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Suspended Mass The next net force problem will examine the picture below.  This setup has one mass, Ma, resting on a table connected to a rope that extends over a pulley and suspends another mass, Mb. Again, the first step in formulating the net force equation is to draw in all of the forces for each mass as well as establishing direction. Let's start with Mass a. Step One:  The first thing you might notice on the picture above is the Fn. That stands for the Normal Force. When an object sets on a flat surface such as a table, the object is pushing down on the surface because of its weight. As a result, the surface pushes back up with a force called the normal force, and it is equal in magnitude to the weight of the object. (This is true for a flat surface. We will deal with an inclined surface later). So, since the normal force and weight are equal but in opposite directions, they cancel. So, the picture can be redrawn as  The only force left is the tension of the rope. Making right positive and left negative, the net force on mass a is a positive T. Now let's look at mass b. Step Two:  Mass b has only two forces acting on it: weight and tension from the string. This is exactly the same tension that acts on mass a because it is the same rope. You cannot have two different forces from the same rope. This will be helpfull later. From the picture, it looks like the two forces also cancel. It turns out that they will not cancel if we neglect the friction between mass a and the table. (We will deal with friction next). So, if we take a look at the whole picture with all of the forces that do not cancel we will have Step Three:  Notice that down is positive and up is negative. This is a direct result of making right positive and left negative. To understand this, point your finger along the string starting from mass a. When you follow the string to the right, which is positive, you end up pointing down when you go over the pulley. Since you started out positive, you have to end up positive. Therefore, down has to be positive. Now we have to set up the net force equations for each individual mass. Step Four:  The goal now is to come up with one expression for the acceleration of this two mass system. The accelerations in both equations are the same because both masses are connected by the same string. So, since the tension is the same in both equations we can set both equations equal to tension first. Step Five:  Fortunately, the first equation is already solved for T. Rearranging the second equation so that T is positive gives you the result above. Since, the T's are equal, we can set the right sides equal to each other. Step Six:  The following steps show how to solve for the acceleration. Step Seven:  There are a couple of things to note about this expression. That last relationship happens because,  That is the process involved in developing an equation for acceleration using Newton's Second Law Net Force Equation. The last two examples entailed different situations because the first example had one mass, and the second example had two masses. Therefore, they do require a different number of steps. I have tried to outline each step of the way to highlight the major parts to each problem. Friction Friction is the result of two surfaces rubbing together. this is not a technical definition because the two surfaces do not actually have to "rub" together in order for friction to be in effect. This will make more sense later. For now, this definition will suffice. Let's take a look at what really causes friction. Consider the picture below.  Even though a service may seem smooth to the touch, a magnified view actually shows quite the opposite. All surfaces have small microscopic irregularities regardless of how "smooth" they may seem. Even a highly polished surface has some slight irregularities. In addition, there are tiny electrical forces at work, which we will cover in a later chapter. Now that we know what causes friction, we can examine the two different types. Static and Kinetic Friction Static friction is the friction between two surfaces before they start moving against each other. This is the resistance that you feel when trying to slide a box across the floor. Before, the box starts moving, there is a resistance due to the static friction between the floor and box. Once the box starts moving, Kinetic friction takes over. kinetic friction is the resistance between two rubbing surfaces. Through experience, most people know that it is harder to get the box moving compared to maintaining its motion. This is because static friction is always larger than kinetic friction. Frictional forces depend on the types of surfaces in question. The equations for static and kinetic friction are show below.  The funny looking u in those two equations is the greek letter mu. The mu with the s stands for the static coefficient of friction and the mu with the k stands for kinetic coefficient of friction. This is how to account for different types of surfaces. The coefficients of friction vary accordingly depending on the nature of the two surfaces together. These coefficients can equal one or zero and anything in between, but they cannot be greater than one or less than zero. This relationship is mathematically illustrated below.  Technically speaking, it is not very practical for a surface to have a coefficient of zero because that means it is frictionless. That is not possible to attain. On the flip side, it is possible to have a coefficient of one. In fact, that is the value for the static coefficient between rubber and dry concrete. It is fortunate for driving that these two surfaces yield the maximum amount of friction possible. However, rubber on wet concrete does reduce the frictional coefficient to values of .7 or .6 depending on conditions. The Fn refers to the normal force. Recall from the last problem, normal force is the upward force that a surface exerts on an object due to the object's weight pushing down on the surface. When the surface is flat, the normal force is equal to the weight of the object given by mg. Let's tie this all together with an example. In doing so, it should be noted that the frictional force will always point in the opposite direction of the applied force. Suppose a wooden crate with a mass of 100.0-kg is resting on a stone floor. Find the force needed to make the crate move, and the force needed to keep it moving once the static friction is overcome. The static coefficient of friction for wood on stone is .5, and kinetic coefficient is .4. The first thing to do is find the static frictional force. Using the equation, That number is the maximum force that the friction can provide before the crate starts moving. In other words, the static frictional force can be anything between zero and its maximum value. It all depends on what the applied force is. Consider the picture below.  When the man pushes with 100 N, the static friction "pushes" back with 100 N. When the man increases his applied force to 400 N, again the static friction increases to 400 N. Even when the man pushes with the maximum value of 490 N, the static friction keeps up. The point is, the static frictional force will match the applied force up until the maximum value. That is the last straw so to speak. Once the applied force becomes greater than the maximum static frictional force, the object will begin to move and kinetic friction takes over. Finding the kinetic frictional force we have, This is the only force that kinetic friction can provide, so it is different from static friction in that respect. So, what happens when the applied force changes? Consider the picture.  If you push the box with a force equal to the kinetic frictional force, the net force would be zero because they cancel. Therefore, the box will move at a constant velocity across the floor. If you push the box with a force greater than the kinetic frictional force, the net force will now be in the direction of the box's motion. As a result, the box will accelerate. In this case, that would mean it would speed up.  If you push the box with a force smaller than the kinetic frictional force, the net force will now be in the opposite direction of the box's motion. Therefore, the box will accelerate , only this time it will slow down.  The following statements sum up what was just covered. Rolling Versus Skidding Which would result in a faster stopping time, rolling to a stop or skidding to a stop for a car? Assume that both situations utilize the same stopping force on the tires. In other words, the car is not simply coasting to a stop when it rolls. The answer: rolling! This seems contrary to common sense, but the explanation should prove other wise. The question here is which situation, rolling or skidding, utilizes static friction as opposed to kinetic friction because static yields a larger force. It is obvious that skidding is kinetic friction. It may not seem so obvious that rolling is actually a result of static friction. When an object skids across the floor, the part of the object that touches the floor is in constant contact and rubs on the floor as it moves.  However, when an object rolls, the part of the object that touches the floor only touches it for a moment as it pushes off. So, the two surfaces never "rub" together. It is really a matter of pushing off one another.  Since rolling utilizes static friction, it is the most effective way of stopping. This is the reason for anti-lock breaks on cars. The anti-lock system keeps the tires rolling. This is also the reason you are taught to pump the breaks on icy or slick conditions rather than slam them. Once you start slidding on ice, your control of the car is gone because the kinetic friction between rubber and ice is very low. So, it is wise to try to maintain static friction whenever possible to maximize the control of the car. Friction complicates every problem because not only are you adding another force, sometimes that force varies which requires calculus to solve. There will be other examples in later chapters that deal with friction in the problem. You could redo the example labeled the suspended mass and include friction on the mass resting on the table. That problem is left for a quiz problem, so I will not divulge the methods here. |