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PHYSICS SEMESTER ONE UNIT 4 UNIT 4: DYNAMICS AND NEWTON’S LAWS DYNAMICS So far we have covered aspects of motion relating displacement, velocity, acceleration and time. We will now look at what causes the motion. Specifically, we will look at what causes the acceleration to get items moving, slow items down and change the direction of the velocity. A force is described as that which causes a change in motion (an acceleration). Force is a vector. The change in motion of an object has both magnitude and direction. At the end of the 2D motion section, we covered the concept of frames of reference and how an object that is moving in one frame of reference can be stationary in another. We can also have accelerating frames of reference. When Bob accelerates around a corner in his car, the air freshener, hanging from the rear view mirror, tilts away from the centre of rotation for the turn. The observer in the car (Bob) is not moving relative to the air freshener. In his frame, he sees the air freshener swing away from vertical without anything pushing on it, so he must invent a fictitious force to explain the fact that the air freshener are not moving (in his frame of reference) yet have the strange tilt. An observer on the ground near the car notes that the velocity of the car must change as it goes around the corner; the car undergoes acceleration (direction change). The tilt of the air freshener can be explained entirely by the forces required to move it around the corner, so no fictitious forces are necessary. Bob’s frame of reference inside of the car is known as a non-inertial reference frame because he had to invent the fictitious force to explain the tilt of the air freshener. The reference frame of the observer on the ground is known as an inertial reference frame because the tilt of the air freshener can be explained due to the motion of the car using known laws of physics. Bob’s non-inertial frame of reference is accelerating relative to the inertial frame of reference of the observer on the ground. Any frame of reference moving at a constant velocity with respect to an inertial reference frame is also an inertial frame of reference. Creative Commons Attribution 3.0 Unported License 1 PHYSICS SEMESTER ONE UNIT 4 We will define an inertial reference frame as a frame of reference where an object that is not interacting with other objects (no net force, no fictitious forces) has zero acceleration. We generally consider anything on the surface of the earth, or moving with a constant velocity relative to the earth as an inertial reference frame. Newton’s First Law of Motion (the Law of Inertia) Under the condition that an object is viewed from an inertial reference frame, an object at rest will remain at rest and an object in motion will remain in motion with a constant velocity (that is, with a constant speed in a straight line), unless the object is acted upon by an external force. The net force on an object, Fnet F F1 F2 F3 ... , is the sum of all forces on the object. Consider an ice rink as our inertial reference frame. We will assume that the ice is frictionless so a puck sliding along the ice surface has no net force on it. The puck will keep sliding with the same speed and direction (same velocity) because the net force is zero. When the net force on the puck is not zero (extra force applied to puck using a stick) the velocity will change. Newton’s Second Law of Motion Under the condition that the object is viewed from an inertial reference frame, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. We will be using Fnet or F to denote the net force on an object. The symbol F indicates that we are making a sum of all of the forces. According to Newton’s Second Law of Motion a F m or a Fnet m This is more commonly written as F ma or Fnet ma Force is a vector quantity so we can split this equation into separate component equations Fx max , Fy ma y and Fz maz Creative Commons Attribution 3.0 Unported License 2 PHYSICS SEMESTER ONE UNIT 4 Each component equation relates the forces on the object in that direction with the acceleration in that direction. A net force in the x direction will cause acceleration in the x direction. We will often just use the x and y directions in 2D. We will skip the vector notation when we are dealing with motion and acceleration in only one direction. With the two or three vector component directions, we have two or three equations that have some variables in common. In a typical 2D problem, there are two unknown variables and the two net force component equations. We can combine the net force equations to eliminate one of the unknown variables and solve for the other. We can then substitute the newly found variable in one of the original equations to find the second unknown variable. More