Vectors

1. Vectors • Vector Operations • Components • Trig Applications

2. Objectives • I can add vector using the head to tail method. • I can subtract vectors using the head to head method. • I can use trigonometric functions to find the components of a vector. • I add components of vectors to find a resultant vector.

3. Vector Addition • Head to tail method • Parallelogram method Suppose 3 forces act on an object at the same time. Fnet is not 15 N because these forces aren’t working together. But they’re not completely opposing each either. So how do find Fnet ? The answer is to add the vectors ... not their magnitudes, but the vectors themselves. There are two basic ways to add vectors w/ pictures: 8 N 4 N 3 N

4. Tip to Tail Methodin-line examples 20 N 12 N Place the tail of one vector at the head of the other. The vector sum (also called the resultant) is shown in red. It starts where the black vector began and goes to the head of the blue one. In these cases, the vector sum represents the net force. You can only add or subtract magnitudes when the vectors are in-line! 16 N 9 N 20 N 16 N 21 N 4 N 9 N 12 N

5. Head to Tail – 2 Vectors To add the red and blue displacement vectors first note: • Vectors can only be added if they are of the same quantity—in this case, displacement. 5 m 2 m Place the vectors head to tail and draw a vector from the tail of the first to the head of the second. Interpretation: Walking 5 m in the direction of the blue vector and then 2 m in the direction of the red one is equivalent to walking in the direction of the black vector. The displacement walked this way is the black vector’s magnitude. 2 m 5 m blue + red

6. Commutative Property red + blue blue + red As with scalars quantities and ordinary numbers, the order of addition is irrelevant with vectors. Note that the resultant (black vector) is the same magnitude and direction in each case. (We’ll learn how to find the resultant’s magnitude soon.)

7. Head to Tail – 3 Vectors We can add 3 or more vectors by placing them head to tail in any order, so long as they are of the same type (force, velocity, displacement, etc.). blue + green + red

8. Parallelogram Method This time we’ll add red & blue by placing the tails together and drawing a parallelogram with dotted lines. The resultant’s tail is at the same point as the other tails. It’s tip is at the intersection of the dotted lines. Note: Opposite sides of a parallelogram are congruent.

9. Comparison of Methods red + blue The resultant has the same magnitude and direction regardless of the method used. Tip to tail method Parallelogram method

10. Opposite of a Vector If vis 17 m/s up and to the right, then -v is 17 m/s down and to the left. The directions are opposite; the magnitudes are the same. v - v

11. Vector Subtraction Put vector tails together and complete the triangle, pointing to the vector that “comes first in the subtraction.” red - blue Why it works: In the first diagram, blue and black are tip to tail, so blue + black = red  red – blue = black. blue - red Note that red - blue is the opposite of blue - red.

12. Vector Components A 150 N force is exerted up and to the right. This force can be thought of as two separate forces working together, one to the right, and the other up. These components are perpendicular to each other. Note that the vector sum of the components is the original vector (green + red = black). The components can also be drawn like this: Vertical component 150 N Horizontal component

13. Finding Components with Trig Multiply the magnitude of the original vector by the sine & cosine of the angle made with the given. The units of the components are the same as the units for the original vector. Here’s the correspondence: cosine  adjacent side sine  opposite side v vsin  vcos

14. Component Example 30.814 m/s 25 14.369 m/s 34 m/s A helicopter is flying at 34 m/s at 25 S of W (south of west). The magnitude of the horizontal component is 34cos 25  30.814 m/s. This is how fast the copter is traveling to the west. The magnitude of the vertical component is 34 sin 25  14.369 m/s. This is how fast it’s moving to the south.

15. Pythagorean Theorem 30.814 m/s 25 14.369 m/s 34 m/s Since components always form a right triangle, the Pythagorean theorem holds: (14.369)2 + (30.814)2 = (34)2. Note that a component can be as long, but no longer than, the vector itself. This is because the sides of a right triangle can’t be longer than the hypotenuse.

16. Component Form Instead of a magnitude and an angle, vectors are often specified by listing their horizontal and vertical components. For example, consider this acceleration vector: a= 10 m/s2 at 53.13 N of W In component form: a= -3, 4m/s2 4m/s2 a = 10 m/s2 53.13 3 m/s2

17. Component Form Some books use parentheses rather than angle brackets. The vector F = 2, -1, 3N indicates a force that is a combination of 2 N to the east, 1 N south, and 3 N up. Its magnitude is found w/ the Pythag. theorem:F = [22 + (-1)2 + 32]1/2 = 3.742 N

18. Finding the direction of a vector x = 5, -2 meters isclearly a position to the southeast of a given reference point. If the reference pt. is the origin, then x is in the 4th quadrant. The tangent of the angle relative to the east is given by: tan =2 m/5 m   = tan-1(0.4) = 21.801 The magnitude of x is (25 + 4)1/2 = 5.385 m.Thus, the vector is 5.385 m at 21.801 S of E. 5 m  2 m

