1. Glencoe Physics Chapter 7 Forces and Motion in Two Dimensions

2. Objectives 7.1
Determine the force that produces equilibrium when three forces act on an object
Analyze the motion of an object on an inclined plane with and without friction

3. Objectives 7.2
Recognize that the vertical and horizontal motions of a projectile are independent
Relate the height, time in air (hang time), and initial vertical velocity of a projectile using its vertical motion , then determine the range
Explain how the shape of the trajectory of a moving object depends upon the frame of reference from which it is observed

4. Objectives 7.3
Explain the acceleration of an object moving in a circle at a constant speed
Describe how centripetal acceleration depends upon the object’s speed and the radius of the circle
Recognize the direction of the force that causes centripetal acceleration
Explain how the rate of circular motion is changed by exerting torque on it

5. 7.1 Forces in Two Dimensions

6. We have already learned about how to find the resultant force of two forces acting on an object, as well as 3 or more forces.
What we want to consider now are cases where Net Force acting on an object is zero, and the object is in Static equilibrium.

7. Static Equilibrium Condition of an object when net forces equal zero
Object is motionless

8. Hanging sign f.b.d. 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. 2 T2 1 T1 mg
Thus, red + green + black = 0.

9. Equilibrant
If you remove one force from a static condition, the system is no longer in static equilibrium
The removed force is called the EQUILIBRANT.
The equilibrant is the force, that if added to other forces, will bring the other forces into static equilibrium.
Equilibrant is always equal to, but opposite, the sum of all the other force vectors

10. FT1
FT2 mg
What force represents the equilibrant of the tensions of the strings?

11. Example problem
What is the tension in the ropes holding the mass in place?
22.5°
Θ = 22.5°
m= 168 N
168 N

12. Solution Fnet =0 (system is in equilibrium) Fax = -Fbx
Fay + Fby – mg = 0
FaSin 22.5° + FbSin 22.5° = 168 N
2Fa Sin 22.5 = 168 N
Fa = 220 N

13. Weight of the Picture? What is the weight of this picture?
Weight is equal to F1y + F2y
So…. 25N + 25 N = 50 N
What is the weight of this picture?

14. If in Equilibrium……..the following would be true
Components & Scalar Equations T2 mg T1 1 2
T1 sin 1
(y component)
T2 sin 2
(Y component)
If in Equilibrium……..the following would be true
T1 cos 1
(x component)
T2 cos 2
(X component)
Vertical:
T1 sin 1 + T2 sin 2 = mg
T1 cos 1 = T2 cos 2
Horizontal:

15. Sample Problem
A mother and daughter are outside playing on the swings. The mother pulls the daughter and swing (total mass 55.0 kg) back so that the swing makes an angle of 40.0° with the vertical (50.0 ° from horizontal)
A. What is the tension in each chain holding the swing seat and the daughter?
B. How hard did the mother have to pull to hold the daughter at that position?
A. 703N
B. 452N

16. How to budge a stubborn mule
It would be pretty difficult to budge this mule by pulling directly on his collar. But it would be relatively easy to budge him by tying the rope to a tree and then pulling up (or pushing sideways) in the middle. Why would this work?????
The tension in the rope has two components, one horizontal and one “vertical” FT
Little Force
FTx
Big Force

17. Sample Problem… Is this a case of equilibrium?
45°
40 °
150N
160N
75N
Is this a case of equilibrium?
Calculate the magnitude of the net force

18. Equilibrium or Motion… along an Inclined Plane
Is the “sled” on the inclined plane in equilibrium?
What are the forces acting on the sled?
Draw a Free Body Diagram of the forces

19. Inclined Plane  m parallel component Fgx perpendicular component Fgymg
perpendicular component Fgy
Note: red + blue = black 

20. Inclined Plane )     m mg mg Sin  mg Cos  m mg mg Sin  mg Cos 
Be careful what  you use in your calculation…..remember you must use the angle, as measured from (+X) (East)
Fgx=
Fgy=
This calculation will look different than what is in your book (p. 152)…but is easier to use if you remember the above caution!!!!!