later … Dimensional Analysis for Force Define the dimensions of mass, length and time as M, L, and T respectively. x dimensions of the variable x The dimensions of mass, acceleration and force in Fnet ma are then m M , a L T2 and Fnet m a M L T2 . The SI unit for force is the newton, 1 N = 1 kg·m/s2. If an object is released near Earth’s surface it will accelerate at g = 9.8 m/s2 due to gravity so we can say that the gravitational force (force due to gravity) on 1 kg of mass is mass times the gravitational acceleration or 9.8 N. In the imperial system, the term pound, lb, is often used to describe both mass and the force of the same mass. The distinction is often made using “pound force” and “pound mass”, though, to be correct, the “slug” should be used instead of the “pound mass”. Example Bob and his car have a mass of 1.20 × 103 kg. The engine and transmission produce a net horizontal force of 1.7 × 103 N. a) Calculate the acceleration of the car. Creative Commons Attribution 3.0 Unported License 3 PHYSICS SEMESTER ONE UNIT 4 Here we have no mention of direction so assume that the net force is in a single direction. Fnet 1.7 103 N, m 1.20 103 kg Newton’s second law of motion for one dimension is Fnet ma Rearranging gives a Fnet m 1.7 103 N The acceleration of the car is 1.4 m/s2. 1.20 10 kg 3 1.42 m/s 2 b) Calculate the time required to accelerate from rest to 95 km/h. There is no escape from kinematics problems in first year physics. They show up as part of most sections. We cover kinematics and dynamics at the start of the course because many of the other sections deal with forces and motion. We need to convert the final velocity to m/s. vi 0, v f 95 km 1000 m 1 h m , 26.39 h 1 km 3600 s s t ? We now know the acceleration, velocities and need to find the time so use v f vi at t v f vi a . 26.39 m/s 0 1.42 m/s 2 18.6 s It takes Bob 19 s to accelerate from 0 to 95 km/h in his car. Creative Commons Attribution 3.0 Unported License 4 PHYSICS SEMESTER ONE y F1 = 456 N UNIT 4 Problem 1: Consider the following diagram showing the F2 = 642 N two horizontal forces applied by two friends as they pull the 1.20 × 103 kg car along an icy road. a) Find the total force (in component form). b) Calculate the acceleration of the car. State 50.0° 75.0° x the answer in magnitude – direction form. Creative Commons Attribution 3.0 Unported License 5 PHYSICS SEMESTER ONE UNIT 4 GRAVITATIONAL FORCE AND WEIGHT When Newton wasn’t playing with forces or inventing calculus, he was, among other things, examining planetary motion and the relationship between the masses, separation distance and gravitational forces. He came up with what is now called Newton’s Law of Universal Gravitation. Newton’s Law of Universal Gravitation Every particle in the Universe attracts every other particle with a force acting along the line joining the centres of mass of the two particles. This force is directly proportional to the product of the masses of the two particles and inversely proportional to the square of the distance between them. If the particles have masses m1 and m2, and their separation is r, the gravitational force has a magnitude Fg of Fg Gm1m2 r2 where G = 6.67 × 10-11 N·m2/kg2 is the Universal gravitation constant. This force applies to both particles. The particle with mass m1 experiences a force of Fg towards the particle with mass m2. The m2 particle experiences a force with the same magnitude Fg towards the m1 particle. The separation distance is measured between the centers of the particles. The term particle can refer to anything from atoms and subatomic particles to planets, stars and galaxies. The masses m1 and m2 are often written as M and m when one item M is much bigger than the other item. M is often used for the mass of the earth or a planet while m is the mass of a small object, like a fish, on the planet. Creative Commons Attribution 3.0 Unported License 6 PHYSICS SEMESTER ONE UNIT 4 We will deal primarily with objects on or near the surface of the earth. The separation distance is the radius of the earth and forces are in the radial direction (along the radius). The gravitational force on an object of mass m on the surface of Earth (mass M) is Fg mg where g GM r2 rˆ is the gravitational field strength or gravitational acceleration. The negative sign in the equation for g accounts for the fact that the gravitational acceleration direction towards (instead of away from) the centre of Earth. We will usually use the scalar forms Fg mg , g GM r2 and know that the direction as “down” or the negative y-direction perpendicular to the surface of the earth. We will assume g 9.8 m/s2 for this course. Those in Canada can check out the Natural Resources Canada website (http://www.geod.nrcan.gc.ca/products-produits/ggns_e.php) for the values in their area. The two significant figure value 9.8 m/s2 is good