applet like this.

In the next lesson we will work with circular motion.

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