Physics 115, Investigating Physical Science Weight and Mass

Transcription

1 Physics 115, Investigating Physical Science Weight and Mass The Mass of an object is defined to be the number of grams (or kilograms) it takes to balance the object on an equal arm balance. 1 kg = 1000 g. In this activity, we will be using electronic balances rather than equal arm balances to measure the mass, and a similar set of scales to measure weight. We will be exploring the relationship between weight and mass in this activity. The Weight of an object is defined to be the reading of a spring scale when the object is hung on or supported by the scale. The units of weight are newtons in the metric system or pounds in the English system. You should start by confirming that a 50 g mass measured on an electronic balance and the same mass on a spring scale are about the same. Note that it is much easier to get an accurate reading on the electronic balance, and the latter is easily zeroed. You are to use a number of objects of different masses ranging from a few grams to nearly 200 g. If you use more than one object for different measurings, their total may be as much as 400 g. Using two different electronics balances (one is set on g and the other on N), determine the weight of each mass in ounces (oz) and newtons (N). You may have to make a number of conversions within each system (and between systems in one case). 1 kg = 1000 g, oz = lb, lb = 4.45 N and oz = g. Each team should take at least 2 or 3 different objects each having masses less than 200 g (the maximum scale capability). After finishing, the class will share values obtained for their objects, and we will fill in the first and third columns below. The first and second columns will require a conversion process from g to kg and g to oz. You will be measuring g and N directly on the electronic balance, but will have to convert to kg, oz and lb. Mass (kg) Weight (oz) Weight (lb) Weight (N) Weight (N) Mass (kg) Make a full page graph with the vertical axis as the weight in newtons (N) and the horizontal axis the mass in kilograms (kg). Plot all ten points. Excel is the preferred method and won t require graph paper. Part of the requirement will be to find the slope of the line. Based on your graph, write a mathematical expression that converts the mass in kilograms to the weight in newtons. Try your relationship on the values in the data table to verify that it works. The data of these 4 columns, a shown conversion for one of your sets of data, and the graph are to be turned in for this activity. 35

2 Clay Boats Floating clay boats is an activity that can be fun for all ages. However, we would like it to be educational as well. You will find at the front of the classroom some lumps of clay, some ceramic tiles, and a tray (or perhaps a bucket) in which you will be able to float the clay boats that you make. Note that the modeling clay to be used is oil based. Obviously you don t want to use a clay that will dissolve or become soggy in water. 1. Start by determining the volume of your lump of clay (by water displacement). You may not want to use all of the clay, as it must be small enough and in a diameter and shape that will fit in the graduated cylinder. Volume = 2. Now dry the clay and determine its mass. Mass = 3. Calculate the density. Density = Mass/Volume = Of course, it s no surprise that the density is greater than 1 g/cm 3, since it sank in the water. We just want to have an idea of the actual density. For the rest of the experiment you may want to use a larger piece of clay. If so, measure the total mass you are going to use. It must be less than 200 g though, since that s all the electronic balances can measure. 4. Mass of clay used = 5. Measure the mass of a 10 or more pieces of ceramic tile on an electronic balance. Calculate the average mass per tile. They won t all have the same mass, but for this experiment we ll assume that each tile has the same mass as the average mass of a larger number of them. We don t want to remass for every different number of tiles we might use in this experiment. 6. Mass of a single tile = 7. Now your challenge is to make a clay boat that will float itself and as many pieces of ceramic tile you can carefully place within the boat without it sinking. You may repeat this a number of times, but record the number of tiles successfully added for each attempt. Largest number of tiles = On the next page, briefly describe the design of the boat that worked best for you. That is, not the possible features that contributed to getting the largest number of tiles into the boat. 36

3 8. You were working with pieces of ceramic and clay that each had densities greater than that of water. How then could you float the clay boats? This involves the concept of Average Density, or if you wish Effective Density. If the total mass of your system (boat plus tiles) is less than the mass of the water that could be displaced by the boat, then the boat floats. That is, the average density determined by dividing the total mass of the boat and its contents is less than the mass of water for the volume of water equivalent to the volume of the boat then the boat will float. 37