anywhere on Earth’s surface. The magnitude of the gravitational force on an object is also called the weight of the object. The weight of mass m is given by the equation w = mg The units of weight are the same as those of force, newtons N. Bathroom scales measure your weight, but report your mass. A 63 kg physics instructor has a weight of 620 N. How can the scales measure weight and report mass? When you step on your bathroom scales, the scales mechanisms rotate a disk inside the scales. When the disk stops, the number under the arrow/line is your mass. Manufactures design the disk with numbers that are your weight divided by 9.8 m/s2 so the number you see is your mass. The value of g on the moon is about 1.6 m/s2. The weight of the 63 kg physics instructor on the moon is w Fg mg 63 kg 1.6 m/s2 101 N Creative Commons Attribution 3.0 Unported License 7 PHYSICS SEMESTER ONE UNIT 4 2 The scales report this weight divided by 9.8 m/s , or about 10 kg. This is false because mass is a measure of the amount of matter inside the teacher, which is still 63 kg. Electronic scales do the division electronically. Example Use the above equation for g, along with g = 9.80 m/s2, and the radius of Earth to find the mass of Earth. Define terms g = 9.80 m/s2, r = 6.37 × 103 km Rearrange equation and substitute with g GM . I will leave that to you. r2 Note that it always a good idea to convert from km to m before you plug the numbers into the formula. The mass of the earth is 5.96 × 1024 kg. Check out the gravitational force examples in your textbook. Creative Commons Attribution 3.0 Unported License 8 PHYSICS SEMESTER ONE UNIT 4 Reduced to their simplest level, all forces are due to the interaction of two items. When you push on a wall, the wall pushes on you. Two magnets repel or attract each other. When we walk, we push on the earth and the earth pushes on us. The gravitational force from planet A on satellite B is the same as that from satellite B on planet A. In each of these cases we have two objects and two forces, one on each object. We can actually show that the force from object A on object B is equal in magnitude and opposite in direction to the force from object B on object A. This is Newton’s Third Law of Motion. Newton’s Third Law of Motion For two interacting objects (interacting = exertion forces on each other), the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1. F12 F21 Example The gravitational forces exerted by two objects on each other. Fg12 Fg21 m1 m2 r The magnitudes of the two gravitational forces are Fg12 Gm1m2 r2 and Fg 21 Gm2 m1 r2 The magnitudes are identical. The forces are in opposite directions. Problem 2 Earth is attracted towards that 70 kg man with a force of 686 N. Why don’t we notice the effect of our gravitational force on Earth? Creative Commons Attribution 3.0 Unported License 9 PHYSICS SEMESTER ONE UNIT 4 More Examples FN = Fdo An object, subscript o, sitting motionless on a horizontal desktop, subscript d, exerts force equal to its weight on the desk Fdo Fgo . The desk exerts a “normal” force FN Fod on the object equal and opposite the force from the object. Fdo Fod Here, the net force on the object is Fod = Fgo Fnet , o Fgo FN Fdo Fod Fdo Fdo ΣF = Fgo + n = 0. 0 The net force is zero so the motionless object on the table will remain motionless. We can use this to equate the magnitudes FN Fg mg In this case the normal force on the object is exactly equal to and opposite the force of gravity acting on the object. The net force is zero so there is no change from the motionless state of the object. A wrecking ball, subscript b, crashing into the wall of a building, subscript w, exerts a force Fwb on the wall while the wall exerts an equal and opposite force on the ball Fbw Fwb Fbw Fwb Creative Commons Attribution 3.0 Unported License 10 PHYSICS SEMESTER ONE UNIT 4 In these examples, the two forces in the action reaction pair are forces on the two objects involved (one on each object). This is an important because if the equal and opposite forces were always on the same object, nothing would move. Accelerations are often compared to the forces and accelerations due to gravity. An object experiencing an acceleration of one G-force (also G, g-force and g-load) is accelerating at 9.8 m/s2. An object accelerating at 5.0 G (or 5.0 g for even more confusion) has an acceleration of 9.8 m s 2 a 5.0 G 49 m s 2 1 G This is often used for spacecraft, fighter jets and roller coasters. Note the confusion that the term Gforce usually refers to an acceleration and not force. We have to multiply by the noted acceleration but the mass to get the net force applied to the object. The net force required to accelerate a 65 kg person at 5.0 G is 9.8 m s 2 a 5.0 G 49 m s 2 1 G F ma 65 kg 49 m s 2 3185 N 3200 N I will leave it to you to show that the net force required to accelerate