(revised 5/17/23 D REQUIRED SUPPLIES AND ANALYSIS TOOLS 0 Meter stick, a clamp with attached hanger to hang meter stick, washers and nuts to serve as weights (you may also use other small objects), spring scales to measure masses, string for hanging weights and for hanging the whole apparatus. 0 Calculator for doing what calculators do. \fLAB GOALS By the end of this lab students will be able to 0 Calculate torque and moment arms for forces perpendicular to the line of action. 0 Calculate the torque required to establish static equilibrium. 0 Identify sources of statistical uncertainty, instrumental precision, and systematic errors. 0 Decide what and how much data are to be gathered to produce reliable measurements. 0 Write a summary of all the decisions made during the experiment including justifications of those decisions. 0 Present data and calculations in a wellorganized fashion including labeled tables and results that are referenced in the summary and/ or conclusion. 0 Write a detailed conclusion that discusses the outcomes of the investigation specifically detailing results of any revisions. INTRODUCTION As you have learned previously, an object is in static equilibrium only if the net force acting on it is zero and thus the object is not accelerating. Of course, the word Ham implies that the object's velocity is zero as well. This is fine for point particles, objects that do not have size. This is the approximation that we have been using so far in class. When you remove that approximation so that you can now consider objects that do have size, like real objects, the net force being zero is not the only requirement for equilibrium. Suppose you have two equal magnitude, but opposite direction forces acting on a wheel as shown below in Figure l. The wheel is in the plane of the page and its axle, or axis of rotation, is perpendicular to the page. Force F1 is applied to the wheel's axle, downward as shown. Force F2 is applied to the edge of the wheel, but upward. The net force on the wheel is zero because these forces are opposite in direction and equal in magnitude. Therefore, the wheel is in tram/42701741 equilibrium and its translational velocity (its updown, leftright velocity) will not change. However, force F2 will cause the wheel to rotate and in fact the rotational speed, the angular velocity, will keep increasing as long as F; is applied. Thus there will be an angular acceleration and the object will not be in rotatiom/ equilibrium. Force F1 will not stop this angular acceleration because you cannot spin a wheel by pushing on its axle as F] does. F1 Figure 1 \\Where could we apply F1 so that it does stop the rotational acceleration? Yes, you're right! You could apply F1 at the edge of the wheel precisely where F2 is being applied as shown below. Ia Figure 2 A force that causes rotational acceleration creates a torque on the wheel. \\X/ e use the Greek letter I (tau) to represent torque. The amount of angular acceleration created depends on the amount of torque that the force provides. The torque depends not only on the strength of the applied force, but also its orientation and distance from the axis of rotation. Consider the oblong shaped object in Figure 3. The actual point on an object where a force acts is called the force's pom; app/z'mz'z'm. A line drawn in the direction the force is pointing, and passing through the point of application, is called the [2m 1y\" 4mm. Finally, a line drawn from the aXis of rotation to the line of action, and perpendicular to it, is called the [ever arm, or moment arm, d, of the force. The vector r is a position vector for the point of application of the force assuming that the origin of the coordinate system is at the object's aXis. The moment arm, d, is equal to the magnitude of the position vector r only when the position vector r and the force F are perpendicular to one another. i ' ~ \\ ,Line of Action Figure 3 The magnitude of the torque created by a force equals the magnitude of the force multiplied by the length of the lever arm, d, or \"E = d F. For the wheel in Figure l the lever arm for F2 is equal to the radius of the wheel, while the lever arm for F1 is zero. Why? Furthermore, torque is a vector and so we can specify its direction. For now we will simply specify the direction by saying which direction the torque would rotate the object if it could. The directions will always be either clockwise (cw) or counterclockwise (ccww). Finally, for an object to be in equilibrium not only must the net force on the object be equal to zero, but the net torque must also be zero. In other words, the net clockwise torque, \"Em, must equal the net counterclockwise torque, Tm. Lg) An example will help. Consider the meter stick in Figure 4 set to balance on a pivot, or fulcrum. Where would you place the fulcrum so that the meter stick balanced? Yes, you are correct again! You would place it at the 50 cm mark (assuming that the meter stick has a uniform mass density throughout). \\X/ e would put the fulcrum at the 50 cm mark because this is where the center 1y\" gravity is. Think of the center of gravity as the balancing point of an object. Figure 4 Why does the stick balance when the fulcrum is at the center of gravity? Consider the meter stick being made up