Vector Applications. Don t think there is anything new here. Let s start easy and don t be too worried about real world complications just yet!

Transcription

1 Vector Applications 3.. Displacement and velocity Don t think there is anything new here. Lets just have a go at the questions! 3.3. The triangle of forces Let s start easy and don t be too worried about real world complications just yet! No friction, no gravity, keep it D. TASK: Get Mat to stand up. Kate, exert a force on Mat. (push him) Now, Talisa, exert a different force on Mat. (push him at the same time as Kate does, but in a different direction) Does anyone else want to you have a go pushing Mat around? Given there are three or more forces on Mat, what is the net result of ALL forces? I hope you recognise that Mat just goes in one net direction, under a single RESULTANT force? Draw a vector diagram of the acting forces, and calculate the Resultant Force! Given everyone is pushing Mat What force would Mr Finney have to exert so that Mat does not move? 1

2 How would you draw Mr Finney s force? So that s the introduction to Force Vectors in Chapter 3. It s all about Resultant Forces being the addition of whatever forces are acting on an object. This is called the Triangle of Forces! And we also consider what force is needed to keep an object from moving. Clearly we are ignoring lots of physics real world problems, but that s OK, it s just Yr 11 J Lets NOT use the Cosine Rule just stick to Vector work let me know which are Cosine Questions and I will delete them from the term planner Humans learn by doing, so what are you waiting for, Stop pushing each other around and go do some practice questions!!!

3 Chapter 3.4 State of equilibrium Have you seen Newton s law on Force in Physics yet? F = ma ( Force = mass multiplied by acceleration ) Firstly lets work out the units of Force. Because Sir Isaac Newton worked this all out first, the unit of Force is in NEWTONS (N), but what exactly is a Newton? Take a look at what makes up a Newton and drill down on what exactly a Newton is. (1 Newton = 1kg per metre per second per second mass is kg and acceleration is in metres per second per second) As soon as we are working in any UNITS, we want to use our general ability to convert units (from yr 7) and change all units BEFORE we start working on the question. So, before you do any calculations, you must convert everything to Kilograms, Metres and Seconds J Under a force, a certain object will experience a certain acceleration. The smaller the object, the greater the acceleration. That s logical, think about pushing Mat with a certain force, and then pushing Talisa with that same force... Talisa would go flying, but Mat wouldn t move much. (sorry Talisa) That was a simplistic way of thinking about acting forces. In this Chapter we are talking about a constant force, giving a constant acceleration. So a push is not something we want to continue thinking about because a push is a force exerted at a particular time, rather than a force continuously acting on a body. So this chapter is Not like pushing Mat. The Forces discussed in this chapter are a constant. Consider an interstellar spacecraft, being propelled by an ion propulsion system. The space ship would have a constant force (thrust) being exerted by its engines, so it would simply get faster and faster and faster and faster and faster... Clearly this situation doesn t happen a lot around here. In a car, no matter how hard you push your right foot down, there is a maximum speed, even though you are applying a force. Clearly FRICTION is in play here! (we will ignore friction for a short time to get our head around things) 3

4 TASK Turn a table upside down. Apply a horizontal force on it. With enough force, it moves! Now someone needs to stand on the table. What force would be required to move the table now? The table surface area is constant, but clearly a larger force needs to be exerted to move the table. So, therefore, Friction is reliant on the weight of the object! Weight is a poor choice of word. Consider the same table, the same carpet, and the same weight on the table. Clearly given the same force, you would expect the table to move faster if it was on the MOON? We will come back to this. TASK Consider the table again. Push gently until it moves. Clearly Friction acts AGAINST your force. What if two or three of you push at different angles on the table. How would friction act then? (against the Resultant Force) TASK So the table on the carpet has Friction, and this Force of Friction acts against anything pushing the table. Stand back. What force is Friction exerting now? We can say that friction only acts as much as it needs to, in order to prevent movement, or can only act up to a maximum resistance force? What if you put the table onto the vinyl? Would this change the amount of force you would need to use to slide the table? It is clear that different surfaces have different frictions? Friction depends on two things the coefficient of friction and the weight of the object 1. We shall call the coefficient of friction Mu or μ pronounced as mue, and the yanks pronounce it moo. (not sure why Yanks think Cows know about the coefficient of friction!). Lets go back to the MOON situation. It s not really about the weight of the object, its about the force in between the table and the carpet that really dictates the friction level between the two surfaces. Weight is really the combination of the mass of an object and the force of acceleration on it (lets call that gravity). So weight is really a Force! 4