an object at n G is n times the weight of the object. Creative Commons Attribution 3.0 Unported License 11 PHYSICS SEMESTER ONE UNIT 4 NORMAL FORCE The normal force ( N or FN ) is a force from solid surface to keep objects from sinking into the surface. If we look close enough, the normal force is created by the stretching and compressing of the material with the surface just enough to support the item on the surface (think of a grapefruit on top of a solid bowl of lime Jell-O). We are not worried about the deformation of the surface, only that it provides just enough force to keep the object from sinking into the surface. For a text sitting on a desk, the surface of the desk pushes back on the book just enough to counter the gravitational force. The normal force is just enough to balance the gravitational force. If the normal force was any less, the book would fall through the table. If it was any more the book would lift off the table. FN Fg The normal force gets it name from the fact that its direction is perpendicular to the desk surface. Normal is another word for perpendicular. (Technically, the force from the text on the table is a normal force. The gravitational pair of forces is between the text and Earth. We will ignore this fact.) All normal forces are perpendicular to the surfaces. All normal forces are exactly the right size to keep the object from sinking into or lifting off the surface. Draw the normal and gravitational forces in the following systems (you may have to print this page). Answer with solution to Problem 3. surface surface Creative Commons Attribution 3.0 Unported License surface 12 PHYSICS SEMESTER ONE UNIT 4 FRICTION Friction covers everything from drag in air or water to friction between an object and its solid surroundings. We will look primarily at friction between two items that are in contact. Chapter 6.4 for the 120 students describes the friction due to motion in fluids. Imagine a large boulder shows up on the horizontal road in front of your residence, and you decide to move it. The boulder is too big to lift and too awkward to roll so you decide to push it. Initially, the boulder ! doesn’t move, so you push harder. Eventually, you push with enough force that the boulder moves. You continue pushing the boulder without letting up because you “know” that it is easier to keep the boulder sliding than it is to let it stop and then start it moving again. Before the boulder starts moving, the friction known as the static friction, Fs . There three other forces acting the boulder: your force F , the gravitational force Fg and the normal force FN from the driveway. The net force on the boulder is zero because the boulder is not moving. Fnet Fg FN Fs F 0 The forces are shown in the following diagram. Force diagrams are incredibly useful for organizing information when doing problems. We will develop more formal force diagrams, later. F FN FN F Fg Fnet Fg FN Fs F 0 Fs Fg Fs Your force and friction are in the horizontal (x) direction while the force of gravity and the normal force are in the vertical (y) direction. In the x-direction Creative Commons Attribution 3.0 Unported License 13 PHYSICS SEMESTER ONE UNIT 4 Fnet , x Fs F 0 Whether you push with a large or small force (that doesn’t exceed the maximum static friction), the static friction exactly balances the force F applied by you so the net force remains zero. Fs F Increasing your effort, the force applied by you eventually exceeds the maximum static friction force and the boulder starts to move. The maximum static friction is proportional to the normal force between the driveway and the boulder Fs ,max s FN Fs Fs ,max where μs is the coefficient of static friction. The coefficient is dimensionless (both forces have the same units). Once the applied force exceeds the maximum static friction, the net force is no longer zero and the boulder starts to slide (grind). The friction for the moving boulder is the kinetic friction, Fk . Kinetic friction is proportional to the normal force between the surface and the object. Fk k FN where μk is the coefficient of kinetic friction. We will assume that kinetic friction is constant for this course. The coefficient of static friction is generally larger than the coefficient of kinetic friction s k so it is harder to get things moving than to keep them moving. If we plot the coefficient of friction, static or kinetic, versus the applied force we typically get the following relationship Ff Fs ,max s FN Fk k FN Fs = F F static, kinetic if moving kinetic only Creative Commons Attribution 3.0 Unported License 14 PHYSICS SEMESTER ONE UNIT 4 The following diagram shows the forces just after you have managed to start moving the boulder. Because you haven’t reduced your applied force yet, the net force is not zero. F FN Fnet F FN Fnet Fg FN Fk F Fg 0 Fk Fg Fk The