4 Weight and Mass Revisited We defined weight as a number measured by a spring scale. Typical units of weight are pounds and ounces, but in the metric system we use newtons (abbreviated as N). Earlier we defined mass as number measured by an equal arm balance in which a comparison is made between a known (standard) mass or set of masses and the mass to be determined. The mass is the number of grams needed to balance the object. So what is the difference between the weight and mass (other than that they are measured using different units)? They might seem like the same thing. For example, a 10 g mass has a weight of N. Why use these two different descriptors? The reason turns out to be that when using something like a spring scale as the measuring instrument, you can get different results in different situations. For example, if you were to take the apparatus and object to the moon to carry out your measurements, you could get a drastically different result. That is, on the moon the gravitational force between the object and the moon is less than the force between the object and the earth when on the earth. You ve all heard that you would weigh less on the moon. Similarly when you accelerate a spring scale and the object attached to it, the reading on the scale will change. This is just like when the elevator is starting up to go upward you momentarily feel an increase in your weight, and when the elevator starts moving down you momentarily feel lighter. On the other hand, the equal arm balance (that is balanced) has a decrease or increase in gravitational force occur on each of the arms, so the net effect is that it remains balanced. Mass does not change under relocation to another point on earth or onto a different planet, but weight does change. Our interpretation is that mass is an indication of how much matter is present, but weight is an indication of the strength of the gravitational force between an object and planet. For our purposes, the planet in question is the earth. No matter where you are, the more mass an object has the more weight it has (if it has any at all). However, when two objects are practically at the same location, if they have the same mass, they also will have the same weight. On earth, one kilogram of mass weight 2.2 pounds or 9.8 N. On the other hand, one kilogram of mass on the moon would be still one kilogram of mass, but would weight less than 6 ounces or about 1.6 N. You ve probably all heard the trick question, What is heavier, a ton of bricks or a ton of feathers? The ton is a unit of weight equivalent to 2000 pounds. If the two different tons weight the same at the same location on earth, then they have the same mass and hence consist of the same amount of matter. Most of the mass in matter come from the nucleons (protons and neutrons) within the atoms. Within any number of significant figures we could practically use, the total number of protons and neutrons in the ton of bricks would be the same as the total number of protons and neutrons in the ton of feathers. Of course, the number of protons would not be the same in the ton of feathers as the ton of bricks, but that s not within the scope of this class. We ve used the term gravitational force in the above discussion, and most of us have a fairly intuitive sense of what force is. We experience forces when we re pushed, or push something 38

5 else. We already know that forces can cause motion, although this will be discussed in more detail later. Newton s first law, which is addressed by EALR (for grades 9 and 10) applies in situations for which forces on an object are balanced. Another way of stating that the forces are balances is to say that the net force on the object is zero. When the net force is zero, the motion of the object is constant. That means that if it is stationary, it remains stationary and if moving it continues to move with a constant velocity. We won t deal with situtations involving constant velocity at this time, except to say that it means both the speed and direction of travel remain constant. An object for which the net force is zero is said to be in equilibrium. Up to now we ve been only discussed forces in static situations. The objects as we study them are at rest. This was true for the objects on our pegboard balances, and our objects floating in water, or at the bottom of our beakers of water. This is true for you standing on the floor or sitting in your chair, or your books or backpack resting on the floor. To understand what is meant by balanced forces or net force, we first have to recall that forces are vector quantities. In physics a vector quantity is something that has direction as well as magnitude (size). Measured quantities like length, mass and speed are consider scalar quantities, since there is no direction associated with their measurement. We could reorient a physical object and its length wouldn t change. The same is true if were to measure the mass of objects oriented differently. On the other hand weight has an implied direction associated with it, since it is a force. We don t usually talk about the direction associated with weight, but it is always directed downward. If no other forces act on an object except the force of gravity (its weight) it will move in the downward direction. Objects never fall upwards or sideways, although we do use the term free fall for objects moving upward (or even at angle from the vertical) with only gravity acting on them. A convient way of representing vectors on a piece of paper (or chalkboard) is with a drawn arrow. The length of the area indicates the magnitude of the force and its orientation represents its direction. Thus, whened view from the side, as opposed to above or below, the force of gravity is represented by a downward arrow. If you look at the weight of two different objects side by side, we would have two downward pointing arrows. If the the weight of one object is twice as much as the other then the arrow for that object must be twice as long as for the other object, etc.. See the figure below. The dots represent the objects in each case, and this kind of m 1 m 2 m 3 figure is know as a free-body diagram (FBD) or more simply stated a force diagram. 50 N 75 N We already know that the gravitational force on a 1.00 kg mass 9.8 N, so if the gravitational force is 50 N, the mass 100 N m 1 must be (50/9.8) kg. We could have shown this by a simple ratio. (9.8 N)/(1 kg) = (50 N)/m 1. Solving for m 1, we obtain m 1 = (50 N)(1 kg)/9.8 N = 5.1 kg. The diagram is free of all representation, except the the body (the dot) and the forces that are directly acting on it. 39