of a bunch of thin slices each of mass m as in Figure 5. The force of gravity on each thin slice is pointing downward and is applied at a point some distance away from the axis of rotation (the fulcrum). m each slice has a torque acting on it. If the aXis of rotation is placed at the center of gravig then all of the torques due to gravity total zero. In other words, the net clockwise torque equals the net counterclockwise torque. For the meter stick this is easy to understand. The slice at the 0 cm and 100 cm marks m equal, but opposite torques on them since they are equal distances from the fulcrum. Similarly the torques are equal and opposite for slices at the 1 cm and 99 cm marks, the 2 cm and 98 cm marks, etc. mg Figure 5 In this lab you will hang weights from various positions on the meter stick so that the stick is in static equilibrium. You will need to determine how much mass and where it must be placed so that the total torque on the meter stick is zero. THE WRITEUP As you work through this lab activity you Should be building your lab report or writeup. The first page must be a cover page with the lab title, date, your name and the names of any classmates who contributed. The second page should be a description of your procedures for making your measurements, photographs of your setup and any information on technique that you found lead to better results. The neXt few pages Should be the data tables, plots, and answers to questions asked within the procedures section below. And finally, the last page should be a thorough conclusion summarizing the results of this experiment and including reflection on your data about errors that might have crept in. Your conclusion should also discuss things that you would have done differently if you had different equipment and what that different equipment might be. PROCEDURES Part I: Fulcrum at Center of Gravity of Meter Stick: Record the mass, M, of your meter stick in the data section. You will be hanging weights from various positions on the meter stick. One way to do this is to use a piece of string looped through a hole in the mass and slid over the meter stick xed into position with a small piece of tape if necessary. There is a clamp that can be used as a pivot support for the meter stick if you slide the clamp onto the meter stick with the screw facing down and numbers on the stick being upright. Take a piece of string and loop it through the hanger on the clamp and hang the whole assembly from something like a doorknob so it can hang freely. (E the photo below.) Position the clamp so that the meter stick balances. Record the distance of the center of the clamp from #96 {em and of the meter stick. We'll call this distance X0. from the zero end of the meter stick. We'll call this distance xo. = i ) (Enter data for this section in the data area under Part 1, subpart i) Measure the mass of a large nut and then, with a piece of string, hang the nut from the 20 cm position of the meter stick. Call this position x1 = 20 cm. Be sure to measure the mass of the nut and include the mass of the string. Call the total mass m, (similarly for all masses you hang, m; will be the hanging mass plus the string's mass). The moment arm of this mass is the distance from Xo to x1; call this r, and record in the data section. Calculate and record the torque created by mass m. Now hang a larger mass (maybe several washere) from ch that the meter stick balances. The torquei) (Enter data for this section in the data area under Part 1, subpart i) Measure the mass of a large nut and then, with a piece of string, hang the nut from the 20 cm position of the meter stick. Call this position X1 = 20 cm. Be sure to measure the mass of the nut and include the mass of the string. Call the total mass m1 (similarly for all masses you hang, mi will be the hanging mass plus the string's mass). The moment arm of this mass is the distance from X0 to X]; call this r1 and record in the data section. Calculate and record the torque created by mass m]. Now hang a larger mass (maybe several washers) from the meter stick such that the meter stick balances. The torque from the second mass now cancels out the torque from mass 1. Record the total mass, m2, its position x2, its moment arm r2 and its torque 1:2. Compare torques 1 and 2 by computing a percent difference. ii) Remove the hanging masses from part i). Hang about 100 g (washers or nuts and string) from the 25 cm position and hang about 200 g (be creative) from the 80 cm position. Calculate moment arms and torques for these masses being sure to use the measured values of the masses. Calculate where a third mass, m3, of about 75 g (including string) should be placed to bring the system into equilibrium. Use the measured mass of m3 in your calculations. Now experimentally determine where my, should be placed and record your measured moment arm in 'Jw the data section. Compute the percent difference between the calculated moment arm and the measured moment arm. iii) Remove all masses from the meter stick. Hang an unknown mass, m1 (maybe a small set of keys) from the 20 cm position. Position a roughly 100 g mass, m2, so that the system is balanced. Compute the mass of the unknown by using the fact that the system is in equilibrium. Show all necessary