GeneralPhysics1-Component of Vector.pptx

Cartesian, Right Triangle

1. Components of Vectors

2. Components of a Vector Motion in two dimensions can be described by using VECTORS such as: a) FORCE (F) b) DISPLACEMENT (d) c) VELOCITY (v) d) ACCELERATION (a)

3. Each VECTORS may be broken down or resolved into two components: X – component (or horizontal component) b) Y – component (or vertical component)

4. Vector X-component Y-component Force, F F X F Y Displacement , d d x d y Velocity , v v x v y Acceleration , a a x a y Representation of Component vectors

5. The magnitude and direction of the components of a vector may be found using the concepts of: rectangular coordinate system

6. b) Trigonometric functions

7. When you draw the component of a vector you should use the tail to head connections .

8. Example: A force of 80.0 N is applied y a janitor on the handle of a mop held at 40 angle with the floor. What force is pushing the mop (a) across the floor and (b) downward to the floor Given: F = 80.0 N θ = 40 Unknown: F X = force that pushes the mop across the floor F Y = force that pushes the mop downward on the floor.

9. Given: F = 80.0 N Unknown: a) F X b) F Y 40 F=80-N F Y = ? F X = ?

10. Solutions: 40 F=80-N F Y = ? F X = ? Cos 40 = F Y / F F Y = F cos 40 = (80.0-N) (0.77) = 61.6 N Sin 40 = F X / F F X = F sin 40 = (80.0N) (0.64) = 51.20

11. An airplane travels 209 km on a straight course at an angle of 22.5 east of north. It then changes its course by moving 100 km north before reaching its destination. Determine the resultant displacement of the airplane. Given: (a) d 1 =209km, 22.5EN (b) d 2 =100km,N (c) d R Vector diagram: N (+Y) S (-Y) W (-X) E (+X) d 1 = 209 km 22.5 N (+Y) S (-Y) W (-X) E (+X) d 2 100 km

12. Finding the components of d 1x N (+Y) S (-Y) W (-X) E (+X) d 1 =209km 22.5 d 1x d 1y Θ = 90 – 22.5 = 67.5 Cos 67.5 = Adjacent side hypotenuse 0.383 = d 1x d 1 0.383 = d 1x 209 km d 1x = (209 km) (0.383) d 1x = 80.0 km

13. Finding the components of d 1y N (+Y) S (-Y) W (-X) E (+X) d 1 =209km 22.5 d 1y =193km d 1y = ? Θ = 90 – 22.5 = 67.5 sin 67.5 = opposite side hypotenuse 0.924 = d 1y d 1 0.924 = d 1y 209 km d 1y = (209 km) (0.924) d 1y = 193 km d 1x = ?

14. N (+Y) E (+X) d 1 =209km 22.5 d 1x = 80.0km d 1y =193.0km Finding the components of d 2 N (+Y) S (-Y) E (+X) W (-X) d 2y = 100.0km Displacement X-component Y-component d 1 + 80.0 km +193.0 km d 2 +100.0 km Σ Σ d x = 80.0 km Σ d y = 293.0 km

15. Displacement X-component Y-component d 1 + 80.0 km +193.0 km d 2 +100.0 km Σ Σ d x = 80.0 km Σ d y = 293.0 km N (+Y) S (-Y) W (-X) E(+X) Σd x = 80.0 km Σd y = 293.0 km d R By the Pythagorean Theorem d R = √ ( d x ) 2 + ( d y ) 2 = 80.0km 2 + 293.0km 2 = 6,400 km + 85,849 km = √ 92,249 km = 304 km =304km θ = 74.7 tan θ = Σ d y ÷ Σ d x tan θ = 293.0 ÷ 80.0 tan θ = 3.66 θ = 74.7 Shift –tan θ =

16. Practice Exercise A man exerts a force of 200 N on a wheelbarrow at an angle of 30 above the horizontal. What are the horizontal and vertical forces exerted on the handle of the wheelbarrow? A`plane flying due north at 100m/s is blown by a 50 m/s strong wind due east. a. Draw a diagram to illustrate the problem b. What is the plane’s resultant velocity? 3. Find the components of the following vectors (present the data in tabular form) a. The initial velocity of a golf is 10 m/s, 32 north of west. b. A force of 120 N pulls a loaded box 40 south of west.