19. 2 N 4 N F2 F1 7 N 3 N Adding vectors in component form IfF1 = 3, 7N and F2 = 2, -4N, then the net force issimply given by: Fnet = 5, 3N. Just add the horizontal and vertical components separately. F2 F1 Fnet

21. Other component pairs vcos vcos   vcos  vsin vsin v v v vsin There are an infinite number of component pairs into which a vector can be split. Note that green + red = black in all 3 diagrams, and that green and red are always perpendicular. The angle is different in each diagram, as well as the lengths of the components, but the Pythagorean theorem holds for each. The pair of components used depends on the geometry of the problem.

23. Scalar Multiplication Scalar multiplication means multiplying a vector by a real number, such as 8.6. The result is a parallel vector of a different length. If the scalar is positive, the direction doesn’t change. If it’s negative, the direction is exactly opposite. x 3x -2x Blue is 3 times longer than red in the same direction. Black is half as long as red. Green is twice as long as red in the opposite direction. x ½

24. Other Operations • Vectors are not multiplied, at least not the way numbers are, but there are two types of vector products that will be explained later. • Cross product • Dot product • These products are different than scalar mult. • There is no such thing as division of vectors • Vectors can be divided by scalars. • Dividing by a scalar is the same as multiplying by its reciprocal.

25. Comparison of Vectors Which vector is bigger? 43 m 15 N 27 m/s 0.056 km The question of size here doesn’t make sense. It’s like asking, “What’s bigger, an hour or a gallon?” You can only compare vectors if they are of the same quantity. Here, red’s magnitude is greater than blue’s, since 0.056 km = 56 m > 43 m, so red must be drawn longer than blue, but these are the only two we can compare.

28. Vectors • Vector Operations • Components • Inclined Planes • Equilibrium • 2-D Force & Motion Problems • Trig Applications • Relative Velocities • Free Body Diagrams

29. Inclined Plane A crate of chop suey of mass m is setting on a ramp with angle of inclination . The weight vector is straight down. The parallel component (blue) acts parallel to the ramp and is the component of the weight pulling the crate down the ramp. The perpendicular component (red) acts perpendicular to the ramp and is the component of the weight that tries to crush the ramp. m parallel component  Note: red + blue = black perpendicular component continued on next slide mg

30. Inclined Plane (continued) The diagram contains two right triangles.  is the angle between black and blue.  +  = 90 since they are both angles of the right triangle on the right. Since blue and red are perpendicular, the angle between red and black must also be . Imagine the parallel component sliding down (dotted blue) to form a right triangle. Being opposite , we use sine. Red is adjacent to , so we use cosine. m mgsin    mgcos mg continued on next slide mgsin

31. Inclined Plane (continued) m mgsin  mgcos mg The diagram does not represent 3 different forces are acting on the chop suey at the same time. All 3 acting together at one time would double the weight, since the components add up to another weight vector. Either work with mgalone or work with both components together.

32. How the incline affects the components m mgsin m mgsin mgcos mgcos mg mg The steeper the incline, the greater  is, and the greater sin is. Thus, a steep incline means a large parallel component and a small horizontal one. Conversely, a gradual incline means a large horizontal component and a small vertical one. Extreme cases: When  = 0, the ramp is flat; red = mg; blue = 0.When  = 90, the ramp is vertical; red = 0; blue = mg.

33. Inclined Plane - Pythagorean Theorem m mgsin  Let’s show that the Pythagorean theorem holds for components on the inclined plane: mgcos mg (mgsin)2 + (mgcos)2 = (mg)2 (sin2 + cos2) = (mg)2 (1) = (mg)2

34. Inclined Plane: Normal Force Recall normal force is perpen-dicular to the contact surface. As long as the ramp itself isn’t accelerating and no other forces are lifting the box off the ramp or pushing it into the ramp, N matches the perpendicular component of the weight. This must be the case, otherwise the box would be accelerating in the direction of red or green. N > mgcos would mean the box is jumping off the ramp. N < mgcos would mean that the ramp is being crushed. N = mgcos mgsin m  mgcos mg

35. Net Force on a Frictionless Inclined Plane With no friction, Fnet = mg + N= mgcos + mgsin + N= mgsin. (mgcos +N= 0 since their magnitudes are equal but they’re in directions opposite. That is, the perpendicular component of the weight and the normal cancel out.) N =mgcos mgsin m  mgcos Therefore, the net force is the parallel force in this case. mg

36. Acceleration on a Frictionless Ramp Here Fnet = mgsin = ma. So, a = gsin. Since sin has no units, ahas the same units as g, as they should. Both the net force and the acceleration are down the ramp. m mgsin  mgcos mg

