Introduction / Theory Newton's Second Law of Motion describes the results of a net (non-zero) force F acting on a body of mass m. F =ma [1] Solving equation [1] for acceleration, we get [2] a =- m Equation [2] suggests two separate relationships: 1. if the total mass is constant, the acceleration is directly proportional to the force ( a .c F ). If Acceleration vs. Force is plotted, then putting equation [2] into "y = (slope)x+b" form yields a = ( _) Fto [3] which means that the slope is the reciprocal of the mass. 2. if the force is constant, the acceleration is inversely proportional to the total mass, which can also be stated as the acceleration is directly proportional to the inverse mass ( a oc 1/m ). If Acceleration vs. Multiplicative Inverse Mass is plotted, then putting equation [2] into "y = (slope)x+b" form yields a = (F ) - to [4] which means that the slope is the total force. Note that directly proportional means that when one = 0, the other = 0, as well as, for example, if one doubles, the other doubles; if one triples, the other triples; .... In this experiment, we will examine both relationships. Figure 1 shows the basic experimental setup. The system under study comprises a low-friction cart connected to a mass hanger via a string that runs parallel to the track and over a pulley. This system is ideal for two reasons: The force on the system is easily quantified, as it is just the weight of whatever is on the end of the string hanging off the table. Figure 1: The Basic Set-up Any masses in the cart or on the hanger become part of the system. Notes: [PEC] = possible error correction. These are quick checks to ensure that an easily corrected problem does not persist through the lab. [*] = a step which can be (but does not have to be) done outside of lab g = 9.798 m/s2 [5] The "A" value should be recorded to at least four decimal places. If the LQ2 does not give you enough decimal places, then assume that the blanks spaces are zeros.Apparatus Triple beam balance, Table, 50g Mass hanger, Low-friction cart, Pulley, Motion Sensor Three load masses (the black String, LabQuest2 (LQ2), bars), Scissors, (potentially) Four slotted masses (in some Dynamics track, combination of ~10g and ~20g) Procedure A: Set-Up 1. Using the triple beam balance, measure the mass ( +0.1g = +0.0001 kg ) of the cart, meant and one of the long load masses, mlong load. (The chosen load mass will be used throughout Procedure B.) 2. Put the track on a level table with the rubber feet (if there are any) hanging off the edge. Make any necessary adjustments to the track so that the track is as level as possible. [PEC: if a cart or a ball rolls on the track when placed on the track without any contact forces impelling it, then the track is not level. However, the cart might move on some parts of the track and not others. ] 3 . Attach the pulley to the hanging end of the track. 4. If the right length of string is not already available, use the scissors to cut off a piece of string (or use an already cut piece of string) which when taut is about as long as the track. Attach one end to the cart and the other to the mass hanger, which will hang off the end when the string is draped over the pulley. Ensure that the when the cart is near its starting point, the string between the pulley and the cart is parallel to the middle of the track and the sides of the track. Adjust the pulley as necessary. 5. Place the motion sensor at the end of the track so that it will record the motion of the cart as it moves away from the sensor and toward the pulley. Procedure B: Acceleration and Force 6. Start with an empty mass hanger. Measure the mass ( +0.1g = +0.0001 kg ) of all four (4) slotted masses and record in Set-up 1 in Table 1. Put all four slotted masses and the load mass on the cart so they don't fall off when the cart is moving. Make sure that the cart does not hit the pulley. Warning: As the cart careens down the track, someone in the group needs to stop it before it damages the pulley. Conduct a trial run to ensure that the motion sensor captures all of the movement. [PEC: Any sharp corners in the plot indicates the cart went out of the sensor's beam. In such a case, adjust the sensor and redo.] 7. Pull the cart back until the top of the mass hanger is just below the pulley (and still not touching the pulley). Based on the motion sensor, note and record ( +0.001 m ) the initial position of the cart, X, cart . If the motion sensor records a position under 18 cm, then move the sensor farther back from the cart. 8. Gently roll the cart down the track until just before the mass hanger touches the floor and then move back 5 cm. Based on the motion sensor, note and record ( +0.001 m ) the final position of the cart, X f cart . 9. Record the masses of the set-up in Table 1. 