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The Inclined Plane Up until now, we have dealt with objects resting on a flat surface. What happens when the surface is not flat like the picture below?  Gravity acts on the object in two different ways. It pulls the object down the incline, and it also pulls the object against the surface. This is illustrated in the picture below.  The vector pointing down the incline is called the parallel force, and the vector pointing agaisnt the surface is called the perpindicular force. (That is why the vectors are labeled Fpar for parallel force, and Fper for perpindicular force). Because those two vectors are a result of gravity, they will add together to make the total weight vector of the object as shown. Keep in mind the vector addition rules.  Notice how the weight vector points straight down as opposed to an angle. This is a rule of gravity. Even though the two components of gravity act in different directions, their resultant has to be straight down. Calculating Fpar and Fper seems difficult, but it turns out to be pretty easy, because we only need one more piece of information, the angle of the triangle. It just so happens that the angle of the incline equals the angle between the weight vector and Fper.  In order to simplify the picture, let's just take a look at the vector triangle by itself.  The angle between mg and Fpar is a right angle, and will always be a right angle for inclined planes. So, we know two angles of this vector triangle, which means we can utilize trig functions to evaluate the components. This is similar to the Vx and Vy equations in projectile motion. The equations for Fpar and Fper become  To see how these equations work together consider the following situations: Fpar = mg(0) = 0 N On the contrary, since all of the object's weight is pressing against the surface, the perpindicual force would be equal to the total weight of the object. This is also verified by the equation because the cosine of zero degrees is one. Fper = mg(1) = mg Fper = mg(0) = m0 N On the other hand, now that the object is basically in free fall because it is not resting on a surface, all of the weight lies in the parallel force. This is again verified by the equation, because the sine of 90 degrees is one. Fpar = mg(1) = mg The Inclined Plane and Friction Recall from the last section that the frictional force is given by,  With the inclined plane, the normal force is no longer simply mg. Since the normal force is Fper, the frictional equations for an inclined plane becomes,  Let's do an example that ties all of these forces into one problem. Keep in mind the steps to setting up net force problems as they are outlined below. The problem is to determine whether or not a wooden box would remain at rest on a 40.0 degree incline if the static coefficient of friction was .90. Step One: This first step is to draw all of the pertinent forces. They are drawn in the picture below.  The normal force and Fper cancel each other out because they are still equal and opposite. They are important, however, because you need the normal force to find the frictional force. Therefore, the picture can be redrawn as,  Step Two: Now we have to determine which way is positive and which way is negative. The usual convention is to make down negative and up positive. So, the picture will become,  Step Three: Now we can set up the net force equation as follows:  Notice after plugging in the equations for friction and Fpar, every term has the mass of the object. This means that you can cancel the mass out of the equation, which also means the mass does not affect the acceleration of an object down an incline. (If you remember, this is exactly what Galileo reasoned when claiming all objects fall at the same rate regardless of mass!) We have just verified Galileo's claim mathematically. So the equation becomes,  Step Four: Here is the deal on how to determine whether or not the block will stay or slide. If the maximum static frictional force is smaller than Fpar, the block will slide. If Fpar is smaller, the block will stay. Remember, the static friction will match the applied force, in this case Fpar, up until the maximum value is reached. So, we need to calculate both and compare. and, Since Fs is bigger, the block will sit there. Technically speaking, the Fs will equal 6.8 N, because it has to match the applied force. When you plug those into the net force equation, they will cancel and you will end up with a zero for the acceleration, which means again that the box will remain at rest. We can now take the same block, and make the incline 50.0 degrees. This will now make the Fpar exceed the Fs, which means if we want the block to stay at rest, we will need to add an additional force. We will use a rope and call that force Tension. The picture, after removing the forces that cancel, will become, Step One and Two:  Keeping the directions the same, we can rewrite the net force equation to accomodate for the extra force. Step Three:  Notice that the mass has to stay in the equation. Since mass does not appear in every term, because of the T, it cannot cancel. This makes sense, because the mass of the object will definitely affect how much weight the rope has to hold. Therefore, making mass 10-kg and plugging in the numbers and making acceleration zero, we can solve the Tension. Solving for T yields 18 N. So, the rope would have to hold 18 N or a little more than 4 pounds. If the rope were to release the block, we can find the acceleration of the block. The equation would now become, This equation was borrowed from the last problem. The T was dropped, which left the 57 from the Fs, and 75, which was Fpar. Solving for a gives 1.8 m/s^2. There are several types of problems that pertain to the inclined plane. You can even set up a suspended mass problem with mass A on an incline, not a flat surface. However, for the purposes of this chapter, we will stop with the main ideas of the problem. Newton's Third Law There is one more law to Newton's laws of motion. I usually hold off on introducing the Third Law until momentum, because of its relationship with momentum conservation. However, for the purposes of this book, all three laws should be under one chapter. Newton's Third Law can be stated as follows: The word action here means force. So you can say for every force there is an equal and opposite force. This law simply means if you punch a wall with a certain force from your fist, the wall "punches" your fist with the same force, only in the opposite direction. This seems obvious and elementary on the surface. But, the consequences of this law when applied to different situations can get confusing. Even Newton had a hard time reconciling this law at times. So, in order to fully grasp this law, we need to consider several scenarios. If you consider a hammer hitting a nail as shown below, it is easy to say that the hammer is the object doing the "hitting."  