6 By directly, we mean the forces from direct physical contact, or forces such as electric, magnetic or gravitational that come from other objects not actually touching the body in question. If you want further clarification on this, ask about it in class. We will restrict ourselves to relatively simple force diagrams. For example, let s consider a simple situation in which you re holding a book in your hand. If your hand weren t there, the book would fall. Since you observe that it s stationary, you know that the net force on the book is zero. The force diagram would appear as shown as below. W B is the weight of the book (i.e., gravitational force on the book) and F H is the force of the hand supporting the book. Note that if F H were greater in magnitude than W B, the book would move (accelerate) upward, and if F H we less in magnitude than W B, the book would accelerate downward. Now let s consider the situation for a floating object. The object floating is stationary, but does that mean that there is no force acting on the object? The answer is no. There are forces, but they combine to produce a zero net force. What are the forces? Well, first of all there is the weight of the object acting downward. We may ask Then what is holding it the object up? Well, the answer to that is the water. If you doubt that water exerts a force, go into a swimming pool. You can feel the force of the water against your body. In fact if you went deeply enough (say hundreds of feet without scuba gear), you could be crushed by the force of the water. The supporting force on the object that is either completely or partially submerged is called a buoyant force. It buoys up the object. Buoyant Force Experiment Now we will investigate the buoyant force in a range of conditions for the object being completely out of the water to the point where it is floating. If it were completely out of the water, the diagram would look like that shown above, with the exception that we would change the notation used in labeling the forces. We re going to do our experiment by supporting a wooden dowel from a spring scale and slowly lowering it into a graduated cylinder that is partially filled with water. We don t want it so full that the water level goes above the top marking of the cylinder when the dowel is floating, so you might want to initially fill the graduated cylinder only half way and see if the dowel floats without touching the bottom of the graduated cylinder or make the water level go up too far. Adjust the initial level accordingly, only if needed. You are now to fill out the table titled Buoyant Force. Filling out the column headed Force of Spring (F S ) should be obvious. Likewise for Water level. Try to reason out what to do for the other columns by yourself, or by discussion with your partner and other people at your table, and only if all else fails ask the instructor or T.A. Hints will only be given if you ve exhausted all reasonable attempts. Note that a force diagram, with the arrows drawn in the correct direction should give you the answer for the Force of Water column. After completing the table, plot on graph paper Force of Water (F B ) vs. Volume of water displaced. Place error bars about your points, and try to draw a best fit straight line. As usual, you may use Excel to accomplish this. 40 F H m W B

7 Buoyant Force Length of Wood Under Water Force of Spring (F S ) Force of Water (Buoyant Force) Water Level 0 cm 2 cm 4 cm 6 cm 8 cm 10 cm Weight in newtons = (mass in kilograms)(9.8 N/kg) 1 kg weighs 2.2 pounds (or 9.8 newtons) on Earth. 1 lb = 4.45 N Volume of Water Displaced Mass of Water Displaced Weight of Water Displaced 41