data and calculations in the data section. Weigh the unknown mass with a spring scale and compare with the calculated value. Part II: Fulcrum Not 1; The Center g The Meter Stick: Now remove all masses from the meter stick and move the pivot clamp to the 30 cm position. This will be your new X'o (which you will adjust shortly). Clearly, the meter stick is unbalanced without any masses hanging from it. There are two ways to understand this: Method a) We can consider the total gravitational force on the meter stick, Mg, to be acting at the stick's center of gravity, X0. (Remember, X] was the position of the fulcrum that balanced the stick all by itself.) Now this force results in an unbalanced torque. The force's moment arm is (X0 X'a) and so the torque it creates is (X0 X'o) Mg. Figure 6 Method b) We can consider the stick to be made up of two pieces, one to the left of X'o of length L and the other to the right of X1] of length L2. The segment on the left has a mass m1 and a center of gravity at X] (halfway between 0 cm and X'()). The segment on the right has mass m2 and has its center of gravity at X2 (halfway between X'() and 100 cm). There are now two torques that do not cancel out. 11 equals mtg multiplied by moment arm r1 = x'o X1, and 1:2 equals ng multiplied by moment arm r2 = X2 X'(]. The question for you is, how do you determine m1 and 1112? X1 X0 mlg ng Figure 7 Now you will apply methods a) and b) above to analyze the torques on a meter stick with a variety of masses hanging from it and with the fulcrum NOT at the stick's center of gravity. 6 i) Tie a piece of string around a roughly 100 gram mass and measure its mass. Tape the string holding the mass to the zero end of the meter stick. Call this mass m and record it and its position X in the data section. Now adjust the position of the pivot clamp, at X'(), so that the stick is balanced. Record the value for the new X'o and compute the moment arm, r, of the hanging mass. Now that the system is balanced, the pivot point is now at the center of gravity of the system and all torques add up to zero. Show this by calculating the various torques using the two methods in a) and b) above. See the data section for the appropriate data that needs to be measured and calculated. ii) With the meter stick set up as in part H(i) with a 100 g mass at the zero position, add another roughly 100 g mass to the 60 cm position. Adjust the pivot clamp to balance the system. The new location, X}, of the pivot clamp is the system's new center of gravity. ReW Move the 100 g mass to the 70 cm DATA (Include this entire section, as-is, in your lab report) Mass of Meter Stick, M = (kg) Part I: XO = (m) 1 ) Make a sketch of your setup: = m = (kg) X1 = (m) n= (m) T = (N m) m, = (kg) X2 = (m 12 = (m) T = (N m) percent difference between t, and to =percent difference between t and 12 = ii) Make a sketch of your setup: m1 kg) x1 = (m) n = (m) (N m) my = kg X2 = (m 12 = (m) D = (N m) my = (kg)_Calculated 13 = (m) Calculated x3 = (m) Measured 13 = (m) Measured X; = (m) Computed percent difference for 1, between calculated and measured = 8Show calculations for 13 below: my = kg x = (m) 12 = (m) Calculated m, = (kg) Measured m, = (kg) E Percent Difference = Show calculations for m, below:Method b) Make a sketch of your setup and the relevant forces. moment arm of gravitational force on meter stick, r1 = (m) torque from gravitational force on meter stick. T1 (N m) torque from gravitational force on hanging mass, \"l? = g Q m) percent difference between torques : Make a sketch of your setup and the relevant forces. Calculations: length of left side of meter stick, L1 = (m) position of the center of gravity of the left side, X1 = moment arm of gravitational force on left side of stick, r1 = mass of left side of meter stick, m1 2 (kg) length of right side of meter stick, L2 = (m) position of the center of gravity of the right side, x2 = moment arm of gravitational force on right side of stick, r2 = mass of right side of meter stick, m2 = (kg) total ccw torque = g Q m) total m torque = (N m) percent difference = (m) (m) (m) l() (m) for roughly 100 g mass at 60 cm. (m) for roughly 100 g mass at 70 cm. (m) for roughly 100 g mass at 80 cm. (m) for roughly 100 g mass at 90 cm. QUESTION AND FURTHER PROCEDURES 1. 2. Suppose your unknown mass was 100 g in Part 1(iii). Could you use a 50 g mass to balance the system? Explain. From your data from part gii) can you predict where the system's center of gravity would be located if you placed the second 100 g mass at the 50 cm mark? Explain and calculate the new center of gravity. \\W hat was the greatest source of error in this experiment? Was this error random, or systematic and how do you know? 4. Were all of your calculated percent differences what you expected? What should they have been? Discuss any results that were not what you expected. For your lab writeup be sure to include a purpose, a general list of procedures for each part, a data/ analysis section that shows your data in the places provided in this handout rather than in your own tables, answers to the questions above, and a conclusion paragraph that summarizes what you did, problems you had, and things that you would have done differently if you were to do the lab activity again