However, it is not so easy to say that the nail hits the hammer back with the exact same force. The reason lies in the fact that the hammer is the one moving, not the nail.  An easy thing to do here is to make the mistake that these forces cancel. After all, they are equal and opposite right? That is true, but they do not act on the same object. The hammer's force acts on the nail, and the nail's force acts on the hammer. You can only cancel forces that act on the same object. The key concept here is to understand that the hammer's force only affects the nail's motion. It does not affect the hammer's motion. Likewise, the nail's force only affects the hammer's motion and not the nail's. To see exactly how they affect each other, let's take a look at the equation.  Because the forces are equal and opposite, we need to include a negative sign to one of them. It does not matter which one. Since force equals ma, we can substitute ma for the hammer and the nail. Furthermore, acceleration equals change in velocity over time, so we can make that substitution as well. Now the question becomes: which one will have the bigger acceleration? Let's analyze this with an actual example. Let's say the hammer's mass is 2.0-kg and the nail's mass is .01-kg. We know that the nail's initial velocity is zero, and let's assume the hammer's final velocity right after impact was slowed to .10 m/s. If the hammer's initial velocity is .50 m/s and the time of impact was .10 seconds, we can calculate the nail's final velocity after impact. If we make down positive and up negative, the equation will become,  80 m/s is a pretty big number, so these assumptions may not be very accurate, but the idea here is more important. It is obvious that the nail had the bigger acceleration. The reason is simple: it has the smaller mass. We can write the Third Law equation in a conceptual form that will help to make that point. (For the concept equations we will only consider the magnitude and not the direction, so the negatives will be left off).  Since both sides have to be equal and the mass of one is way bigger than the other, the acceleration of the smaller mass has to bigger than the bigger mass to equal out the equation. This is expressed in conceptual form above. This idea is another key concept of Newton's Third Law. Even though the forces are equal between two different masses, the acceleration of each mass will not be equal. We can now apply this concept equation to any situation to solidify these key ideas. Consider the picture below.  If a car has a head on collision with a massive truck, even the impact force is exactly the same, the car will have a much bigger acceleration. In fact, the truck would most likely just plow through the car and maintain its current direction with a smaller velocity. The car on the other hand would most likely reverse direction after the impact. Whenever an object is traveling in one direction and a force causes it to travel in the opposite direction, the acceleration is very large.  The "kick" from a gun is a consequence of the Third Law. Not only does the gun push the bullet, the bullet pushes the gun back. However, because the gun is way more massive than the bullet, the gun's acceleration back is very small, which is good for the shoulder.  Rockets fly into the air because of Newton's Third Law as well. When I was younger, I used to think that the rocket's boosters shot the fire out of the bottom and the fire pushed off of the ground to lift the rocket. A little thought would have made that impossible, because how did the rocket keep going once it was a considerable distance from the ground? A rocket pushes gas out of the thrusters, and in turn, the gas pushes the rocket in the opposite direction. This is why it takes a massive amount of fuel. The large mass of fuel times the large acceleration yields a force large enough to lift the rocket. What happens when an object pulls on another object? The other object pulls back. Imagine two people sitting on chairs with wheels and they are holding onto each end of a rope as shown below.  Even if the man on the left pulls on the rope and the man on the right just holds on, the man on the right still "pulls" the other man. How do you prove that? When the man on the left pulls, they will both move towards each other. What would make the man on the left move if the man on the right didn't actually move his arms to pull? The only thing that causes objects to move is a force. That force is a result of the Third Law. Now if they have equal masses, they will have the same acceleration, which means they would meet in the middle. (This is assuming no friction which is why they are on chairs with wheels).  What happens if the man on the left has a larger mass? Two things, the acceleration of the man on the left will be smaller, and they will meet at a point closer to him.  Newton's Third Law pops up everywhere. You couldn't walk without it. When you walk, you push against the floor, and in turn the floor pushes against you, which makes you move forward. Try walking on a surface with little friction like ice. Without the friction, there would be no equal and opposite force available to push you forward, so you would literally be walking in place if you were on a perfectly frictionless surface. Trying to walk on ice has a similar result. Even though this chapter is finished, this is not the last time Newton's three Laws will be mentioned. They show up all over physics, and help to explain many physical phenomenon. That is a testament to the brilliance of such simple laws. |