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Team: ____________________ ____________________ Force and Motion 2 In the first Force and Motion lab, you studied constant forces and friction-free motion. In this sequel, you will study forces that depend on time and position. You will also explore the force of friction. Part I. Force Curves A force curve is the graph of the force F on an object as a function of time t. The way in which F depends on t tells you everything about the dynamical behavior of a system. Here, you will use the force sensor to measure the force function F(t) for different systems. “Reading” the shape of force curves is an integral part of the “art” of Newtonian mechanics. A. Bungee Jump What kind of forces do you feel during a bungee jump? Build the following small-scale replica of a bungee-jump system: force sensor rods 3 rubber bands 500-gram weight table ground Open the file ForceProbe. Set the switch on the sensor to the ‘10N’ range. Zero the sensor when it is in the vertical position and nothing is attached to the hook. Start with the person (weight) high above the ground (floor, not table) so that the bungee cord (rubber bands) is relaxed (unstretched). Caution: Before releasing the heavy 500-gram mass, make sure it will not hit your foot. Activate the force sensor. Release the person. The force sensor will record the tension force F(t) exerted by the bungee cord on the person. Change the scales of F and t on your F(t) graph (try the Autoscale) in order to magnify all features and clearly display the shape of the force curve during the entire up-down-up-down- … motion of the bungee jump. PRINT your force curve. 1 1. Mark the two points on your printed force curve where the person is momentarily at rest at the extreme bottom (closest to ground after being released) and at the extreme top (farthest from ground after the first ‘bounce’) of the bungee jump. Label the points “Bottom of Jump” and “Top of Jump”. 2. What is the Apparent Weight of the person at these two extreme points? Note: The force sensor is a “hanging weight scale” just as a bathroom scale is a “standing weight scale”. The weight of any object hanging from the force sensor is simply the value (F) recorded by the sensor. If the object is not accelerating, then the sensor reads the “actual weight” (F=W). If the object is accelerating up or down, then the sensor reads the “apparent weight” ( F>W or F<W). Write the values of the two apparent weights next to the corresponding two extreme points on your force curve. 3. Mark the value of the actual weight of the person on the Force (Newton) axis. Draw a horizontal line at this value across the entire graph. Write “Actual Weight” on the line. Summary: Actual Weight = _______ N . Apparent Weight (top) = _______ N . Apparent Weight (bottom) = ________ N . At the bottom of the bungee jump, the jumper feels ______ times heavier/lighter (circle one) than their actual weight. At the top of the bungee jump, the jumper feels ______ times heavier/lighter (circle one) than their actual weight. B. Collision What does the force of impact look like during a collision? Collisions are everywhere. A ball hits a bat. A foot kicks a ball. A car hits a car. A ball bounces off the floor. An asteroid hits the earth. An electron scatters off a proton. A rusher tackles a quarterback. An asteroid collides with a planet. The dynamical trademark of all collisions is that the force between the colliding objects is an Impulsive Force − a LARGE FORCE acting over a small time. Place the force sensor against the end of the track. Start the cart at the other end. Zero the force sensor. Activate the force sensor. Give the cart an initial velocity. Let it coast along the track and collide with the force sensor (hook). force cart sensor 2 The force curve recorded by the force sensor should have a “triangular shape” and clearly show the two dynamical features that characterize all collisions: 1. Maximum Force Fmax (height of curve). 