Honors Lab 4.5 Freefall, Apparent Weight, and Friction

Transcription

1 Nae School Date Honors Lab 4.5 Freefall, Apparent Weight, and Friction Purpose To investigate the vector nature of forces To practice the use free-body diagras (FBDs) To learn to apply Newton s Second Law to systes of asses To develop a clear concept of the idea of apparent weight To explore the effect of kinetic friction on otion Equipent Virtual Dynaics Track PENCIL Explore the Apparatus Before starting this activity, watch the introductory video in the ebook at There s soe inforation that doesn t apply to this Honors lab, but you ll need to watch it all anyway. Sorry. Open the Virtual Dynaics Lab on the website. You should see the low-friction track and cart at the top of the screen. At the botto you ll see a roll of assless string, several asses and a ass hanger. Roll your pointer over each of these to view the behavior of each. Note the values of the asses. Also note that the epty hanger s ass is 50 g. If you ove your pointer over the cart you ll see that its ass is 250 g. Let s take a trial run with the apparatus. Reeber, everyone needs to take a turn at this. Turn on the cart brake (stop sign) and then ove the cart to the iddle of the track. Drag the roll of assless red string up to the cart and release it with your pointer (ouse button up) soewhere just to the right of the cart. (Fig. 1a.) You should see a short segent of string connecting the cart to the pointer. (Fig. 1b.) Without pressing any ouse buttons, ove your pointer to the right. The string will follow. Figure 1a 1b 1c 1d 1e 1f 1g Continue until your pointer passes the pulley by a bit. (Fig. 1c) Now ove downward. When the scissors appear, click with your ouse. (Fig. 1d) The string will extend downward and a loop will appear at the end. (Fig. 1e) Drag the ass hanger until its curved handle is a bit above the loop and release (Fig. 1f). The hanger will attach. (Fig. 1g) Drag the cart back and forth. It s alive! Everything should work just as you d expect. Now turn off the brake and explore the cart s behavior. It even bounces a little off the ends of the track. We refer to this group cart, string, and hangers as a syste. You can actually call any group of objects a syste solar syste, digestive syste, etc. In this case, since the objects ove together as a group, we ll focus on their coon otion and look at the as a syste. Add soe ass to the cart as follows. Turn on the brake. Drag the largest ass, 200 gras, and drop it when it s soewhat above the base of the spindle of the cart. If you iss, just try again. We now have a cart with a total ass of 450 g, a 50-g ass hanger, and a assless string connecting the. Now observe the ore assive syste s otion. You should notice that it oves with less acceleration than before when its ass was 200 g less. We now have the (weight of the) sae 50-g ass, oving a larger total ass (500 g.) Figure 2 Honors Lab Dynaics 1 February 21, 2012

2 We ll now use this and other arrangeents of asses to look at various systes, how they ove, and why. The why part will always coe fro Newton s second law,, where is the net, external force on the syste, is the ass of the syste, and is the acceleration of the syste. I. Drawing force vectors A picture s worth a thousand points. We ve found that writing down what we know and want to know is essential to successful proble-solving. But with forces, there s the added directional inforation that s best described with a figure. Figure 2 is a nice representation of what our apparatus looks like, but it doesn t show us the forces. Drawing force vectors can ake it easier to identify the forces acting on systes. They also guide us in setting up the equation ΣF = a. Click on Reove Masses to reove the 200-g ass fro the cart. Turn the brake on and ove the cart near the left buper. Click to turn on Dynaic Vectors. The vectors in the iddle of the track represent the forces on the cart as well as its velocity and acceleration. (You ll generally want to leave Dynaic Vectors on ost of the tie in this lab.) In the vertical you should see the noral force, N, and the weight, W, which are equal and opposite. Thus there is no vertical acceleration by the cart, and there won t ever be with a level track. (N and W aren t to the sae scale as the horizontal forces because they re so uch larger.) In the horizontal you should see the friction force, F(f), and the tension in the (right) string, Tr. They are also equal but opposite. Thus there is currently no acceleration in the horizontal. In the horizontal there are also vectors for velocity, Fnet, a, W(x), and Tl. Since these are all zero, you just see their naes. The forces on the ass hanger are also shown. You just see the opposite and equal forces Tr, and Wr. Wr is the weight of the right ass hanger. For siplicity, let s turn soe of the off. Click the check boxes on the left side of the screen to ake it look like Figure 3. You should now have soething like this. Figure 3 Forces The friction force, F(f), is only present because of the brake. When you turn it off for an actual run the vectors should like Figure 5. This eans the brake s force will not be involved in any of what follows. Figure 4 Figure 5 Honors Lab Dynaics 2 February 21, 2012