coefficients of friction are generally less than one. This makes sense because, if we can push with a force larger than the normal force, we can lift the object off the surface, eliminating the normal force. Typical Friction Coefficients Note: the values shown here are approximate, and are intended for demonstration purposes only. Check your textbook, on-line, and reference books to find more accurate values between a wider array of materials. contact μs μk rubber on concrete 1.05 0.90 waxed wood on wet snow 0.20 0.10 lubricated metal on metal 0.15 0.05 Example FN A boulder on a horizontal surface has a mass of 520 kg. The coefficients of static and kinetic friction are 0.80 and F 0.60 respectively. Find the maximum static friction and the kinetic friction. Fk or Fs Fg Creative Commons Attribution 3.0 Unported License 15 PHYSICS SEMESTER ONE UNIT 4 Define terms: m 520 kg, s 0.80, k 0.60 The normal force and force of gravity are the only vertical forces acting on the stationary boulder. Fnet , y Fg FN 0 We will define the positive y-direction as “up”. This makes the force of gravity in the negative direction. mg FN 0 FN mg 520 kg 9.8 m/s 2 5096 N The formula for the maximum static friction is Fs ,max s FN 0.80 5096 N 4080 N The formula for the kinetic friction is Fk ,max k FN 0.60 5096 N 3060 N The maximum static force is 4100 N and the kinetic friction is 3100 N. What is the direction of the friction? The direction of the static friction is opposite the applied force. The direction of the kinetic friction is opposite the direction of motion. Friction can be both good and bad. tire: road friction – good car: air friction, drive train friction – bad Creative Commons Attribution 3.0 Unported License 16 PHYSICS SEMESTER ONE UNIT 4 Check out http://phet.colorado.edu/new/simulations/sims.php?sim=Forces_in_1_Dimension. This applet shows all of the forces acting on an object with motion in the horizontal direction. Try setting the applied force to a maximum, letting the object accelerate, and then reverse the direction of the applied force so the object comes close to but does not hit the wall. Also note that the friction force is just big enough to counter the applied force until the object starts moving. When the object is moving, the friction is the constant kinetic friction. Also play with the ramp http://phet.colorado.edu/new/simulations/sims.php?sim=The_Ramp. When you start the applet, the parallel force graph is the only graph shown. This shows the components of the force in the direction parallel to the ramp (forces perpendicular to the ramp are not shown). If you vary the ramp angle, you will see that the size of the different components varies with angle (the component of the gravitational force is Fgx mg sin . The force of kinetic friction is F fx FN Fg cos mg cos Both of these forces are functions of the ramp angle θ. If the angle is too low, the friction will be static, and will only be large enough to counter the applied and gravitational forces. Creative Commons Attribution 3.0 Unported License 17 PHYSICS SEMESTER ONE UNIT 4 EQUILIBRIUM A system where the net force is zero is said to be in equilibrium. Most of the problems you will do in this course will involve objects in equilibrium. An object is in equilibrium when its acceleration is zero. Under equilibrium conditions: F ma 0 The sum of the forces on the object is zero. The sum of the forces in the x-direction is zero. Fx max 0 The sum of the forces in the y-direction is zero. Fy ma y 0 Can a moving object be in equilibrium? Yes, as long as the acceleration is zero. As noted, equilibrium shows up often in dynamics question. Creative Commons Attribution 3.0 Unported License 18 PHYSICS SEMESTER ONE UNIT 4 FREE-BODY DIAGRAMS When solving problems with several forces on a single object, it is often useful to use a free-body diagram (FBD) to help visualize all of the forces acting on the object. FBDs help you organise your problem solving process. Check out the following PDF file on FBDs. Creating FBDs Select an Object - usually there is only one or two, and the question will often indicate the objects - in complex systems, it helps to sketch the whole system before drawing individual object FBDs Draw All Forces Acting on the Object - this is all forces noted in the question plus any additional forces that you can assume are present (interaction forces, force of gravity, friction if coefficient noted …) - gravitational force and normal force - usually you will be told whether to include friction Choose a Convenient Set of Directions - this is BIG, it can be the difference between a simple and difficult problem - Can you use the regular y-vertical, x-horizontal directions? - Is the motion in an obvious direction? Choose this as the either the positive x- or y-direction, and choose the other direction perpendicular to the