37. Incline with friction at equilibrium At equilibrium Fnet = 0, so all forces must cancel out. Here, the normal force cancels the perpendicular component of the weight, and the static frictional force cancels the parallel component of the weight. N = mgcos fs = mgsin m mgsin  mgcos mg continued on next slide

38. Incline with friction at equilibrium(cont.) fs sN = smgcos. Also,fs = mgsin (only because we have equilibrium). So, mgsin  smgcos.Since the mg’s cancel and tan  = sin/cos, we have s  tan. N = mgcos fs = mgsin m mgsin  mgcos mg continued on next slide

39. Incline with friction at equilibrium(cont.) Suppose we slowly crank up the angle, gradually making the ramp steeper and steeper, until the box is just about to budge. At this angle, fs = fs,max = s N = smgcos . So now we havemgsin = smg cos, ands = tan. N = mgcos fs = mgsin mgsin  (Neither of these quantities have units.) mgcos An adjustable ramp is a convenient way to find the coefficient of static friction between two materials. mg

40. Acceleration on a ramp with friction In order for the box to budge, mgsinmust be greater than fs,max which means tanmust be greater thans. If this is the case, forget about sand use k. fk = kN = kmgcos.Fnet = mgsin - fk = ma. So, mgsin - kmgcos = ma. The m’s cancel, which means a is independent of the size of the box. Solving forawe get: a = gsin - kgcos. Once again, the units work out right. N = mgcos fk = kmgcos mgsin  mgcos mg

41. Parallel applied force on ramp In this case FAand mgsin are working together against friction. Assuming FA + mgsin> fs,maxthe box budges and the 2nd Law tells us FA + mgsin - fk = ma.Mass does not cancel out this time. N fk FA mgsin  If FAwere directed up the ramp, we’d have acceleration up or down the ramp depending on the size of FA compared to mgsin. If FAwere bigger, friction acts down the ramp and a is up the ramp. mgcos mg

42. Non-parallel applied force on ramp Suppose the applied force acts on the box, at an angle  above the horizontal, rather than parallel to the ramp. We must resolve FA into parallel and perpendicular components (orange and gray) using the angle +. FAserves to increase acceleration directly and indirectly: directly by orange pulling the box down the ramp, and indirectly by gray lightening the contact force with the ramp (thereby reducing friction). FAsin(+) FA N fk   FAcos(+) mgsin  mg mgcos continued on next slide

43. Non-parallel applied force on ramp (cont.) FA sin ( + ) Because of the perp. comp. of FA, N < mgcos. Assuming FAsin(+) is not big enough to lift the box off the ramp, there is no acceleration in the perpendicular direction. So, FAsin(+) + N = mgcos. Remember, Nis what a scale would read if placed under the box, and a scale reads less if a force lifts up on the box. So, N = mgcos - FAsin(+), which means fk = kN = k[mgcos - FAsin(+)]. FA N fk   FAcos(+) mgsin  mgcos mg continued on next slide

44. Non-parallel applied force on ramp(cont.) Assuming the combined force of orange and blue is enough to budge the box, we have Fnet= orange + blue - brown = ma. FAsin(+) FA N fk   FAcos(+) mgsin  Substituting, we haveFAcos(+) + mgsin-k[mgcos - FA sin(+)] = ma. mgcos mg

45. Hanging Sign Problem Support Beam T2 T1 1 2 mg The Hanging Sign Problem continued on next slide

46. Hanging sign f.b.d. T2 T1 1 2 mg Free Body Diagram Since the sign is not accelerating in any direction, it’s in equilibrium. Since it’s not moving either, we call it Static Equilibrium. Thus,red+green+ black = 0. continued on next slide

47. Hanging sign force triangle T2 mg T1 Fnet= 0 means a closed vector polygon ! As long as Fnet = 0, this is true no matter many forces are involved. Vector Equation: T1+ T2 + mg = 0 continued on next slide

48. Hanging sign equations Components & Scalar Equations T2 T1 T2sin2 T1sin1 1 2 T1cos1 T2cos2 Horizontal: T1cos1=T2cos2 mg Vertical: T1sin1+ T2sin2 = mg We use Newton’s 2nd Law twice, once in each dimension:

49. Hanging sign sample Support Beam T1 T2 35 62 75 kg Sample Problem Answers: T1 = 347.65 N T2 = 606.60 N Accurately draw all vectors and find T1 & T2.

50. Vector Force Lab Simulation Go to the link below. This is not exactly the same as the hanging sign problem, but it is static equilibrium with three forces.Equilibrium link • Change the strengths of the three forces (left, right, and below) to any values you choose. (The program won’t allow a change that is physically impossible.) • Record the angles that are displayed below the forces. They are measured from the vertical. • Using the angles given and the blue and red tensions, do the math to prove that the computer program really is displaying a system in equilibrium. • Now click on the Parallelogram of Forces box and write a clear explanation of what is being displayed and why.