2. Collision Time ∆t (width of curve). Change the F and t scales on your F(t) graph so that the triangular shape of your collision curve is clearly displayed, i.e. “zoom in” on the collision event − that narrow window of time ∆t during which the cart and the sensor are in contact (F ≠ 0). The height and width of your “collision triangle” should be obvious just by looking at the F-axis and the t-axis of your F(t) graph. PRINT your force curve. Mark the value of “Fmax” on the F axis. Mark the time interval “∆t” on the t axis. Report the values of these collision parameters here: Fmax = __________ N . ∆t = __________ s . Area Under F(t) Curve = “Impulse” Note that that ½ Fmax estimates the average force Fav acting on the cart over the interval of time ∆t during which the impulsive force acts. The “ product of Fav and ∆t ” is an important dynamical quantity in physics because it determines the “ change-in-momentum m∆v ” of the cart, according to Newton’s Law: Fav∆t = m∆v. The force-time product Fav∆t is called Impulse. Calculus says Fav∆t is equal to ∫Fdt. It also says ∫Fdt is equal to the area. Thus Impulse = Area Under F(t). Use the computer to find the impulse acting on your cart during the collision. Here’s how: Highlight the “collision triangle” (click and drag mouse across the time interval ∆t) and then click on the area icon [π]. Impulse = Area under F(t) = _________________ N•s . Calculate the area by hand. Approximate the F(t) curve as a triangle (area = ½ base × height). Show your hand calculation directly on the graph. Compare to the computer calculation. 3 Part II. Position-Dependent Force − A Case Study “Feel the Force”. How does F(x) depend on x ? Many forces in nature depend on position. The gravitational force on a planet depends on its distance from the sun. The magnetic force on a mag-lev train depends on its distance above the tracks. The electric force on an electron in an atom depends on its distance from the protons in the nucleus. In this experiment, you will encounter one of the most important position-dependent forces in physics: the force due to a spring. The mechanical system that you will study consists of a cart connected to two springs moving on a metal track: x=0 The origin of the x-axis is chosen to be the point where the net force on the cart is zero. The cart will remain at rest at x = 0. At this so-called “equilibrium point”, the left spring pulls on the cart with a force that exactly balances the pull of the right spring. Thus F = 0 at x = 0. With your hand, pull the cart away from x = 0 to another position, such as x = +5 cm and x = −5 cm, and feel the force. The force always restores the mass back toward x = 0: F F −5 0 +5 As you pull the cart further away from x = 0, the restoring force back to x = 0 becomes stronger due to the increasing expansion and compression of the springs. Thus the magnitude of F is not constant: F depends on x. F F 0 x x Since the force F on the cart depends on the position x of the cart, the physical quantity F is represented by a mathematical function F(x). Given any value of x, the function F(x) specifies the value of the force at that particular x. Based on your qualitative study so far, all you know about this function is that F(0) = 0 and that the magnitude of F(x) is an increasing function. Your experimental goal is to find the precise form of this force function. Does F(x) have the form F = 3x , or F = 5x2 , or F(x) = ln(1+x)? You will find F(x) two different ways: (1) Dynamically, by observing the back-and-forth motion of the cart. (2) Statically, by using a spring scale. 4 1. Discovering the Law of Force In physics, a force law is a mathematical rule that specifies how force depends on position. From the previous Force and Motion Lab, you learned how physicists discover the force laws of nature: Observe the Motion , Deduce the Force. In this experiment, your quest is to Deduce how the force F(x) depends on x from a measurement of how the position x(t) depends on t. For a constant force (independent of time), the graph of x(t) is a parabola , i.e. x(t) = ½ at2. For the non-constant force due to springs, what is the “shape” of x(t)? Exercise: Suppose you measure x(t) for a mass m = 3 and find x = 4t2 . What is the net force acting on the mass? Hint: F = ma or F = md2x/dt2 . F = ____________ . Sinusoidal Motion Level the track using the “steel ball test”. Place the wooden block covered with felt on top of the cart. This block acts as a “flag” for the motion sensor to see. In order for the motion sensor to see the cart, you will probably have to elevate the sensor by placing it on top of blocks on the table. Open Logger Pro file Moving Along. Pull the cart from x = 0 to about x = 20 cm and release. Use the motion sensor to record the position x(t) of the cart as a function of time. Does your x(t) curve look like a “sine curve” ? A sine it is! There is a deep connection between force and geometry: Gravitational Force ⇔ Elliptical Motion