10. Once ready (including that the cart is at X, cart ), conduct three good trials for the set-up. To get a good trial, start recording with the motion sensor (i.e. hit the green triangle in the lower left corner.) Release the cart. Stop the cart before it collides with the pulley and stop the recording. [PEC: If the main part of the Position vs. Time plot does not look like a nice parabolic curve, then rerun the trial. ] From the 2Graph screen, highlight the part of the curve which is a nice parabola. [PEC: The minimum position should not be less than X; can , and the maximum position should not be greater than *, can .] Touch the "Analyze" drop down. Select "Curve Fit" > Position. [When the next screen appears, there should be an open and close bracket on the graph to the left. The position of the brackets should match the edges of the area you highlighted on the previous screen.] Touch the drop down arrow beside "Fit Equation:" and select "Quadratic". Record the coefficient of the quadratic term (the "A" value") [PEC: When the curve fit occurs, there will be a black line plotted on the graph on the left. It should closely match the points on the graph between the brackets. If it does not, then highlight a different portion of the parabola and repeat. ] 11. Transfer one of the slotted masses from the cart to the mass hanger and repeat Steps 9 - 10. Repeat until the three trials have been run for each of the five set-ups. [PEC: As more mass is added to the hanger, the value of "A" should increase.] 12. [*] Calculate the total mass (= mass of the cart + load mass + slotted mass on cart + hanging mass), the total force (= weight of hanging mass), and the acceleration (= 2 * the average "A" value) of each Set- Exel Up. Record in Table 2. 13. [*] Graph Acceleration vs. Force. Force Acc Procedure C: Acceleration and Mass There will be four sets of load masses used in this section: a) no load masses b) 1 short load mass c) 1 long load mass d) 1 short and 1 long load masses e) 2 short and 1 long load masses 14. Remove all masses from the hanger and cart. No mass will be added to the hanger for this entire procedure. 15. Pick one set-up not already tested, above, use the triple beam balance to measure the mass ( +0.1 g ) of any new bar, record the total mass of the load mass(es). Put the load mass(es) on the cart. 16. Run the trial three times, recording the data in Table 3. 17. Repeat the previous two steps until all five set-ups have been run three times. [PEC: As more mass is added to the cart, the "A" value should should decrease.] 18. [*] Calculate the displacement, the total force (= weight of hanging mass), the total mass (= mass of the cart + total mass on cart + hanging mass), the inverse mass (1 / the total mass), and the acceleration (= 2 * the average "A" value) of each Set-Up. Record in Table 4. When recording the inverse mass in Table 4, please present the numbers in decimal form, not fractional form. 19. [*] Graph Acceleration vs. Inverse Mass. Data m car ( kg ) : 0.5 109 m shortload ? ( kg ) : 0. 2488 [Proc B,C] mlongload ( kg ) : 0. 4998 [Proc B,C] * i can ( m ) : 0. 167 m shortload ( kg ) : 0, 2493 [Proc B,C] * f can ( m ) : 0. 687 wSet-Up Slotted Mass Hanging "A" for Trial 1 "A" for Trial 2 "A" for Trial 3 on Cart (kg) Mass (kg) (m/sz (m/s') (m/s') 1 0.05 92 0.050 0.0329 0. 2295 0. 131 1 2 0,0388 0.0704 0. 2744 0. 2107 0. 2254 3 10.0184 10. 0908 0.5488 0. 1813 0. 3087 4 10. 0081 10.1041 0. 2401 0. 3920 0. 2401 5 1- 0. 0023 10. 1415 0. 3773 0. 3945 0. 3773 Table 1: Data for N2L, Procedure B Set- Total Mass on Hanging "A" for Trial 1 "A" for Trial 2 "A" for Trial 3 Up Cart (kg) Mass (kg) (m/sz) (m/s') (m/s') 1 10 020500 o. uusg 0.3993 0 65 17 2 10. 2493 0.0500 0. 2787 0. 2853 0. 2628 3 10. 49gg 0,05 00 0. 1857 0. 2058 0 2005 4 10 . 7 492 20500 0, 1609 0. 1602 0. 15 72 5 10. 898 20500 0. 1346 0. 1371 10. 1352 Table 3: Data for N2L, Procedure C Results Total Mass (kg) Set-Up Total Force on System (N) Acceleration (m/s?) 10.48 89 0. 13 12 . 2 = 0. 2624 2 1. 6181 0.6898 0. 2368 . 2 = 0. 4736 3 0 . 8897 0. 3463 . 2 = 0.6926 4 10 . 98 06 0. 2807 .2 = 0.5894 5 1.0925 0. 38 30 . 2 = 0. 7660 Table 2: Results for N2L, Procedure B 0. 0 5 X P Total Force on Set-Up Total Mass (kg) Inverse Mass (1/kg) Acceleration (m/s?) System (N) 0. 5609 1.7828 0. 4989 . 2 = 0.9878 0. 4899 2 0. 8 102 1. 23 43 0 . 2756 . 2 = 0. 5 5 12 3 1. 06 08 0. 94 27 0. 2007 . 2 : 0. 4014 4 1. 34 01 0. 76 33 0. 15 94 . 2 = 0. 3 /18 % 5 1, 5589 0. 64 15 0 . 13 36 . 2 = 0. 2672 Table 4: Results for N2L, Procedure CAnalysis B 1. Based on equation [3], what is the ideal value of the slope (including units)? What is the actual slope (including units)? What is the % error? Ideal slope - 76181 = 0.6180 90 Error + 10.6406 -061801, 100%= 3.65 0 .6180 Actual slope = 0. 6406 60 Co Error = 3.656% 2. Does the graph confirm the relationship between acceleration and force? Why or why not? Analysis C 3. Based on equation [4], what is the ideal value of the slope (including units)? What is the actual slope (including units)? What is the % error? Ideal slope = 0,4898 N % Error = 10.7653- 0.4899 0. 4899 100% = 56.216% Actual slope= 0.7853 N. 8% Error = 56, 216% 4. Does the graph confirm the relationship between acceleration and mass? Why or why not? 5. What consequence would frictional forces have on the net force, F ner ? In this experiment normal force and weight cancel each other, so the net force will have dew the same magnitude as tension. since friction has the opposites direction to tension, frictional forces would decrease the magnitude of Fnet 6. Would Newton's 2nd Law of Motion F=ma still be correct with frictional forces acting on the system? Justify your answer. Ui