3 The hanger s weight, Wr, is gravity s pull downward on the hanger, so it s an external force. That is, it s a force exerted on the syste fro outside. (Gravity is a very ysterious force that s still not clearly understood, but we speak of it as a force acting on the hanger by the earth below it.) The two tension forces, Tr, are internal to our cart-string-hanger syste, so they don t affect its otion. Got that? The syste includes the string. The pulley changes the direction of the tension force but otherwise doesn t affect it. So the only external force acting is Wr. So the net force, Fnet is equal to Wr. Note that their lengths are equal. Masses The ass being accelerated is the su of the cart s ass and the hanger s ass. After all, it s all accelerating. Free-body diagra We can clarify this a bit with a free-body diagra (FBD) which represents the ass being accelerated as a dot and each force as a vector arrow. We ll draw an FBD for each object and then one for the whole syste. The vector arrows on our apparatus actually create the diagras for us on each individual part of the syste. Since the pulley just changes the direction of the force we can siplify our drawings by treating the syste is if it were entirely horizontal with Wr acting to the right on the hanger. (Both representations are shown in figure 6b.) Note: We will ignore forces not acting along the direction of otion. The direction of otion is in the horizontal for the cart, and in the vertical for the hanger. a. The Cart. Once the brake is released, the only force acting on the cart is Tr, the tension in the right string. Hence its acceleration will be to the right. b. The Hanger. The two forces on the hanger, Wr, and Tr, exert opposite, but not equal forces on the hanger. As a result its acceleration is in the direction of the larger force, Wr. c. The Syste. The net force acts on the total syste ass. The pulley just changes the direction of the otion of the different parts of the syste. So the net force doesn t have a specific direction, but in our siplified, linear representation the net force is just to the right. Fro Newton s second law,, we have or ΣF = (Wr T) + T = Wr Wr = ( hanger + cart ) a hanger g = ( hanger + cart ) a (Actual) (Siplified) (Siplified) Figure (6a) (6b) (6c) Substituting in.050kg for h,.250 kg for c, and 9.80/s 2 for g, we have a syste acceleration of If we now add the extra 200 g to the cart we get = 1.63 /s 2 =.98 /s 2 which as we observed is saller than before. That s because there s the sae external force, W h accelerating ore ass. Don t panic! That s all just a = F/ where F is the weight of the hanger and is the total ass, including the hanger. And does the.98 /s 2 ring any bells? That s g/10. The syste is accelerating at 1/10 the acceleration it would have if you just dropped it. That s because the weight of only 1/10 of the syste s ass (.050 kg g) is pulling the entire syste (500 g). What if you just snipped the string and left the cart behind? That s called freefall. Honors Lab Dynaics 3 February 21, 2012

4 IIa. Freefall Acceleration We ve found that all bodies in free fall have the sae downward acceleration, g. Let s use our apparatus to find the value of g for a falling ass hanger. We need a syste of one object the ass hanger. H. You can t have a ass hanger without a string to attach it to. And all position easureents are associated with the cart, not the hanger. So we need a cart, but then again we don t. We need a cart that won t interfere with the falling hanger; a sort of vacant, epty, soulless, assless cart. What we need is a Zobie CART! Or as Dr. Seuss would call it, little cart Z. Set up the dynaics track with an epty cart and hanger. Drag the cart to the left and let it go. That looks OK, but as noted, the hanger is not really free-falling? It s dragging the cart behind it. Put the brake on and ove the cart near the left end of the track. Now click on the Z-cart icon below the track. The cart will becoe translucent indicating that it has becoe assless. Zoo! Clicking the Z-cart icon again will restore its ass. Repeat in and out of Z-ode to convince yourself that the whole syste sees to ove with a greater acceleration, possibly equal to g, when the cart s ass is eliinated. That s because the syste here is really just the ass hanger. So the Z-car is doing just what we want it to do. Try this. Turn on the otion sensor and record the otion in noral ode (250-g cart) in one color. Then select another line color and try a Z-ode run. Do the graphs show different accelerations? They should. 1. You ay have noticed that the brake doesn t work in Z-ode. Check it out. Why doesn t it work in Z-ode? 2. Since we can eliinate the ass of the cart with z-ode, we can now look at the otion of the freely falling hanger. Find its acceleration as follows. a. Set T ax to 5 seconds. b. Set Recoil to zero to eliinate bouncing. c. [Noral ode] (Not Z-ode.) Epty cart and hanger. Cart near the left end of the track with the brake on. d. [Motion Sensor on] e. [Z ode] Click on the Z-icon. The cart should quickly accelerate to the right end of the track. f. [Motion Sensor Off] 3. Let s find the approxiate acceleration using one of our kineatics equations. The cart and hanger travel with identical otions. The pulley just puts a 90 turn in the iddle. We ll find the acceleration of the cart which is the sae as that of the hanger. For the cart we can say Δx = v o t + ½ a Δt 2 But if we choose our initial tie, t 1, as just about when we turn off the brake, v o 0. So Δx = ½ a Δt 2 Equation 1 We can get Δx and Δt fro the on-screen data table. Look at the position values. They re all the sae until you turned off the brake. Find the first data pair after the first change in position. Ex. 0.14, 0.14,,0.14, 0.15 that s the one we want. This will be our initial value, when v o 0. The cart s otion actually started between this instant and the one before it, but we ll get a fairly accurate nuber if we ignore this error. a. Record this first data pair as x 1, and t 1 in the table. b. Scroll through the data to find a point soewhat before the cart hit the right buper. The first position reading greater than 1.8 is a good spot. Record the data for this point as x 2, and t 2 in the table. c. Calculate and record Δx, and Δt. d. Calculate and record the acceleration in Table 1 using Equation 1. Honors Lab Dynaics 4 February 21, 2012

