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Vectors Part II Ch. 6.

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Presentation on theme: "Vectors Part II Ch. 6."— Presentation transcript:

1 Vectors Part II Ch. 6

2 Introduction Although we already talked a little about vectors, there’s still quite a bit to learn about them (I’m sorry ). By studying the vector nature of forces, the concept of equilibrium will be developed. This study of the forces acting on an object that is in equilibrium is called statics.

3 6:1 Review In adding 2 vectors, place the tail of one vector at the head of the other vector. Next, add a third vector going from the tail of the first vector to the head of the 2nd vector. This is the resultant. The direction is expressed as an angle measured clockwise from north (0°).

4 6:2 Independence of Vector Quantities
Vectors act independently. One vector does not change another. Example: A motorboat heads due east at 8.0 m/s across a river flowing due south at 5.0 m/s. The boat will travel 8.0 meters due east in 1 second. It will also travel 5.0 meters due south in that same second. These velocities cannot change one another. What is the resultant vector? When the vectors are added together, the resultant velocity is: This means the same thing as saying the boat traveled east 8.0 m/s and south 5.0 m/s at the same time.

5 6:3 Vector Addition on Forces
Force vectors are added in the same way as velocity vectors. Forces that act on the same point at the same time are called concurrent forces. Example: A force of 45 N and a force of 65 N act concurrently on point P. The smaller force acts in the direction of 30° and the larger force acts at 90 °. Draw the original, then move the vectors to make a triangle. You may have to do the measuring of this with a protractor… 

6 Guided Practice/HW: Let’s do #1-4, 6 on pg. 97 and pg. 98, # 9.

7 6:4 Vector Addition- Mathematical Method
The Pythagorean Theorem can be used to find the length of the 3rd side of a right triangle. Review: a2 + b2 = c2. Example: Consider an airplane that is flying due east at 90.0 km/h and is being blown north at 50.0 km/h. What is the resultant velocity of the plane?

8 Non-right triangles: The addition of vectors doesn’t always result in the creation of right triangles. If this is the case, you must use the Law of Cosines and the Law of Sines.

9 Law of Cosines: When we know two sides of a triangle and their included angle, then the Law of Cosines enables us to find the third side. Thus, if we know sides a and b and their included angle, then the Law of Cosines states: c2 = a2 + b2 – 2ab cos C

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11 Law of Sines: Allows us to solve triangles that are not right-angled, and are called oblique triangles. The sides of a triangle are to one another in the same ratio as the sines of their opposite angles.

12 Practice! #14-17 on pg. 100

13 6:5 Addition of Several Vectors
Quite often, 3 or more vectors act concurrently on the same point. You follow the same procedure as with adding 2 vectors.

14 6:6 Equilibrium Sometimes, the sum of these forces equals 0.
If this occurs, the object is in equilibrium and the object has no acceleration.

15 6:7 The Equilibrant An equilibrant force balances the forces acting on a point to produce equilibrium. The equilibrant is equal in magnitude to the resultant, but opposite in direction. The equilibrant is the force that is equal and opposite to the resultant. It produces equilibrium. Equilibrium does not mean that there is no motion! There is no CHANGE in motion.

16 Practice! Pg. 102, #20-22

17 6:8 Perpendicular Components of Vectors
As we’ve seen in the past, it is possible to regard a single vector quantity as the resultant of 2 vectors, each acting in directions other than the original vector. These vectors are called components. The process of finding the effective value of a component in a given direction is called vector resolution.

18 Example: A 58-N force is being exerted on a rope held at an angle of 30 degrees with the horizontal. What are the vertical and pulling forces being exerted on the rope?

19 One thing that we find is that as the direction of the force changes, the magnitude of the components also change. In adding non-perpendicular vectors, first find the components of those vectors. Then add the vertical and horizontal components separately. Finally combine the sums vectorally to form the resultant vectors.

20 Example: Resolving a Velocity Vector into its Components
A wind with a velocity of 40.0 km/h blows at 60.0 degrees. What’s the North component of the wind’s velocity? What is the East component of the wind’s velocity?

21 Practice! Pg. 105: #25-27

22 6:9 Gravitational Force and Inclined Planes
The gravitational force acting on an incline is directed toward the center of the Earth. This means that the object’s weight, W, must act perpendicular to the surface of the Earth. The weight of the trunk FW can be resolved into 2 components that are perpendicular to each other.

23 The parallel force and force of friction are parallel to the surface
The parallel force and force of friction are parallel to the surface. The weight goes straight down to the earth’s surface. The perpendicular force is perpendicular to the surface of the incline. The normal force is numerically equal to the perpendicular force.

24 Example: A trunk weighing 562 N is resting on a plane inclined at 30.0 degrees from the horizontal. Find the magnitude of the parallel and perpendicular components of the weight.

25 Practice! Pg. 107: #30, 32

26 6:10 Nonperpendicular Components of Vectors
A vector doesn’t always have to be resolved into 2 components at right angles. It can be resolved into components that lie in any direction as long as their vector sum is equal to the original vector. The direction of the component depends on the direction of the object exerting the force.

27 Example A sign weighing 40 N is supported by ropes A and B.
There are 3 forces acting on this sign. Force in rope A, force in rope B, and the weight of the sign. Because the sign is in equilibrium, the forces in the 2 ropes must produce a resultant of 40 N straight up to balance the weight of the sign. This makes an equilateral triangle, so the force on ropes A and B must also be 40 N.


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