8 Archimede s Principle In the buoyant force experiment we just completed, we should have drawn a force diagram (or FBD). Since the object with the combination of forces acting on it was in equilibrium the net force on the object is zero. Regard the three situations illustrated below: Object Completely out of Water Object Partially Immersed, but not Floating Unaided Object Floating without help of Scale F S F S F B F B W W W Larger forces are represented by longer arrows. Note that the weight (W) of the object is the same in each case. The force of gravity acting on the object doesn t change as it s lowered into the water. Also note that for all three situations, the downward force (W) and total upward force (F S + F B ) have the same magnitudes. For the situation on the left, the buoyant force (F B ) is zero and not shown, since the object isn t in the water. For the situation on the right, the spring force (F S ) is not shown because it is zero (i.e., the string has gone slack). In the middle situation, both the spring and water are helping support the object. Let s assume that we partially filled the graduated cylinder and read its initial depth as 50, meaning the volume of water was 50 ml. With the wooden dowel suspended from the spring scale the reading was 0.38 N. This is just the weight (W) of the dowel. As we lowered the dowel into the water, we found the spring force decreased. At a depth of 2 cm, we read 0.33 N, and the water level rose to 55 ml. Of course, this didn t mean that there was 55 mm of water in the graduated cylinder, but that 5 mm of water had been displaced by the dowel. The data obtained is illustrated in the Table below. Length under water 0 cm 2 cm 4 cm 6 cm 8 cm 10 cm Force of Spring (F S ) 0.38 N 0.33 N 0.29 N 0.24 N 0.18 N 0.14 N Force of Water ( F S ) 0 N /-.005 N /-.005 N /-.005 N /-.005 N /-.005 N Water Level (ml) 50 ml 55 ml 60 ml 65 ml 70 ml 75 ml Volume of water displaced (Level - 50) 0 5 +/-.5 ml 10 +/-.5 ml 15 +/-.5 ml 20 +/-.5 ml 26 +/-.5 ml We also estimated the uncertainty in these data and decided that we could read the spring scale to +/ N or a half division on the scale, and the volume level of the water in the graduated cyclinder to +/- 0.5 ml, or half a division. The obvious conclusion frome these data is that as the volume of the displaced water increases, the upward force of the water also increases. In making a graph of the relationship it becomes obvious that within the uncertainties in our 42

9 measurements the relationship is linear. In the graph below, the uncertainties are shown as error bars, and the straight line resulting is a best fit straight line that goes through the origin of the graph (0, 0) and the data points. We conclude that the buoyant force (F B ) is proportional to the volume of water displaced. In this statement, proportional to means that if the volume of displaced water is doubled (or tripled, etc.) the buoyant force is also doubled (tripled, etc.). The graph consistent with the table on the previous page is shown below. 0.3 F B (N) Volume of Water Displaced (ml) We see that even with error bars, the line did not pass through all the points. Either error was underestimated, one of the data points was measured less carefully, or the equipment didn t work consistently. If we go back to the data on page 26, it should be apparent that the buoyant force column and weight of water displaced column were very close in value. In fact, it can be shown theoretically (not a topic for this course) that the weight of the water displaced will alway equal the buoyant force. This is true not only for water and other liquids, but all fluids (including gases). We so state: Achimede s Principle: The buoyant force is equal to the weight of the displaced fluid. Now we can state our sinking/floating rule in force terms. If an object can displace a weight of liquid equal to its own weight, it will float. If it cannot, it will sink. Consider an object that is completely submerged in a fluid. The buoyant force is pushing upward and the weight force is pulling downward. If the buoyant force (FB) is larger, when released the object will move upward. If the fluid is a liquid, it will rise to the surface and stop moving upward when the volume of the object below the surface is such that the weight of the displaced liquid equals the weight of the object. In the case of a helium balloon floating it air, it will continue upward until it encounters air that is less dense and light, or until it pops. In any case, if the weight is greater than the buoyant force, the object will sink. 43

10 Force, Mass and Acceleration Today we ll investigate the acceleration of a cart rolling down a nearly frictionless sloping track. Such a sloping ramp is often referred to as an inclined plane. This will be a group activity in the same sense as the experiments with the balls rolling down the incline, but we ll assume we already know that if the acceleration is constant (as would be shown by Galileo s odd number rule and square rule), then the displacement of an object when starting from rest is given by x = at 2 /2. We will time our carts rolling down our incline and use the measured time and corresponding distance to calculate the acceleration in various situations. Our goal is to determine a relationship between the mass of the object, the net force acting on the object, and its acceleration. Our equation is simply a = 2 x/t 2. We can measure the mass of our different carts by placing them on an electronic balance, and measure the force needed to keep them stationary on our inclined planes using the spring scales. Since we ve kept the object stationary using the force provided by the scale, the assumption is that when that scale is removed the net force remaing has the same magnitude as the force read by the scale. Let s start by measuring the force on, mass and acceleration of carts having different masses on an incline having the same slope. We ll repeat this for several different angles of the incline. We won t carefully measure the angles, but will use the amount by which one end of the incline is higher than the other. We ll note the length of the incline (which will be the same for all of the activity), and place the altitude (H) by with one end is elevated in our first column. The second column will be for the mass of the cart, the third for the acceleration calculated, and the last for the force measured by the spring scale. We ll repeat for 3 different masses at each of 3 angles of incline. Be certain to include the uncertainty for the time and spring scale force. H(m) x(m) m(kg) t(s) a(m/s 2 ) F(N) F/m(N/kg) Hopefully, you will have found an apparent relationship between the column for acceleration and the column for force/mass. As one last exercise, multiply the ratio of length of the track to the difference in altitude bewteen the two ends times your acceleration and see what you get. Can you guess why? 44