5 Table 1 Free-fall acceleration Mass, (kg) Position, x () Tie, t (s) Δx () Δt (s) a (/s 2 ).050 x 1 = t 1 = x 2 = t 2 =.100 x 3 = t 3 = x 4 = t 4 = 4. For your syste (ass hanger), = kg, ΣF = g = N, predicted a = F/ = /s 2 Your experiental and predicted values should correspond fairly well. Actually the experiental value in the table ay be as uch as 15% larger than the theoretical value since t 1 was recorded a bit after the cart started, so Δx is a bit sall. But the iportant thing to notice is how well the two trials copare. You should find that the acceleration is independent of ass. IIb. Freefall the effect of the weight of the falling object on its acceleration Throughout ost of recorded history it was believed that the heavier a body is the faster it will fall. (Those of that opinion also didn t quibble over the definition of the ters fast and slow. ) F = a would see to bear their theory out since the acceleration is directly proportional to F. Let s test this old theory. In part IIa) we observed just the fall of the ass hanger. Let s ake it heavier to see if it goes faster. (Greater acceleration.) 5. Double the weight of the hanger by adding a 50-g ass to it. (You ll have to turn off Z-ode teporarily to allow you to put ass on the hanger.) a. Change graph colors so that you can see the new graph along with the previous one. Take your data and record it as before but in the.100-kg part of Table 1. Calculate and record the free-fall acceleration of the 100-g ass. b. H. Doubling the force did not see to double the acceleration. If a F, we ust be issing soething. The of a body is directly proportional to the net force acting on it. So if you double the force on it, its acceleration should. When you doubled the ass in this case, the acceleration. This is because doubling the weight also doubled the. This is a very special relationship that is still not clearly understood. Why should the force of gravity on a body be directly proportional to its ass (inertia)? Or vice versa? Perhaps this will be solved in your lifetie. Stay tuned! III. Apparent Weight In Figure 6b we had two opposing forces. Are they an action-reaction pair? No! And not just because they aren t equal. Newton s 2 nd Law: ΣF = a where ΣF is the su of the forces on a body with ass. Newton s 3 rd Law: Body 1 exerts a force on body 2. Body 2 exerts an equal but opposite on body 1. FBD s are about the 2 nd Law: It shows the external force on one body or a syste. In figure 6b, the dot represents the weight hanger. There are two external forces on it. the weight, Wr is the force that the earth exerts on the weight hanger. The reaction force is the weight hanger s force on the earth which is not shown since it doesn t act on the hanger. F(Earth on hanger), Wr = -F(hanger on Earth) A-R pair #1 Likewise the string exerts an upward tension force on the weight hanger and the weight hanger acts equally downward on the string. (6b) F(string on hanger), Tr = -F(hanger on string) A-R pair #2 Honors Lab Dynaics 5 February 21, 2012