5 TASK lets think about how weight affects friction The difference in how hard the table was to push was due to the different weights on the table, and because the table did not have any vertical movement, we can assume the floor was able to apply an equal and opposite force to the table weight and whoever stood on it. It is this equal and opposite force acting against the weight that determines Friction. Clearly the upwards force acting on the table is acting at a Normal to the horizontal floor. So the floor is exerting a Force Vector on the table at a Normal to the surface. Lets call that Normal Force vector N. So Friction is a force too, lets call it F f (the force of friction), and is determined by the coefficient of friction and the Normal force being exerted between the two objects. F f = μn *** remember F = ma now we have another F formula, so we need to NOT get them confused, hence we denote the force of Friction as F f. *** We really need to do some diagrams on the White Board!!! This will clarify the Friction vector and the Normal vector. What about rubbing your hand down a wall. Clearly Friction happens when there is a force between any two objects. It doesn t have to be a horizontal force! Lets consider this further because there is a lot of friction happening between surfaces that are Not horizontal. TASK Now go outside and stand on the hill. In what direction is gravity acting now? I hope you know that Gravity always pulls directly Downwardly? It may be hard to feel it, but the Earth is not pushing directly upwards if you are standing on a sloped surface Although gravity is pulling you directly Downwardly, the Earth s force on you acts on your feet at a NORMAL to its surface. Just think, if the Grass had a coefficient of friction of Zero, then you would be sliding down the hill and into the bottom drain. Therefore, if gravity acted down, and the Earth just pushed directly UP against gravity, you wouldn t move, even if there is no friction. (eg, a ball wouldn t roll down a hill) 5

6 I think we should do a diagram on the white board again, as this is important. Make sure you get Mr Finney and do some diagrams to show this. So the fact that things slip down a slope is PROOF that the reactive force to gravity, from the Earth, acts at a Normal to the surface. Gravity is still the same, so our downward Weight force vector remains straight down, and since you are just standing there, all forces MUST be in balance. So, if we split up ALL Force vectors into up/down components, and sideways components (in the direction of the slope), all components must equal out. In this way we can find the component of force wanting to make us slide down the hill, and therefore, we can also see what the Friction vector is as it is just the opposite of that J Everything depends on the Mass x gravity Vector. Again we need to draw some diagrams on the white board. This is important, ensure we do some diagrams all together on the white board. We will mostly be considering the forces acting on bodies that remain stationary, where all forces are in balance! We can talk about the friction on a slope that prevents the movement of an object or we could talk about the Tension of a rope that is holding something in a stationary position or where there is a constant velocity, this is again where all forces are in balance, as there is no acceleration happening! OK so your take out is the FRICTION formula: F f = μn And, our mass x gravity vector is calculated using F = ma, where a =gravity, and the acceleration of gravity is 9.8 m/s/s W = mg 6

7 Look carefully If the question indicates the object is on a SMOOTH surface, then ignore friction, otherwise, you will need to take friction into consideration Look carefully! Another way that objects can be held still is by using a rope to keep them from slipping, or from swinging. In this case, it s not necessarily friction that is preventing the movement of the object, it is Tension on the rope, also measured in Newtons (N). It really all does revolve around you working out the Force that gravity is exerting on the object, and playing with Force Diagrams ensuring that the Resultant Force is ZERO. When ropes are involved, rather than thinking about the Normal Force acting at a normal to the surface against Gravity, you will have a Rope pulling in a certain direction. So it is essentially the same diagram strategy, but we would not label the force a N, but as T (for Tension). Strategy: Everything depends on the Mg downwardly acting vector call it vector W From this, Find your N generally N = W cos θ Find your F f generally F f = W sin θ *** Be careful with what θ is *** Then because the system is in balance, - up/down must equate - right/left must equate - or along the sloped surface each way must equate *** the book sometimes refers to i and j components for the resultant force. Don t expect that terminology in your exam as I find it quite unclear. I won t mark you down if you follow it, but I would prefer to work Logically and simply work in components that act at a normal to the surface, and along the surface, and just be clear in your setting out! *** 7

8 3.5 Relative Velocity Starting out simply You are riding your bike in perfectly calm weather. If you are riding at 5 metres per second, then the wind is hitting your face at 5 metres per second. Now, consider you are riding along a road at 5 metres per second, and you are riding into the wind which is travelling against you at metres per second. Now, the wind is hitting your face at 7 metres per second or, you are travelling through the air at 7 metres per second or, we could say your Velocity Relative to the Air is 7 metres per second! Now we have the idea lets go one step further Picture a plane flying in a crosswind. The plane is flying in the air, and the air is moving does it make sense that the ground speed is equal to the speed of the plane in the air PLUS the speed of the air? or V <=>?@ A<@@B >CDEAA FDEG?B = V <=>?@ A<@@B HIDEGI >JD + V >JD A<@@B >CDEAA FDEG?B V <=>?@ D@=>HJL@ HE FDEG?B = V <=>?@ D@=>HJL@ HE >JD + V >JD D@=>HJL@ HE FDEG?B Lets substitute p=plane a=air g=ground and the formula becomes V </F = V </> + V >/F and we can also see the formula abstractly in the general case, as: V </M = V </N + V N/M Read as Velocity of a particle (p) relative to an object A is equal to the velocity of the particle (p) relative to object B, plus the velocity of object B relative to object A. The book gives a different arrangement of the formula, but I like mine better, because I can think of it as a plane, and it all just makes sense! 8