motion. - Do most forces lie along one of two perpendicular directions? choosing the directions can save work when calculating the vector components Using FBDs Check for hints about the net force on the object - Is the object accelerating? - Is the object is stationary or moving at a constant velocity in one or more directions? In this case, Newton’s 1st Law tells us the net force in those directions is zero. If you are asked to find the acceleration, you will probably need to use Newton’s 2nd Law. In most cases, you will need to break the forces into components and use Newton’s Laws to solve for unknown variables. Creative Commons Attribution 3.0 Unported License 19 PHYSICS SEMESTER ONE UNIT 4 Example 1 Tossing a ball into the air. If you toss a 0.10 kg ball into the air with a force applied by your hand is 5.0 N, what is the acceleration of the ball? F +y draw FBD, pick directions (only one in this case) Is there acceleration? – yes, so looking to find the net force and then acceleration Fg Define Terms: m 0.10 kg, Fg mg 0.10 kg 9.8 m/s2 ˆj 0.98 N ˆj, F 5.0 N ˆj According to Newton’s Second Law the acceleration is proportional to the net force and inversely proportional to the mass. We need to find the net force in order to find the acceleration. Fnet F Fg 5.0 N ˆj 0.98 N ˆj 4.02 N ˆj The acceleration is then a Fnet m 4.02 N ˆj 0.10 kg 40.2 m/s 2 ˆj The acceleration is 40. m/s2 up. Creative Commons Attribution 3.0 Unported License 20 PHYSICS SEMESTER ONE UNIT 4 Problem 3 Find the acceleration of a 1600 kg car that is being pushed horizontally with a force of 5200 N to the right, assuming that the coefficient of kinetic friction is 0.25. Is there acceleration? Draw FBD, define directions Define terms: m 1600 kg, F 5200 N ˆi, k 0.25, a ? Can you find the acceleration using only one of the component equations? If not, can you use the other component equation to help find it? Creative Commons Attribution 3.0 Unported License 21 PHYSICS SEMESTER ONE UNIT 4 TENSION Tension is another common force that shows up in dynamics problems. Tension is another form of force used to describe the force from/on a rope. In this part of the course, we assume that the tensions on both ends of the rope are equal in magnitude, pulling towards the other end of the rope. Consider a tug-of-war with a very light rope where the rope is not moving. We can ignore the force of gravity on the light rope, and then assuming we have the force FL on the left end and FR on the right end, the FBD for the rope, looks like this FL FR The rope is not moving so the net force is zero. Fnet FL FR 0 FL FR , the magnitudes FL FR We can now apply Newton’s third law and see that the tension on each end is equal and opposite to the force on it. TL FL TR FR TL TR The tensions on the ends are then equal in magnitude and opposite in directions. TL TR T We often write the magnitudes as T at both ends. This is true even if the rope goes over a light, frictionless pulley. The case where the pulley is not light and frictionless will be covered later when we look at rotational motion. T T Creative Commons Attribution 3.0 Unported License 22 PHYSICS SEMESTER ONE UNIT 4 Finally, there are several cases where we only have to consider one end of the rope. In these cases, the tension is just another force. Instead of stating that the applied for is 5 N, we have a rope attached with a tension of 5 N. Problem 4 A boat is secured to a lakeside pier with two horizontal ropes. A wind is blowing off shore. The tension in the ropes are T1 = 48 N [16° N of E] and T2 = 48N [16° S of E]. Assuming that the net horizontal force on the boat is zero, what is the force of the wind on the boat? There is no movement or tension in the vertical direction so we will only look at the horizontal plane. Draw FBD, draw directions Terms defined in question. Is the object accelerating? What is the net force on the boat? The sums of the force components in both directions must equal _______ . Break the tension forces into components T1E T2E T1N T2N Use the net force to find the components of the wind force Fw . The force of the wind on the boat is Creative Commons Attribution 3.0 Unported License 23 PHYSICS SEMESTER ONE UNIT 4 Problem 5 Imagine, instead of pushing the car in Problem 3, you are tow the same car (mass 1600 kg, coefficient of friction 0.25) with a tow line that makes an angle of 37° with the horizontal. The tension on the tow line is 5200 N. What is the acceleration of the car? Problem 6 A child pulls a sled up a snow-covered hill at a constant velocity with a force parallel to the hillside. If the sled has a mass of 9.5 kg, the hill slope is 12°, and the coefficient of friction between the sled and the snow is 0.20, what is the force applied by the child? Problem 7 Sleds A and B are connected by a horizontal rope, with A in front of B. Sled A is pulled