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Download report-9-archimedes-principles.pdf and more Exams Acting in PDF only on Docsity! Report 9: Archimedes Principles Title: Archimedes Principles Laboratorial Report 9 Created to fulfill the assignment for Mechanic and heat EN222 subject By: Debby Syefira Damavara I Wayan Surya Aryana Lecturers: Lina J Diguna Fatih Sampoerna University Performed: Jakarta, June 26th, 2015 Report 9: Archimedes Principles Abstract As the Archimedes concept-strengthening activity, the ninth experiment is done by undergoing a sort of force measurement done in an object. By consider the Archimedes Principle is work perfectly regardless to the spring error and human error, the experiment is, besides force, aiming to find the specific gravity affecting object as a result of the buoyance force and weight of the object. These two main measurement data are gotten primarily through the data gained by weighting object in air and water. Besides measuring, the experiment is also providing a question sheet to be answered; by providing a reasonable answer and suit conclusion/recommendation, the experiment is closed. Keywords: Archimedes Principle, Weight, Buoyance Force, Specific Gravitation. Report 9: Archimedes Principles e. Determine the density of alloy from your measurements, see Equation (4). Step 2: Now find the density, specific gravity and buoyant force acting on an object that normally floats. a. Attach a cord to the top and bottom of the wooden object. b. Attach the bottom cord to the brass cylinder, and the top cord to the spring balance. c. Submerge both objects into water, as shown in Figure 1and Figure 4. d. Record the tension in the string attached to the spring balance. e. Calculate the buoyant force acting on the wooden object. Figure 5 shows you will need to apply Newton’s second law to two objects simultaneously to calculate this buoyant force. You will also need to think about the following: How does the buoyant force acting on the submerged brass cylinder compare to the buoyant force that acted on it during the first part of the experiment? Hint: What does Archimedes’ principle tell us about the buoyant force? f. Calculate the specific gravity of your wood Report 9: Archimedes Principles CHAPTER III RESULT AND DISCUSSION 3.1 The Principles of Archimedes. “When an object is placed in a fluid, the fluid exerts an upward force we call the buoyant force. The buoyant force comes from the pressure exerted on the object by the fluid. Because the pressure increases as the depth increases, the pressure on the bottom of an object is always larger than the force on the top - hence the net upward force. The buoyant force is present whether the object floats or sinks. Let's consider a floating object, but the analysis is basically the same for a submerged object. We'll also consider a rectangular block, although a similar (more complicated) analysis leads to the same result for funny-shaped objects.” (The Bouyant Force, 2015) Figure 2. Theory Figure 3. Theory The density of any object, ususally abbreviated as 𝜌, is defined as the ratio of an object’s mass o its volume. Specific gravity is the ratio of the density of an object to the density of some other substance that has been chosen to act as a standard. Waer, at room temperature, is generally chosena as this standrad. Report 9: Archimedes Principles If a solid body is submerged in a fluid and kept stationary as in figure 1, it will be acted upon by three forces: 1. The weight of the body W, which can be expressed in terms of the object’s density as 𝑊 = 𝑚𝑏 ∗ 𝑔 = 𝜌𝑏 ∗ 𝑉𝑏 ∗ 𝑔 Where mb is the mass of the body, Vb is the volume of the body, 𝜌b is the density of the body and g is the acceleration due to gravity. 2. The buoyant force B; wich can be expressed as 𝐵 = 𝑊 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝐵 = 𝜌𝑓 ∗ 𝑉𝑓 ∗ 𝑔 𝐵 = 𝜌𝑓 ∗ 𝑉𝑏 ∗ 𝑔 Where W fluid displaced is the weight of the fluid displaced, 𝜌 f is the density of the fluid, V f is the volume of the fluid displaced, Vb is the volume of the body. Notice that the last equality comes from the fact that a submerged solid object will displace a volume of fluid equal to its own volume. 3. The tension T in the string. Since the body is equilibrium, T + B – W = 0. We can then compute the buoyant force from the equation, 𝐵 = 𝑊 − 𝑇 That is, form the difference of the object’s weight in air to its weight in fluid. It can also be shown that the ratio W/B is equal to the specific gravity of the immersed substance relative to the fluid. As mentioned above, for a completely submerged object the volume of the body immersed (call them both V). Then, 𝑊 𝐵 = 𝑉 ∗ 𝜌𝑏 ∗ 𝑔 𝑉 ∗ 𝜌𝑓 ∗ 𝑔 = 𝜌𝑏 𝜌𝑓 = 𝑆. 