Spring Force ⇔ Sinusoidal Motion Electromagnetic Force ⇔ Helical Motion As the cart moves back and forth along the x axis, the motion repeats itself. The cart “oscillates”. The position function x(t) is a periodic function of time t. Oscillatory motion is everywhere in Nature: pendulum, tuning fork, guitar string, sound waves, light waves, radio waves, electron in antenna, atomic clock, molecular vibrations (heat), nuclear magnetic resonance (NMR), pulsating stars (pulsars), oscillating universe (big bang-big crunch) … The mathematical equation describing the x-t curve recorded by your motion sensor is x(t) = A sin (2πt/T) . This sinusoidal worldline is graphed below. Note how the parameters A and T uniquely define the motion function x(t) and determine the overall shape of the worldline. Qualitatively speaking, A and T specify the “height” and the “width” of the sine curve. 5 x T 2A t More precisely, the amplitude A is the maximum displacement of the cart away from the equilibrium point. The period T is the time it takes the cart to complete one oscillation (one back-and-forth motion) Note that the function x(t) = Asin(2πt/T) is bounded, −A ≤ x ≤ A, and periodic, x(t+T) = x(t). Examine the x-t curve recorded by your motion sensor. Read off the values of A and T that characterize the sinusoidal motion of your cart: A = ______________ m . T = ______________ s . The Period of Motion T Determines the Law of Force F(x) Amazing Fact: If you know the time T, then you can figure out the force F(x). Here are the key steps that allow you to deduce the force function F(x) from the motion function x(t): F = m a = m dv/dt = m d2 x /dt2 = m d2 Asin(2πt/T) /dt2 = m [−(2π/T)2 Asin(2πt/T) ] = m [−(2π/T)2 x ] Can you justify each step (equal sign)? This derivation illustrates the power of Newton’s PHYSICS and CALCULUS in analyzing nature. The last line in the derivation reveals precisely how F depends on x. In summary, by inserting the motion function x = Asin(2πt/T) into Newton’s law, F = m d2x/dt2 , we have deduced the following force function: F(x) = − (4π2m/T2) x . 6 Physicists write this force law in the form F = − kx , where the force constant k is k = 4π2m/T2 . The magnitude of the force F on the cart increases in direct proportion to the distance x that the cart is displaced from the equilibrium point (x=0). If you double x , then you double F. The constant k is the proportionality constant between F and x. The minus sign in F = −kx indicates that the force is always opposite to the displacement. A cart on the right (left) side of x=0 experiences a force that is directed to the left (right). Physicists say that the spring force is a “Linear Force” because the force function F = −kx is a linear function of x. In contrast, the force of gravity and the force of electricity are “Inverse- Squared Forces” (F = k/x2 ). Whereas the spring force gets stronger as the distance increases, the forces of gravity and electricity get weaker as x increases. If you know the value of the force constant k, then you know everything about the spring force in your system. Different spring systems have different values of k. Stiff springs (car suspensions) have large values of k. Weak springs (toy slinkys) have small values. What is the value of k that characterizes the chemical bond – “the electromagnetic spring” − between the vibrating oxygen atoms in an O2 molecule? Note that the relation “k = 4π2m/T2 ” says that “k goes like 1/T2 ”: strength of force ∼ inverse square of period . Why should the force be related to the inverse of time? This relation makes intuitive sense. A stiff spring (large k) vibrates faster (small T) than a weak spring. If the cart moves back-and forth rapidly (small T), then the velocity changes rapidly (large a), and thus the force (ma) must be strong (large k). Measuring the Force Constant with a Stopwatch Bottom Line: To find the law of force F(x) = − (4π2m/T2)x , you need to measure the mass m of the system and the period T of the motion. Measure m of system (cart + block). m = ________________ kg . Measure the period with a stopwatch. More specifically, measure the time it takes the cart to complete five oscillations (five back-and-forth motions). Divide this five-cycle time by five to obtain the period. Why five? Ten would be good too. Because of the uncertainty in starting and stopping the watch and counting the oscillations, it is more accurate to measure a long time rather than a short time. For example, if the uncertainty