21. Fn on block Fgx The next 3 slides show the effect
of increasing the angle.
Fgx

22. Fn on block
F gx
Fg on block
CD 5.3 p-22
CD 5.3 p 21

23. Fn on block
F gx
Fg on block
CD 5.3 p-22
CD 5.3 p 21

24. Fe on block A B C Speed increases, acceleration:
acceleration greatest at:
net force greatest at: A A A
decreases B C
Fe on block
CD 5.3 p 22

25. Example p. 153 of your book….
A 62 kg person on skis is going down a hill sloped at 37°. The coefficient of kinetic friction between the skis and the snow is 0.15.
A) What is the Horizontal component of the skier’s weight?
B) What is the Vertical component of the skier’s weight?
C) What is the Normal force acting on the skier?
D) What is the Frictional force acting on the skier?
E) What is the Net Force acting on the skier?
F) What is the acceleration of the skier?
G) How fast is the skier traveling after 5.0s, starting from

26. 7.2 Projectile Motion in Two Dimensions:

28. Fountain at ¡Explora! Science Museum, Albuquerque, NM

29. What is Projectile Motion?
Any object moving horizontally, with only the force of gravity acting on it, will exhibit “projectile motion”. (not moving straight up or down)
An object displaying projectile motion will follow a “trajectory”, or a curved, parabolic-shaped path.

30. Projectile Motion
At a given location on the earth and in the absence of air resistance, all objects fall with the same uniform acceleration. Thus, two objects of different masses, dropped from the same height, will hit the ground at the same time.
An object’s horizontal and vertical motion are independent. Object projected horizontally will reach the ground in the same time as an object dropped vertically. No matter how large the horizontal velocity is, the downward pull of gravity is always the same.

31. Why does a projectile follow a parabolic path?
What happens to the horizontal displacement during each successive time period?
What happens to the vertical displacement during each successive time period?

32. Horizontal motion….

33. …+ Vertical = parabolic path

36. Horizontal Vs. Vertical Velocities
We can see from the diagrams that the horizontal velocity of a projectile remains constant (with no air resistance)
We can also see that that vertical velocity is constantly changing, at a rate of 9.8 m/s2.

37. Horizontal Vs. Vertical Velocities
The horizontal and vertical velocities are also independent of each other, that is, they have no effect on each other.
The only thing the horizontal and vertical motions have in common is that each motion takes exactly the same time to occur.

38. Equations of Projectile Motion

39. Working Projectile motion problems…
Separate the problem into two problems…vertical motion and horizontal motion
Vertical motion is exactly that of an object dropped or thrown straight up or down. Use free-fall equations (or accelerated motion equations)
Horizontal motion is constant velocity (Newton’s 1st Law), work this part using constant velocity equations.
Both Vertical and horizontal motion are tied together by TIME. (Tv=Th)

40. Sample Problem….
A rock is thrown horizontally off a tall cliff at 20.m/s. If the cliff is 70.m high…
A) How far horizontally from the bottom of the cliff does the rock land?
B) What is the magnitude of the rock’s total velocity at the instant before landing?
Another example in book on page 157

41. Monkey Hunter
A young native hunter in the Amazon is hunting monkeys with a bow and arrow. The hunter see a monkey sitting on a branch, and since the bow does not have a sight, the hunter pulls back the bow and points the arrow directly at the monkey. As the hunter releases the arrow the monkey notices the movement and drops off the branch. What happens next?

42. Throw at monkey with no gravity…

44. Throw above monkey with gravity…

45. Throw at monkey fast with gravity…

47. Horizontal Range…What is the horizontal range of the projectile?
First…find time using vertical motion..
Then, use that time to find range…!
ΔVy =at
t= Δvy a
dx = vxt
dx = (v0 cos Θ)t
Notice nice equation to use to find vertical time in air!!!
=- V0Sin25.0 ° -V0Sin25.0 °
-9.8m/s2
dx = (v0Cos Θ) (V0 Sin Θ/4.9)
=2.58s
dx = (30.0 m/s Cos 25.0°)(30.0 m/s Sin 25.0°)/4.9
25.0°
V=30.0 m/s
dx = 70.4 m

48. Alternate Range equation…same thing….
Range (dx) =V02 Sin 2 Θ g

49. Max height & hang time depends only on initial vertical velocity
Each initial velocity vector below has the a different magnitude (speed) and the projectiles will have different ranges (green the greatest), but each object will spend the same time in the air and reach the same max height. This is because each vector has the same vertical component.

50. Max Range at 45
Over level ground at a constant launch speed, what angle maximizes the range, R ? First consider some extremes: When  = 0, R = 0, since as soon as the object is launched it’s back on the ground. When  = 90, the object goes straight up and lands right on the launch site. The best angle is 45, smack dab between the extremes.
All launch speeds are the same; only the angle varies.
76
45
38