11 We should have observed a strong similarity between the column for acceleration (a) and the column for F/m (net force/mass). In fact, the only difference between these two columns are the units. We can make reasonable arguments from our own experience why acceleration should be proportional to the applied force, but inversely proportional to the mass. Try to think of some examples and write them below. We generally write Newton s second law as F = ma. F and a are vectors, and acceleration is always in the same direction as the net force. However, it s important to remember that acceleration depends on force and mass. In other words, force and mass are the controlled variables, and it is the acceleration that is determined by them. Of course, we can plan to have a particular acceleration by careful picking of our mass and net force. You may have noticed the coincidence of numbers in the 9.8 N being the weight of a 1.0 kg mass, and the acceleration of an object in free fall being 9.8 m/s 2. Well, it s more than a coincidence. The newton has been defined as the net force required to produce a 1.0 m/s 2 acceleration of a 1.0 kg mass. Thus, a one kilogram mass weighs 9.8 N. Its weight is the accelerating force and the resulting acceleration is thereby 9.8 m/s 2. Another way of stating our definition for the newton is to say 1 M = 1 kg m/s 2. Thus, the units in our acceleration column are consistent with those in the F/m column. 45

12 More on Force Diagrams Let s take another look at force diagrams in a non-static situation. Consider an object moving on a flat, level surface, acted on by two horizontal forces, which we label F a and F b. In the diagram below, the velocity direction is shown by the arrow. In this case, F a is in the direction of the velocity, and F b is opposite the direction of the velocity. There are two vertical forces acting on the object. W is the weight force exerted by the earth, and N (meaning normal force) is is the upward push of the table on the object. Note that the velocity (v) is not part of the force diagram, since it isn t a force. It is only used to indicate the direction of motion. v F b M F a Rules: W N A. Consider the horizontal forces. Our law of motion (Newton s second law) states: F a - F b = ma 1. If F a and F b are equal, the forces are balanced, a (acceleration) = 0 and the velocity is constant. 2. If F a is larger than F b, there is an unbalanced force in the direction of the velocity. Therefore, there is an acceleration in the direction of the velocity, and the velocity increases. 3. If F b is larger than F a, there is an unbalanced force opposite to the velocity. Therefore there is an acceleration opposite the velocity and the velocity decreases. If the condition continues long enough, the velocity will go to zero, and then go negative, picking up speed in the opposite direction. B. Consider the vertical forces. We know there is no vertical motion, so the vertical velocity is zero. That is, the object is at rest vertically. The two forces, W and N, must balance, so there is no vertical acceleration. 46

13 Acceleration on an Inclined Plane Note that the discussion below is beyond the scope of this class, but it is hoped that some of you will follow it. When an object is placed on a frictionless incline the force diagram can be draw as shown below. Since there is no friction, the incline can only exert a force perpendicular the the incline. That s the normal force N shown in the diagram. Note that the word normal means perpendicular in this context. The only other actual force acting on the object in this case is the weight (W). You should note that there is a dashed line drawn in the opposite direction of the normal force. This the part of the weight that cancels the normal force. The evidence for this cancellation is the fact that the object doesn t sink into the incline or leave it. The other dashed line is parallel to the incline and is the part of the weight that results in the acceleration (a) of the object down (along) the incline. It is this F Net that is equivalent to Ma. In this figure, L is the distance of the lowest edge of the object (line A) to the bottom of the incline (line B) and H is the vertical distance from the level line below the lowest edge of the object up to the lowest edge of the object. By similar triangles, H/L = F Net /W. Since the weight is W = Mg, and F Net = Ma, F Net /W = Ma/Mg = a/g. The the acceleration is given by (H/L)g. Obviously if H = L, the acceleration is g (9.8 m/s 2 ). M N A W L F Net H B 47