in measuring the time with the 7 stopwatch is 0.5 seconds, then a “short time” measurement, such as 5 ± 0.5 seconds, is a 10% error, whereas a “long time” measurement, such as 50 ± 0.5 seconds, is only a 1% error. Report the value of the period of your oscillating cart. T = _______________ s . Is this stopwatch value of T consistent with (within 5% of) the graphical value of T that you obtained from the “width” of your x-t curve? If not, see your instructor. Based on your measured values of m and T, what is the force constant that characterizes your spring system? Show your calculation. k = _________________ N/m. Given this value of k, compute the magnitude of the force (F = − kx without the minus sign) at those values of x listed in the following table. Force deduced from the Motion x (m) 0.00 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 F = kx (N) 0 2. Measuring F(x) with a Spring Scale Now that you have deduced the force function F(x) from the motion data x(t), you can check your result by directly measuring F(x) using a spring scale. This may seem like an easy method to find the force, but remember, someone had to calibrate the force meter to convert a length measurement (stretch of spring) into a force quantity (Newtons). Finding force from F = ma is the gold standard for measuring force. First make sure the spring scale is “zeroed”. When the scale is held in a horizontal position and nothing is connected to the hook, the force should read “zero”. Start with the cart at rest at x = 0. The two springs should still be connected to the ends of the cart and the track. Attach the spring scale to the right end of the cart. While keeping the scale as horizontal as possible (parallel to the track), pull on the scale case and displace the cart from x = 0 to x = 0.02 m. Hold the cart at 0.02 m and record the scale reading Fs in the table below. Next pull the cart to 0.04 m, hold, and record the reading, etc. Note that since the cart is at rest at each location, the force exerted on the cart by the scale is equal and opposite to the force exerted 8 on the cart by the two attached springs. The subscript “s” on Fs denotes that this force is measured with the spring scale. Force measured with the Spring Scale x (m) 0.00 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Fs (N) 0 Use the program Graphical Analysis to plot Fs versus x (Fs on the y axis). Your data points should fall on a line. Analyze the graph to find the best-fit line through the points. The equation of the line is Fs = ksx. Report the value of the slope: ks = _________________ N/m . On the same graph, plot the force function F = kx that you discovered from your analysis of the oscillatory motion (see the Table Force deduced from the Motion). [ How do you graph a second set of data on the same graph? Click on “Data” to enter Data Set 2. Click on the “y” label of the graph to plot Data Set 2 ] Note that since the minus sign in the force law F = −kx has been omitted in both graphs, these graphs display the magnitude of the force and not the direction. PRINT your graph On the printed graph, label the F = kx line with the name “Force deduced from the Motion”. Label the Fs = ksx line with the name “Force measured with the Spring Scale”. Compare Your Measured Force Laws The force function F(x) = − kx deduced by analyzing the oscillatory motion is F(x) = − _________ x . The force function Fs(x) = − ksx obtained by directly reading the spring scale is F(x) = − _________ x . The percent difference between the force constants k and ks is ______ % . 9 Part III. Exploring Friction The force of kinetic friction acts on any object sliding across a surface. This friction force f obeys the empirical law: f = µN , where N is the normal force and µ is the coefficient of kinetic friction. The coefficient µ depends on the nature of the two surfaces in contact. For steel on steel, µ = 0.5. For Teflon on steel, µ = 0.04. For rope on wood, µ = 0.3. For shoes on ice, µ = 0.05. For rubber on dry concrete, µ = 0.8. For rubber on wet concrete, µ = 0.3. Measuring the Coefficient of Friction You goal is to measure the value of µ for felt on aluminum. Attach the spring scale to the wooden block covered in felt. Practice pulling the block across the aluminum track at a constant speed. Judge the speed (qualitatively) by covering a constant amount of distance every second of time. Try a speed of about 10 cm per second. Then try a speed of about 20 cm per second. To a very good