𝐺 (𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦) Report 9: Archimedes Principles e. The specific gravity of the wood. 𝑊 𝐵 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑊 𝐵 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 3 𝑁 2 𝑁 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 1.500 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 Explanation: Based on the data above, the student could examine the theory of Archimedes. Theoretically, if a substance or material that put on the liquid, the substance that has higher density rather than the liquid will sink. Contrastingly, if the substance or material that put on the liquid has smaller density rather than the liquid will float. On the experiment, the alloy has weight of 2 N in the air and 1.7 N on the water. As a result, it has buoyant force of 0.3 N and gives a result of specific gravity of 6.667. On the water, the alloy did not float. The wood has weight of 3 N in the air and 1 N on the water. As a result, it has buoyant force of 2 N and gives a result of specific gravity of 1.500. On the water, the wood did float. As natural, the water has density of 1000 kg/m3. If those substance is compared with the water, the alloy has much bigger density than the water. Theoretically it will sink where in the fact it was sinking. However, even though it was sinking, the weight of alloy on the water is less than the weight of the alloy on the air. It would give conclusion that the water cause the tension become smaller. The wood has smaller density than the alloy and has almost the same density as the water. Theoretically if the density of the substance is smaller than the water that will give result of the floating of the substance on the water. However, on the field, the wood which has almost same density with the water is half floating. Also, as well as what happened to the alloy, the tension caused by the wood on the water is smaller than on the air. It would support the conclusion that the water cause the tension become smaller. Report 9: Archimedes Principles Moreover, as well as the calculation, the wood get bigger buoyant force than the alloy get. The alloy only get 0.3 N whereas the wood get 2 N of buoyant force. So then, the wood get so much force to float. As the result the wood will float further than the alloy did. 3.3 Answering the question 1. You are told that an unknown material has a specific gravity in water of 0.95. Will this object float in water? How does the density of this material compare to the density of water? If we compare with the material that we used during the experiment, the alloy and the wood, that unknown material with specific gravity in water of 0.95 will float completely above the water. As what happened on the experiment, the alloy was not floating. That was because the specific gravity of the alloy is much larger than 1. As the result, the object density surely will be bigger that the density of the water. Contrastingly, the wood with specific gravity of almost 1 is half floating on the water. As the result, the density of the wood is almost the same as the water. In conclusion, any object that has density either almost the same or less than the fluid will float on the fluid. Therefore, the unknown material with specific gravity of 0.9 will has lower density than the water, so then it will float on the water. 2. What did you determine for the specific gravity of alcohol relative to water? If you mix alcohol and water, what fluid would settle on top? Alcohol has density of 800 kg/m3. Water has density of 1000 kg/m3. Thus, alcohol has smaller density than water. So then, the alcohol will float on the water. Therefore if the water and alcohol is mixed together, the alcohol will settle on top. 3. By direct measurement of its dimensions (radius, height, etc.), determine the volume of the alloy. Using its weight, which you measured earlier, calculate its density. How does this compare with the density you obtained using Equation (4)? In which answer do you have the greatest confidence? You may wish to think about this in terms of how well you could measure the dimensions, given its shape, of the alloy. Report 9: Archimedes Principles The volume of alloy is 22 mL = 22 cm3 = 2.2 x 10-5 m3 The mass of the alloy is 0.2 kg So then, the density of the alloy = 𝑚𝑎𝑙𝑙𝑜𝑦 𝑉𝑎𝑙𝑙𝑜𝑦 = 0.2 𝑘𝑔 2.2 𝑥 10−5 𝑚3 = 9091 kg/m3 The density that was determined by the previous equation is 6667 kg/m3 The student feel more confident with the result of the second calculation by using the volume and mass of the alloy due to the second equation has higher intensity to be used globally rather than the previous one.