6 Newton s second law relates the two forces acting on the hanger, the force on the left sides of the equations. Tr Wr = r a Now suppose that instead of the ass hanger, you were clinging to the end of string. You are the dot in the figure. Gravity pulls you downward with a force Wr. The string pulls up with a force Tr. That force Tr is what we call your apparent weight. If you were standing on a scale in an elevator it would work the sae way. It s the force upward on you needed to hold you at rest, ove you up or down at constant speed, or accelerate you up or down. And it changes depending on the acceleration. Let s explore this with our apparatus. To ake it ore vivid, iagine yourself as the right ass hanger and attached asses. You re holding the string to keep fro falling. To ake you accelerate up and down we need a hanger on the left side. 3. Set up the syste shown in Figure 7. You ll want a string and epty hanger on each side. You ll want Z-ode for this activity. With Dynaic Vectors on you ll notice a set of check boxes on the left side of the apparatus that allow you to show and hide vectors. The colun of arrows identifies the by color. This colun is also a toggle to show/hide the check boxes. You ight want to hide the. Add 50 gras to each hanger. Your ass ( r ) is 100 g! Figure 7 4. Let s look at the initial situation - hanging fro the rope at rest. That is, v o = 0. a = 0 So Tr = Wr Note the length of these vectors on the right side. They re equal since a = Now give the cart a good shove in either direction. Go at Vo ight be better. (It s a bit jerky. Vectors are slow to draw.) 5a. How does your apparent weight (Tr) at rest copare to Tr when oving at a constant, non-zero velocity? Tr(rest) < Tr(constant velocity) Tr(rest) > Tr(constant velocity) Tr(rest) = Tr(constant velocity) 6. Now try an upward acceleration. Add 100 gras to the left hanger. You should now have a total of 100 g on the right and 200 g on the left. Drag the cart to near the right end of the track. Note that Tr = Wr when the cart is being held. 6a. When you release the cart, what happens to Tr and Wr? For each, state increases, decreases, or reains the sae. 6b. When you release the cart, what happens to your apparent weight? (Sae choices.) 6c. When you release the cart, what happens to your actual weight? (Sae choices.) 7. Now try a downward acceleration. Adjust the asses so that the right side is still 100 g and the left side is an epty 50-g hanger. Drag the cart to near the left end of the track. Note that Tr = Wr when the cart is being held. 7a. When you release the cart, what happens to Tr and Wr? 7b. When you release the cart, what happens to your apparent weight? 8. When you accelerate upward your apparent weight. 9. When you accelerate downward your apparent weight. 10. When you are at rest or oving at a constant speed your apparent weight. Honors Lab Dynaics 6 February 21, 2012

7 IV. Kinetic Friction And finally, let s have a quick look at kinetic friction. For this study we ll just be working with the cart. The only force affecting its otion will be kinetic friction. So click [Reove All] to reove all the asses and hangers. So far the only friction force we ve encountered is the one supplied by the cart brake. It s uch too strong a force for our purposes. Instead we ll use a friction pad like the brake. But this one is adjustable. In Figure 8 you see a snapshot of the cart traveling to the right at velocity v o. A friction force acts to the left, thus slowing the cart down. In Figure 9 you see the friction pad control. When wheels is selected this control show the failiar Vo controls. By clicking on friction pad the controls change to allow you to adjust μ k, and μ s. We ll use only kinetic friction which is controlled by the μ k stepper. We ll start with it set just as it is in the figure. After one trial you ll try to deterine an unknown kinetic friction coefficient. 1. Set the friction control as shown in figure Turn on the ruler. 3. Move the cart to x = 10.0 c. (This is the position of the cart ast.) 4. Turn on the dynaic vectors: 5. Click wheels and set Vo to 200 c/s. Then switch back to friction pad. Figure 8 Figure 9 6. [Go at known Vo] The cart should launch to the right and slow to a stop before reaching the right buper. If not, retry the preceding instructions. Repeat the preceding step as needed to answer the following. 7. Clearly the cart slows down. What about the friction vector, F(f), tells you that why it slows down. 8. What about the velocity vector tells you that it does slow down. The cart slows down due to the friction force acting on it. Fro Newton s second law we know ΣF = a F f = a F f = -μ k F N = -μ k g -μ k g = a μ k = -a/g Equation 2 So if we knew a, we could divide by g to deterine μ k. How can we find the acceleration, a? If we launch the cart down the track, letting it coe to a halt we know v o = 2.00 /s v f = 0.00 /s Δx can be found with our ruler a can be found fro a single kineatics equation. You figure that part out. Honors Lab Dynaics 7 February 21, 2012

8 9. Got it! Take the data you need and record it in Table 2. If your value for μ k is not close to.12, then you ve done soething wrong with your technique or your calculations. You should get a negative value for the acceleration. Table 2 Deterination of known μ k =.12 g = 9.80 /s 2 v o = 2.00 x 1 =.100 x 2 = Δx = /s a = /s 2 μ k(experiental) = (no units) Show calculations below for a and μ k 10. Let s try an unknown friction coefficient. Change the μ ks? value to 1 using the nueric stepper. You ll quickly find that your 2.00 /s is not a good choice this tie. You can change it to whatever you like. But letting the cart go ost of the way along the track will give better results than a short run. Note also that the new μ k should be uch saller than the previous value. Table 3 Deterination of known μ ks? (1) g = 9.80 /s 2 v o = 1.30 x 1 =.100 x 2 = Δx = /s a = /s 2 μ k(experiental) = (no units) Show calculations below for a and μ k Honors Lab Dynaics 8 February 21, 2012