approximation, the force of friction is independent of speed. Why pull at constant speed? If the speed of the block is constant, then its acceleration is zero, which means that the net force on the block is zero, which means that the scale reading F (spring force) equals the magnitude of the friction force f . In the vertical direction, N equals mg. See the force diagram below. This force analysis assumes that the track is level so that there is no component of mg along the track direction. N ⇒ v = constant f (friction) F (scale reading) mg Record the value of the friction force f for five different values of the normal force N. Load the block with five different brass weights (0, 100, 150, 200, 250 grams) in order to change the value of N = mg. Note: m = mass of wooden block + mass of brass weights. m (kg) N (Newtons) f (Newtons) Use the program Graphical Analysis to graph f versus N (f on y axis, N on x axis). Include the point (N, f) = (0, 0). Find the best-fit line through your data points. If you compare the mathematical equation of a line “y = mx+b” with the physical equation of friction, written in the form “f = µN+0”, then it follows that the coefficient of friction µ is equal to the slope (m) and the y-intercept (b) should be zero. Indeed, µ = f/N is the rise over the run of the line. PRINT your graph showing the best-fit line and the value of the slope. Report your measured value of the coefficient of friction: µ = ________________ ( felt on aluminum ) . 10 Part IV. Experimental Design: Lowering a Weight 3/5 m in 9/5 s. Your goal is to lower a “heavy” weight (100 gram brass cylinder) over a pulley so that the weight, starting from rest, falls to the ground – a distance of 0.60 meters – in a time of 1.8 seconds. Here is the schematic of the mechanical system: Load mL mB Block Felt Aluminum Hanging Mass mH = 100 grams 0.60 m 1.8 s The Theory 1. Calculate the value of the acceleration a of the hanging mass as required by the design specs (falls 0.60 m in 1.8 s). a = ___________ m/s2. 2. Draw a free body diagram for the hanging mass mH. Set up Newton’s equation of motion Fnet = ma for this body. Plug the numerical values of mH and a into Newton’s equation and solve for the value of the tension T in the string. T = _____________ N . 3. Treat the Block-plus-Load system as one body of mass m ≡ mB + mL . Draw a free body diagram for the mass m. Set up Newton’s equation Fnet = ma for m. Plug the numerical values of T , µ , a , and g into Newton’s equation and solve for the value of m. Then compute the mass of the load using the relation m = mB + mL . m = _______________ kg . mL = _______________ kg . 11 * Have your instructor check your theoretical work before you move on to the experimental stage of the design project. The Experiment Check that the track is level – again! This design experiment is sensitive to track tilt. So use the “steel ball test” to make sure that the track is “exactly” horizontal. Also make sure that the string is horizontal (if it is not horizontal, then adjust the tilt of the pulley). Add the amount of load mL to the block as predicted by your theory. Get the mass accurate to within one or two grams. Distribute the load mass evenly over the top of the block. Start the hanging mass at the specified height (0.60 m) above the ground. Release the system from rest and measure the time it takes (using a stopwatch) for the hanging mass to hit the ground. Repeat the measurement five times and compute the average value t. Time trials: ________ ________ ________ ________ _________ t = ____________ seconds . % difference between the required time 1.8 seconds and the actual t is _______ %. Sensitivity to Load Mass If your % difference is greater than 5 % , then add or remove enough load mass so that your system lowers the weight in the required time of 1.8 seconds. The total mass that now sits on top of your block is the actual load mass – the load mass that insures the design specs are satisfied in the actual experiment. % difference between the actual load mass = _____________ kg and the theoretical (calculated) load mass = _____________ kg is _______ % . 12

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the Force, Force and Motion, Force & Motion, Forces and Motion, newton's laws, Roller Coaster, Simple Machines, force of gravity, kinetic energy, Newton's Laws of Motion

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posted: | 4/10/2011 |

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