Wednesday, October 28, 2009

1. Wednesday, October 28, 2009 Ch 7.1 Forces in two dimensions

2. Equilibrium • An object is said to be in a state of equilibrium when the net force acting on it is ZERO • When in equilibrium, the object is at rest or moving with a constant velocity…in other words it is not accelerating.

3. Equilibrium • In the case of an object moving in two dimensions; an object is in equilibrium when the resultant of three or more forces equals a net force of ZERO.

4. Three forces acting on an object in equilibrium We can use vector sums to tell if an object is in equilibrium if it forms a closed system. Which means that the forces cancel each other out.

5. Example • A 150 N painting is supported by two motionless ropes. Each rope makes an angle of 25 degrees with the horizontal. What is the tension in the rope? Known Fg = 150 N (force of mass due to gravity) Angle = 25 Unknown Tensional force on both ropes

6. Example Cntd. • Because the object is motionless, we can say that it is in equilibrium. Or that the three combined forces is equal to zero. • Fnet,x = 0 and -FAx + FBx =0 0=-FAcos(25) + FBcos(25) Fa Fb Fa = Fb • Fnet,y=0 so FAy+FBy- Fg=0 FAsin(25) + FBsin(25)-Fg = 0 Fg 2FAsin(25) = Fg Fa = 177.5 N ) 25 +x

7. Practice Problems Page 151 #1-2

8. Inclined Planes In all cases gravity is pulling objects towards the center of the earth. • On an inclined plane the force of gravity pulls down, the normal force pushes perpendicular to the plane and friction opposes the object’s motion.

9. Label the Forces represented by each number

10. Example Problems • A cart weighing 562 N is resting on an inclined plane at 30.0 degrees above the horizontal. Find the components of the weight force parallel and perpendicular to the plane

11. Parallel is the frictional force acting against the cart • Perpendicular is the normal force of the object on the plane Both of these components will be negative because we are finding their weight forces and the Fgy is pointing down (-) and Fgx is opposing friction which is to the left (-).

12. Draw the diagram • Find the x component • Find the y component Hint ** the weight of the box is already in Newtons so it is already in the form of its mass times gravity. If the mass of the cart were in grams we would multiply by 9.8 m/s2 to get Newtons before doing the problem

13. Fgx = - fg sin θ Fgx = -562Nsin(30) = -281 Fgy = -fg cos θ Fgy = -562Ncos(30) = -487 **The values have to be less than Fg in order for it to be realistic because the object is at rest**

14. Work on Practice Problems 5-8 • Equations from Ch. 6 to remember Fk = μkFN Fs = μsFN F = ma in some cases F – f = ma FN = mg

15. Tuesday November16, 2010 Ch 7.2 Projectile Motion

16. Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

17. Projectile Motion It can be understood by analyzing the horizontal and vertical motions separately.

18. Projectile Motion The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

19. Solving Problems Involving Projectile Motion Projectile motion is motion with constant velocity in the horizontal direction and constant acceleration in the vertical direction. These are the equations from p101 for each direction.

20. Solving Problems Involving Projectile Motion • Read the problem carefully, and choose the object(s) you are going to analyze. • Draw a diagram. • Choose an origin and a coordinate system. • Examine the x and y motions separately. • List known and unknown quantities. • Choose equation(s).

21. Example Problem pg. 158 Practice Problem 9 Put in your notes…then we will go over.

22. Projectiles launchedat an angle • When objects are launched at an angle the initial velocity has both a vertical and horizontal component. • When an object is launched upwards it rises with a slowing speed, reaches the top of its path and then descends with an increasing speed.

23. Projectile motion • Maximum Height – The height of the projectile when the vertical velocity is zero, and the projectile has only its horizontal velocity component. • Range – Horizontal distance the projectile travels. • Flight Time – The time the projectile is in the air.

24. Projectile Motion Object launched at an initial angle of θ0 with the horizontal

25. Y direction Vyo = (Vo)(sin θ) X direction Vxo = (Vo)(Cos θ) Commonly Used Equations

26. Example Problem An arrow is shot at 30.0 degrees above the horizontal. Its velocity is 49 m/s and it hits a target. • What is the maximum height the arrow will attain? • The target is at the height from which the arrow was shot. How far away is it?

27. There is an Example Problem on pg. 159 Homework: Practice Problems 12, 13 pg. 160 Don’t try to read the x and y values from the graph – calculate them.

28. Tuesday November 3, 2009 Ch 7.3 Circular Motion

29. Kinematics of Uniform Circular Motion Uniform circular motion: motion in a circle of constant radius at constant speed Instantaneous velocity is always tangent to circle.

30. Kinematics of Uniform Circular Motion

31. Kinematics of Uniform Circular Motion This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.

32. Dynamics of Uniform Circular Motion For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so can immediately write the force: (5-1)

33. Dynamics of Uniform Circular Motion We can see that the force must be inward by thinking about a ball on a string: We can see that the force must be inward by thinking about a ball on a string:

34. Dynamics of Uniform Circular Motion Centrifugal force is a myth. The natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off tangent to the circle.

35. Highway Curves, Banked and Unbanked When a car goes around a curve, there is a net force towards the center of the circle. If the road is flat, that force is supplied by friction.

36. An object sliding down an incline has three forces acting on it: the normal force, gravity, and the frictional force. • The normal force is always perpendicular to the surface. • The friction force is parallel to it. • The gravitational force points down. If the object is at rest, the forces are the same except that we use the static frictional force, and the sum of the forces is zero.

37. Rigid Rotating Object – mass that rotates around its own axis. Examples: merry-go-round, spinning washing machine, and a revolving door. Lever Arm – perpendicular distance from the axis of rotation to a line along which the force acts. Torque – The product of the force and the lever arm is called the torque. The greater the torque, the greater the change in rotational motion.

38. Torque Torque can stop, start or change the direction of rotation. Torque = lever arm (distance)* force applied

39. Torque problem examples • How much torque is applied to a wrench with a handle 0.24 m long when 4.0 N of force is applied tangent to the handle? • How much force is applied to a wrench with a handle 0.65 m long that has .5 Nm of torque.