forward with a horizontal rope with a tension of magnitude 29.0 N. The masses of sleds A and B are 6.7 kg and 5.6 kg, respectively. The magnitudes of kinetic friction on A and B are 9.0 N and 8.0 N respectively. B A TAB Creative Commons Attribution 3.0 Unported License Tpull = 29.0 N i 24 PHYSICS SEMESTER ONE UNIT 4 Problem 8 +y A loonie (mL = 6.99 g) and a dime (mD = 2.09 g) are attached to the ends of a thread. The thread lies over a smooth horizontal bar. Initially, the coins are held motionless. When released, the coins +y +y start to move. The friction between the thread and bar is negligible (zero). a) Find the acceleration of the coins. b) Find the tension in the thread. *** There are as many different examples of dynamics problems are there are problems needing forces. Check out other examples and problems in your textbook, especially those with springs, pulleys, inclines and blocks that are side-by-side. *** Creative Commons Attribution 3.0 Unported License 25 PHYSICS SEMESTER ONE UNIT 4 Example Highway Curve Angles What angle do you bank a curve in order to minimize the chance of sliding off? Each highway curve is designed for a particular speed. If driven at the specified speed, a car on a curve will only under go centripetal acceleration around the curve. The car will not accelerate sideways up or down the curve even under icy conditions. Consider a 1600 kg car moving at a constant speed around an ice covered curve with a radius of curvature of 150 m. What angle is necessary for the bank to ensure that a car moving at 85 km/h will not slide up or down the curve? The forces and accelerations are: the normal force, the gravitational force and the centripetal acceleration (ice = no friction). +y +x n Draw FBD θ Important: The normal force is always perpendicular to the surface. mac Choose Directions: This is very important too. The direction of the centripetal acceleration is horizontally to the left. Any vertical Fg acceleration will cause the car to slide up or down the bank. Choose regular x-horizontal and y-vertical directions as shown. This way, only the normal force needs to be split into components. Define terms: m = 1600 kg, v = 85 km/h = 23.6 m/s, r = 150 m The forces on the car are Fnet Fg FN mac In the x-direction Fnet , x FNx mac Creative Commons Attribution 3.0 Unported License 26 PHYSICS SEMESTER ONE UNIT 4 The centripetal acceleration is ac v2 r 2 23.6 m/s 150 m 3.72 m/s 2 The x-component of the normal force is then FNx mac 1600 kg 3.72 m/s 2 5950 N In the y-direction, there is no acceleration so the net force is zero (no movement up or down the ramp) Fnet , y Fg FNy 0 Rearranging gives FNy Fg mg mg to account for direction = 1600 kg 9.8 m/s 2 15680 N The normal force is perpendicular to the bank so we can use the angle of the normal θ force relative to horizontal to find the angle between the surface of the road and n horizontal. tan FNx FNy tan 1 tan 1 θ FNx FNy 5950 15680 20.8 The highway must be banked at an angle of 21° to avoid sliding up or down at 85 km/h under icy conditions. Creative Commons Attribution 3.0 Unported License 27 PHYSICS SEMESTER ONE UNIT 4 SUMMARY: Newton’s Laws of Motion and Universal Gravitation Newton 1: v constant if Fnet 0 Newton 2: Fnet ma , Fnet , x max , Fnet , y ma y , Fnet , z ma z Newton 3: F12 F21 Gravity: Fg Gm1m2 r2 g gravitational force GM r2 gravitational acceleration The gravitational force and acceleration are radial, directed towards the centre of the other object. Friction Static: Fs Fs ,max s FN Kinetic: Fk k FN Equilibrium The sum of the forces on the object is zero. F ma 0 The sum of the forces in the x-direction is zero. Fx max 0 The sum of the forces in the y-direction is zero. Fy ma y 0 Free Body Diagrams Draw all forces acting on an object. Choose the coordinate directions (x and y axes) to simplify calculations. Tension - The tension is always from the end of the rope in the direction along the rope. - Most ropes are considered ideal in that have no mass. Ideal ropes they don’t stretch so the ends of the rope move with the same velocity and acceleration. - If the rope changes direction, you can modify your coordinate systems (one set for each end of the rope) so both ends have the same velocity and acceleration vectors. Creative Commons Attribution 3.0 Unported License 28 PHYSICS SEMESTER ONE UNIT 4 - A rope that goes over and ideal pulley (no mass, no moment of inertia) or a frictionless surface has the same tension magnitude at both ends. NANSLO Physics Core Units and Laboratory Experiments by the North American Network of Science Labs Online, a collaboration between WICHE, CCCS, and BCcampus is licensed under a Creative Commons Attribution 3.0 Unported License; based on a work at rwsl.nic.bc.ca. Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges. Creative Commons Attribution 3.0 Unported License 29