9 Getting this correct can be trickier than you think Have a play with the River boat problem Projectiles/Riverboat-Simulator/Riverboat-Simulator-Interactive Now let s do another easy example before we move on. Think of a bike rider travelling at m/s north, consider that the wind is blowing from the north east at 3 m/s. What is the velocity of the wind through the bike rider s hair? Establish your variables : Rule: V O/F = bike velocity over ground = 0, V O/> = bike velocity through air =? V >/F = wind velocity over ground =.1,.1 V </M = V </N + V N/M In this case our particle is the bike, so p à bike, a à air our formula in this case becomes:, g à ground, so V O/F = V O/> + V >/F by substitution: 0, = V O/> +.1,.1 V O/> = 0,.1,.1 V OJh@ L@=ECJHi HIDEGFI >JD =.1, 4.1 velocity is the magnitude of the vector, so, and if needed; velocity = V = k.1 l l = 4.63 θ = tan qr = 6.77t 9

10 Some of these are basic right angle triangles, so there s lots of simple trig and Pythagoras in here isn t there J The text book also uses the Sine or Cosine rule but I prefer my way that doesn t need those rules! The Specialist Mathematics Syllabus wants us to use VECTOR TECHNIQUES, not sine and cosine rules! 10

11 When you get to Q4, it is different and you will want to look at the text book worked example tip: don t! You could also do this intuitively but again, tip: don t! This is Specialist Mathematics, and you need to use your Vector Techniques to solve! v v>h@d EL@D FDEG?B = v v/f = 6,0 v OE>H EL@D FDEG?B = v O/F = 0, k *** are you wondering where the k came from the horizontal component must be zero as the boat has to head directly across the river, and our northerly component is?, so set to k *** v OE>H EL@D v>h@d = v O/v = 0 cos θ, 0 sin θ OK, so you can try to think conceptually about how this goes into the relative velocity formula, but all we need to think about is the subscript pattern and it just falls into place! and substitute: v O/F = v O/v + v v/f 0, k = 0 cos θ, 0 sin θ + 6,0 0, k = 0 cos θ + 6, 0 sin θ and from here we can Equate terms to get: 0 = 0 cos θ + 6 cos θ = 6 0 θ = 7.54 E So, yes you can get this solution just using some common sense and some trig but that won t get you too far in the exam please ensure you implement Vector techniques in your solutions J The question doesn t ask it, but we could then go on to find how fast it travels directly across the river as that is what the k represents J 11

12 Question 7, 8 and 9 in the chapter are quite nice they all use a similar technique I know there are solutions for the text book, but here is how I would look at Question 8. Consider North journey v OJh@/FDEG?B = 15j v vj?b/ojh@ = A cos 3π 4 i + A sin 3π 4 j Consider South journey v O/F = 15j v v/o = B cos 5π 6 i + B sin 5π 6 j v v/o = 1 1 A i A j v v/o = 3 B i + 1 B j Formula v v/f = v v/o + v O/F Formula v v/f = v v/o + v O/F and v v/f = 1 A i 1 A j + 15 j v v/f = 1 1 A i + }15 A~ j v v/f = 3 B i + 1 B j 15 j and v v/f = 3 B i + }1 B 15~ j Clearly, Equate x components: and Equate y components: v v/f = v v/f 1 3 A = B A = 6 B }15 1 A~ = }1 B 15~ substitute in A, to get: B = 1 B B = 1 B B = B 30 3B B = B = B = B =

13 ** We only need to find A or B so now we have found B, we are all done! Just use B in our southern journey to find the Wind Vector From southern journey, sub B into v ƒ formula becomes v v/f = 3 B i + }1 B 15~ j v v/f = i + } ~ j there s some nice manual algebra in there. Remember, if you use your calculator in the exam, I can t give you full marks. But if you need to, its better to get part marks, than none at all J So use your calculator WISELY! And you should end up with: v v/f = i 15 3 j Magnitude, just use the Pythagoras formula For the direction, use tan qr θ = i Clearly with these numbers you will not get an Exact angle, but sometimes you will J OK it s possibly too easy to just go do this question NOW while the solution is so fresh in your mind, so make sure you write a note and do it again yourself at some later stage! Humans learn by doing so go DO! 13