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CHAPTER - 7 Forces and Motion in Two Dimensions

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1 CHAPTER - 7 Forces and Motion in Two Dimensions
7.1 Forces in Two Dimensions Equilibrium and the Equilibrant

2 Equilibrium = An object is in equilibrium when the sum of all the forces acting on an object is zero.

3 Equilibrium Cont. When an object is in equilibrium NO acceleration can occur b/c there are NO net forces (Fnet) acting on the object.  The object either remains at rest or continues to move with the same velocity (speed and direction)

4 What is the sum of vectors A, B, & C.
Sum = Resultant SKETCH on W.B. or (Transparency) The vectors can be moved to any position as long as their magnitude and orientation (direction) remain the same.

5 Equilibrant = A force, that when added to two or more forces, produces equilibrium in the system.  The equilibrant is equal in magnitude to the resultant, but opposite in direction. Fnet = ∴ 0 = F1 + F2 + F3 + … + FE Example on board.

6 Example #1 A 200N rock is attached to two ropes, each rope makes a 36.7o angle with the horizontal, the rock does not move. What is the tension in the ropes?

7 Draw sketch & FBD Fnet = 0 Fnet in the “x” direction (Fnetx) = 0 Fnet in the “y” direction (Fnety) = 0 Forces in the “X” Direction Fnetx = 0 Fnetx = -F1x + F2x 0 = -F1cos + F2cos F1cos36.7o = F2cos36.7o  F1 = F2

8 Forces in the “Y” Direction
Fnety = 0 Fnety = F1y + F2y -200N 0 = F1sin + F2sin - 200N 200N = F1sin + F2sin Since F1 = F2 substitute F1 for F2  200N = F1sin + F1sin 200N = 2F1sin

9 200N = 2F1sin36.7o 100N = F1sin36.7o 100N / sin36.7o = F1 100N / = F1 167.3N = F1 F1 = F2 = 167.3N

10 Example; Handout of object on an incline plane. Draw the FBD.

11 Motion Along an Incline Plane
Book Example page “Handout” 1. Create a coordinate system for reference, the x-axis should be parallel to the surface of the plane. 2. Label the coordinate system and indicate all forces acting on the object with the values.

12 Example #2 “Skier on Slope”
A 97.6kg skier is traveling down a ski slope. The slope has a 49o angle and the coefficient of kinetic friction, μk, is Beginning from rest, how fast is the skier going after 6.7 seconds.

13 Draw sketch and FBD List table of known/unknown μk = m = 97.6kg  = 49o vi = 0 vf = ? t = 6.7s a = ?

14 Equation = ???????? From chapter-5 v = vi + at v = 0 + (__)(6.7s) Solve for acceleration (a = ?) Solve on Board and/or Power Point

15 “y” Direction Fnety = 0 FN – Fgy = 0
FN = Fgy which = adjacent side of triangle ADJ = (HYP)cos ADJ = (mg)cos FN = Fgy = mgcos FN = mgcos (will be substituted in “x” dir)

16 “x” Direction Fnetx = max Fgx – Ff = max opp – μkFN = max
mgsin - μkmgcos = max gsin - μkgcos = ax g(sin - μkcos) = ax

17 g(sin - μkcos) = ax 9.8m/s2[sin49o – (.1739)(cos49o)] = ax 9.8m/s2[.7547 – (.1739)(.6560)] = ax 9.8m/s2[ ] = ax 9.8m/s2(.6406) =ax 6.28m/s2 = ax Substitute ax value into original equation

18 Original Equation v = vi + at v = 0 + (6.28m/s2)(6.7s) v = m/s

19 7.2 Projectile Motion Independence of Motion in Two Directions

20 Projectile = 1) Any object sent through the air by the application of some force. 2) Any object that is shot or thrown at some positive, upward angle .

21 Which ball hits the ground first?
One is dropped, the other thrown with a horizontal velocity of 100m/s.

22

23 Hits ground at same time
If you would throw an object exactly parallel to the Earth's surface, the sideways motion of the object would have no effect on how gravity acts on it. In other words, the object would drop at the same rate as an object dropped from the identical height. The time it would take either object to hit the ground would be the same.

24 Drop a bullet vs shoot a bullet, which will hit the ground first?

25 Hits ground at same time
If you would throw an object exactly parallel to the Earth's surface, the sideways motion of the object would have no effect on how gravity acts on it. In other words, the object would drop at the same rate as an object dropped from the identical height. The time it would take either object to hit the ground would be the same.

26 White Board Activity Draw the velocity (motion) diagram for a ball dropped and thrown horizontally at 50m/s. Disregard air resistance. Draw the velocity diagram for five consecutive and equal time intervals.

27 Projectiles Launched Horizontally
A projectile launched horizontally has NO initial velocity in the vertical direction.  Its motion is identical to a dropped object in the vertical direction.

28 EQUATIONS: “Y” Direction
Vy = -gt y = yi + vyit – 1/2gt2 Vy = velocity in the “y” direction g = acceleration due to gravity (9.8m/s2) y = final vertical distance yi = initial vertical distance vyi = initial velocity in the “y” direction t = time in seconds

29 EQUATIONS: “X” Direction
x = xi + vxit x = horizontal distance xi = initial horizontal distance t = time in seconds vxi = initial horizontal velocity ( the horizontal velocity remains constant disregarding air resistance)

30 Example Problem Randy Johnson throws a baseball horizontally at 42m/s from a 57m high bridge. a) How far from the bridge will the ball land in the water? b) How fast is the ball moving just before it hits the water?

31 Projectiles Launched at an Angle
WHITE BOARD ACTIVITY “A” Draw the velocity (motion) diagram for a ball dropped from a height of 50m. “B” Draw the velocity diagram for a ball thrown straight up that reaches a height of 50m. “C” Draw the velocity diagram for a ball thrown at an angle from center field to home plate that reaches a height of 50m.

32 “C” center field to home plate:

33 Maximum Height = The height of the projectile when the vertical velocity is zero and the projectile has only horizontal motion. Range = The horizontal distance the projectile travels.

34 More Equations “Y” Direction vyi = visin vy = vyi – gt
y = yi + vyit – 1/2gt2 ymax = vyit – 1/2gt2

35 “X” Direction vxi = vicos vx = vxi velocity in x direction is constant disregarding air resistance x = xi + vxit

36 Projectile Range “R” vi2sin2i R = g

37 Trajectories Depend Upon the Frame of Reference
Question: Object “A” is a cannon ball that weighs 200N, object “B” is a cannon ball that weighs 100N. Both have an initial velocity of 100m/s and are shot at an angle of 21o. Which cannon ball will have the greater range?

38 Answer: They will both have the same range. WHY? An object’s R (range) is independent of an object’s mass, mass is not part of the equation.

39 Calculate the R of cannonballs 1-6 if all have an initial velocity of 59m/s.
#1 is shot at an angle of 20o #2 is shot at an angle of 40o #3 is shot at an angle of 45o #4 is shot at an angle of 50o #5 is shot at an angle of 70o #6 is shot at an angle of 90o Draw a sketch of all of the paths of the trajectories.


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